OP Malhotra Class 12 Maths Solutions Chapter 17 Differential Equations Exercise 17 (F)

S Chand Class 12 ICSE Maths Solutions Chapter 17 Differential Equations Ex 17(f)

Solve the following equations :

 

Question 1. Find the integral factor of the following differential equations :
(i) \( \frac{d y}{dx}+\frac{y}{\sqrt{x}}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} \)
(ii) \( (1 + y²) + (2xy – \cot y)\frac { dy }{ dx } = 0 \)
Answer:
(i) The given differential equation is: \( \frac{d y}{dx}+\frac{y}{\sqrt{x}}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} \).
This is a linear differential equation of the first order in \( y \).
Comparing it with \( \frac { dy }{ dx } + Py = Q \), we find that \( P = \frac{1}{\sqrt{x}} \).
The integrating factor (I.F.) is calculated as \( e^{\int P dx} \).
\( \text{I.F.} = e^{\int \frac{1}{\sqrt{x}} dx} \)
We know that \( \int \frac{1}{\sqrt{x}} dx = \int x^{-1/2} dx = \frac{x^{1/2}}{1/2} = 2\sqrt{x} \).
So, \( \text{I.F.} = e^{2\sqrt{x}} \).
(ii) The given differential equation is: \( (1 + y²) + (2xy – \cot y)\frac { dy }{ dx } = 0 \).
We can rewrite this equation to make it linear in \( x \).
\( (1 + y²) = (\cot y - 2xy)\frac{dy}{dx} \)
\( \implies \frac{dx}{dy} = \frac{\cot y - 2xy}{1 + y^2} \)
\( \implies \frac{dx}{dy} = \frac{\cot y}{1 + y^2} - \frac{2xy}{1 + y^2} \)
\( \implies \frac{dx}{dy} + \frac{2y}{1 + y^2}x = \frac{\cot y}{1 + y^2} \) ... (1)
This is a linear differential equation in \( x \).
Comparing it with \( \frac { dx }{ dy } + Px = Q \), we have \( P = \frac{2y}{1 + y^2} \).
The integrating factor (I.F.) is \( e^{\int P dy} \).
\( \text{I.F.} = e^{\int \frac{2y}{1 + y^2} dy} \)
Let \( t = 1 + y^2 \), then \( dt = 2y dy \).
So, \( \int \frac{2y}{1 + y^2} dy = \int \frac{1}{t} dt = \log|t| = \log(1 + y^2) \).
Therefore, \( \text{I.F.} = e^{\log(1 + y^2)} = 1 + y^2 \).
In simple words: To find the integral factor, we first rearrange the equation into a standard linear form. Then, we identify the 'P' part of the equation and calculate \( e \) raised to the power of the integral of P with respect to the independent variable. This special factor helps us solve the differential equation easily.

🎯 Exam Tip: Remember to identify whether the equation is linear in x or y before finding the integrating factor. The choice of variable for integration (dx or dy) depends on this.

 

Question 2. \( \frac { dy }{ dx } + y = e^{-x} \)
Answer: The given differential equation is: \( \frac { dy }{ dx } + y = e^{-x} \).
This equation is a linear differential equation of the first order in \( y \).
We compare it with the standard form \( \frac { dy }{ dx } + Py = Q \).
Here, we can see that \( P = 1 \) and \( Q = e^{-x} \).
First, we calculate the integrating factor (I.F.). It is given by \( e^{\int P dx} \).
\( \text{I.F.} = e^{\int 1 dx} = e^x \).
Next, the general solution is given by the formula: \( y \cdot (\text{I.F.}) = \int (Q \cdot (\text{I.F.})) dx + c \).
\( y \cdot e^x = \int (e^{-x} \cdot e^x) dx + c \)
\( \implies y \cdot e^x = \int e^0 dx + c \)
\( \implies y \cdot e^x = \int 1 dx + c \)
\( \implies y \cdot e^x = x + c \).
Finally, we solve for \( y \):
\( y = (x + c)e^{-x} \). This is the required solution.
In simple words: This equation shows how something changes. We find a special number called the integrating factor, which helps combine parts of the equation. Then we use this factor to solve for \( y \), which tells us what \( y \) is equal to.

🎯 Exam Tip: Always make sure to isolate 'y' after integration to present the solution in its simplest and most direct form.

 

Question 3. \( x\frac { dy }{ dx } – ay = x-y^2 +1 \)
Answer: The given differential equation is: \( x\frac { dy }{ dx } – ay = x-y^2 +1 \).
First, we need to rewrite this equation into the standard linear form \( \frac { dy }{ dx } + Py = Q \).
Divide the entire equation by \( x \):
\( \frac { dy }{ dx } – \frac{a}{x}y = \frac{x-y^2+1}{x} \)
\( \implies \frac { dy }{ dx } – \frac{a}{x}y = 1 - \frac{y^2}{x} + \frac{1}{x} \)
The provided solution however uses a different transformation. Let's follow the solution given in the OCR, which seems to imply a different initial question or a non-linear approach in disguise of a linear solution structure. Re-evaluating the OCR, the provided solution for Q3 shows: `\frac{d y}{d x}-\frac{a}{x} y=\frac{x+1}{x}`. This suggests the original question was `x\frac { dy }{ dx } – ay = x+1`. I will proceed with the OCR's *solution's implied question* for consistency with the provided steps, as per IRON RULE 6.
So, assuming the given differential equation is \( x\frac { dy }{ dx } – ay = x+1 \).
Then, rewriting it in standard form: \( \frac { dy }{ dx } – \frac{a}{x} y = \frac{x+1}{x} \).
This is a linear differential equation of the first order in \( y \).
Comparing with \( \frac { dy }{ dx } + Py = Q \), we have \( P = -\frac{a}{x} \) and \( Q = \frac{x+1}{x} \).
Now, we find the integrating factor (I.F.) using the formula \( e^{\int P dx} \).
\( \text{I.F.} = e^{\int -\frac{a}{x} dx} \)
\( \implies \text{I.F.} = e^{-a \int \frac{1}{x} dx} \)
\( \implies \text{I.F.} = e^{-a \log|x|} = e^{\log x^{-a}} = x^{-a} \).
The general solution is then \( y \cdot (\text{I.F.}) = \int (Q \cdot (\text{I.F.})) dx + c \).
\( y \cdot x^{-a} = \int \left(\frac{x+1}{x}\right) x^{-a} dx + c \)
\( \implies y x^{-a} = \int (x^{1-a} + x^{-a}) dx + c \).
Now, integrate term by term:
\( \int x^{1-a} dx = \frac{x^{1-a+1}}{1-a+1} = \frac{x^{2-a}}{2-a} \) (provided \( a \neq 2 \))
\( \int x^{-a} dx = \frac{x^{-a+1}}{-a+1} = \frac{x^{1-a}}{1-a} \) (provided \( a \neq 1 \))
So, \( y x^{-a} = \frac{x^{2-a}}{2-a} + \frac{x^{1-a}}{1-a} + c \).
Multiplying by \( x^a \) to solve for \( y \):
\( y = \frac{x^{2-a}}{2-a} x^a + \frac{x^{1-a}}{1-a} x^a + c x^a \)
\( y = \frac{x^2}{2-a} + \frac{x}{1-a} + c x^a \). This is the required solution.
*In simple words: We take the given equation and rearrange it into a standard format. Then we calculate a special multiplying factor using the 'P' part of the equation. Finally, we use this factor to integrate and find the expression for \( y \).

 

Question 12.
(i) \( (1 + x²) \frac { dy }{ dx } + y = e^{\tan ^{-1} x} \)
(ii) \( (1 + x²) \frac { dy }{ dx } + y = \tan ^{-1} x \)
(iii) \( \frac{d y}{d x}+\frac{y}{x}=\cos x+\frac{\sin x}{x} \)
(iv) \( y' + y = \frac{1+x \log x}{x} \)
Answer:
(i) Given differential equation: \( (1 + x²) \frac { dy }{ dx } + y = e^{\tan ^{-1} x} \)
Divide by \( (1+x^2) \) to get the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \):
\( \frac{d y}{d x}+\left(\frac{1}{1+x^2}\right) y=\frac{e^{\tan ^{-1} x}}{1+x^2} \)
Here, \( P = \frac{1}{1+x^2} \) and \( Q = \frac{e^{\tan ^{-1} x}}{1+x^2} \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int \frac{1}{1+x^2} dx} \)
\( \implies I.F. = e^{\tan ^{-1} x} \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot e^{\tan ^{-1} x} = \int \frac{e^{\tan ^{-1} x}}{1+x^2} \cdot e^{\tan ^{-1} x} dx + c \)
\( \implies y \cdot e^{\tan ^{-1} x} = \int \frac{e^{2\tan ^{-1} x}}{1+x^2} dx + c \)
To solve this integral, let \( t = \tan ^{-1} x \). Then \( dt = \frac{1}{1+x^2} dx \).
So, the integral becomes:
\( \int e^{2t} dt + c = \frac{e^{2t}}{2} + c \)
Substitute back \( t = \tan ^{-1} x \):
\( y \cdot e^{\tan ^{-1} x} = \frac{e^{2\tan ^{-1} x}}{2} + c \)
Now, divide by \( e^{\tan ^{-1} x} \) to find \( y \):
\( y = \frac{e^{2\tan ^{-1} x}}{2 e^{\tan ^{-1} x}} + \frac{c}{e^{\tan ^{-1} x}} \)
\( \implies y = \frac{e^{\tan ^{-1} x}}{2} + c e^{-\tan ^{-1} x} \)
This is the general solution for the differential equation. Linear differential equations are often used in physics to model various systems, like circuits or population growth.
In simple words: First, we change the equation into a standard form. Then, we find a special multiplier called the integrating factor. We use this multiplier to solve the equation and find 'y'. The final answer shows how 'y' changes with 'x'.

🎯 Exam Tip: Remember to express the given equation in the standard form \( \frac{dy}{dx} + Py = Q \) before finding the integrating factor. Don't forget the constant of integration \(c\)!

 

(ii) Given differential equation: \( (1 + x²) \frac { dy }{ dx } + y = \tan ^{-1} x \)
Divide by \( (1+x^2) \) to get the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \):
\( \frac{d y}{d x}+\left(\frac{1}{1+x^2}\right) y=\frac{\tan ^{-1} x}{1+x^2} \)
Here, \( P = \frac{1}{1+x^2} \) and \( Q = \frac{\tan ^{-1} x}{1+x^2} \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int \frac{1}{1+x^2} dx} \)
\( \implies I.F. = e^{\tan ^{-1} x} \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot e^{\tan ^{-1} x} = \int \frac{\tan ^{-1} x}{1+x^2} \cdot e^{\tan ^{-1} x} dx + c \)
To solve this integral, let \( t = \tan ^{-1} x \). Then \( dt = \frac{1}{1+x^2} dx \).
So, the integral becomes:
\( \int t e^{t} dt + c \)
We use integration by parts for \( \int t e^{t} dt \), where \( u=t \) and \( dv=e^t dt \), so \( du=dt \) and \( v=e^t \).
\( \int t e^{t} dt = t e^t - \int e^t dt = t e^t - e^t = e^t (t-1) \)
Substitute back \( t = \tan ^{-1} x \):
\( y \cdot e^{\tan ^{-1} x} = e^{\tan ^{-1} x} (\tan ^{-1} x - 1) + c \)
Now, divide by \( e^{\tan ^{-1} x} \) to find \( y \):
\( y = (\tan ^{-1} x - 1) + c e^{-\tan ^{-1} x} \)
This is the general solution for the differential equation. Integrating factors help simplify complex differential equations into an easily solvable form.
In simple words: We first put the equation in a standard form. Then we find an integrating factor. After that, we use a special method called integration by parts to solve the remaining problem. The final result tells us what 'y' is in terms of 'x'.

🎯 Exam Tip: When using substitution for integrals, make sure to change the limits of integration if it's a definite integral, or substitute back the original variable if it's an indefinite integral, as done here.

 

(iii) Given differential equation: \( \frac{d y}{d x}+\frac{y}{x}=\cos x+\frac{\sin x}{x} \)
This equation is already in the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = \frac{1}{x} \) and \( Q = \cos x + \frac{\sin x}{x} \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int \frac{1}{x} dx} \)
\( \implies I.F. = e^{\log x} = x \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot x = \int \left(\cos x + \frac{\sin x}{x}\right) \cdot x dx + c \)
\( \implies yx = \int (\cos x \cdot x + \sin x) dx + c \)
\( \implies xy = \int x \cos x dx + \int \sin x dx + c \)
For \( \int x \cos x dx \), use integration by parts with \( u=x \) and \( dv=\cos x dx \), so \( du=dx \) and \( v=\sin x \).
\( \int x \cos x dx = x \sin x - \int \sin x dx = x \sin x - (-\cos x) = x \sin x + \cos x \)
So, substituting this back:
\( xy = (x \sin x + \cos x) + (-\cos x) + c \)
\( \implies xy = x \sin x + \cos x - \cos x + c \)
\( \implies xy = x \sin x + c \)
This is the required solution. The method of integrating factor is powerful for solving first-order linear differential equations.
In simple words: This equation is already in a good form. We find a special multiplier (integrating factor) which is 'x'. We then multiply the whole equation by 'x' and solve it. We use integration by parts to help solve one part, and in the end, we get the equation for 'y'.

🎯 Exam Tip: Carefully apply integration by parts, especially with signs. Remember that \( \int \sin x dx = -\cos x + C \).

 

(iv) Given differential equation: \( y' + y = \frac{1+x \log x}{x} \)
Rewrite \( y' \) as \( \frac{dy}{dx} \):
\( \frac{dy}{dx} + y = \frac{1+x \log x}{x} \)
This equation is already in the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = 1 \) and \( Q = \frac{1+x \log x}{x} = \frac{1}{x} + \log x \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int 1 dx} \)
\( \implies I.F. = e^x \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot e^x = \int \left(\frac{1}{x} + \log x\right) e^x dx + c \)
\( \implies y e^x = \int \frac{e^x}{x} dx + \int e^x \log x dx + c \)
We can use integration by parts for \( \int e^x \log x dx \). Let \( u = \log x \) and \( dv = e^x dx \), so \( du = \frac{1}{x} dx \) and \( v = e^x \).
\( \int e^x \log x dx = e^x \log x - \int e^x \cdot \frac{1}{x} dx \)
Substitute this back into the main equation:
\( y e^x = \int \frac{e^x}{x} dx + \left(e^x \log x - \int \frac{e^x}{x} dx\right) + c \)
Notice that the term \( \int \frac{e^x}{x} dx \) cancels out:
\( \implies y e^x = e^x \log x + c \)
Now, divide by \( e^x \) to find \( y \):
\( y = \log x + c e^{-x} \)
This is the required general solution. Recognizing patterns for integration by parts, like \( \int e^x (f(x) + f'(x)) dx = e^x f(x) \), can simplify calculations significantly.
In simple words: We rewrite the equation in a standard form. Then, we find the integrating factor, which is \( e^x \). We multiply both sides of the equation by this factor. A special integration trick helps us solve the right side, and many terms cancel out, leaving a simple answer for 'y'.

🎯 Exam Tip: Look for the pattern \( \int e^x (f(x) + f'(x)) dx = e^x f(x) \) when \( e^x \) is present in the integral. Here, \( f(x) = \log x \) and \( f'(x) = \frac{1}{x} \).

 

Question 13. Find the general solution for the differential equation:
\( y \log y \frac { dx }{ dy } + x – \log y = 0 \)
Answer:
Given differential equation: \( y \log y \frac { dx }{ dy } + x – \log y = 0 \)
We rearrange this to the standard linear differential equation form for \( x \) as the dependent variable: \( \frac{dx}{dy} + Px = Q \).
First, move the \( x \) and \( \log y \) terms to different sides:
\( y \log y \frac { dx }{ dy } = \log y - x \)
Now, divide by \( y \log y \):
\( \frac { dx }{ dy } = \frac{\log y}{y \log y} - \frac{x}{y \log y} \)
\( \implies \frac { dx }{ dy } = \frac{1}{y} - \frac{x}{y \log y} \)
Rearrange to the standard form \( \frac{dx}{dy} + Px = Q \):
\( \frac { dx }{ dy } + \frac{1}{y \log y} x = \frac{1}{y} \)
Here, \( P = \frac{1}{y \log y} \) and \( Q = \frac{1}{y} \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dy} = e^{\int \frac{1}{y \log y} dy} \)
To solve \( \int \frac{1}{y \log y} dy \), let \( t = \log y \). Then \( dt = \frac{1}{y} dy \).
So, \( \int \frac{1}{t} dt = \log |t| = \log |\log y| \).
\( \implies I.F. = e^{\log (\log y)} = \log y \)
The solution is given by \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + c \):
\( x \cdot \log y = \int \frac{1}{y} \cdot \log y dy + c \)
To solve \( \int \frac{1}{y} \log y dy \), again let \( t = \log y \), so \( dt = \frac{1}{y} dy \).
The integral becomes \( \int t dt = \frac{t^2}{2} \).
Substitute back \( t = \log y \):
\( x \log y = \frac{(\log y)^2}{2} + c \)
Now, divide by \( \log y \) to find \( x \):
\( x = \frac{(\log y)^2}{2 \log y} + \frac{c}{\log y} \)
\( \implies x = \frac{\log y}{2} + c (\log y)^{-1} \)
This is the required solution. Differential equations can often be transformed to a linear form, which makes them easier to solve using integrating factors.
In simple words: We change the given equation so that it becomes "linear in x". Then, we find a special multiplier called the integrating factor, which turns out to be \( \log y \). We use this to solve the equation, and after some steps and integration, we find the formula for 'x'.

🎯 Exam Tip: When the equation is linear in x (i.e., of the form \( \frac{dx}{dy} + Px = Q \)), the integrating factor is \( e^{\int P dy} \), and the general solution is \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + c \).

 

Question 14. Find the general solution for the differential equation:
\( \frac { dx }{ dy } + y \sec x = \tan x \)
Answer:
Given differential equation: \( \frac { dx }{ dy } + y \sec x = \tan x \)
*Self-correction/Interpretation based on solution steps*: The problem statement appears to have a variable mismatch, as the solution proceeds to solve a linear differential equation in \(y\) with \(x\) as the independent variable. Therefore, we assume the intended question was \( \frac { dy }{ dx } + y \sec x = \tan x \).
This equation is already in the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = \sec x \) and \( Q = \tan x \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int \sec x dx} \)
\( \implies I.F. = e^{\log |\sec x + \tan x|} \)
\( \implies I.F. = \sec x + \tan x \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot (\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx + c \)
\( \implies y (\sec x + \tan x) = \int (\tan x \sec x + \tan^2 x) dx + c \)
We know that \( \tan^2 x = \sec^2 x - 1 \).
\( y (\sec x + \tan x) = \int (\tan x \sec x + \sec^2 x - 1) dx + c \)
\( \implies y (\sec x + \tan x) = \int \tan x \sec x dx + \int \sec^2 x dx - \int 1 dx + c \)
\( \implies y (\sec x + \tan x) = \sec x + \tan x - x + c \)
Now, we can rearrange the equation to solve for \(y\). The solution for such equations often involves trigonometric identities.
\( \implies y = \frac{\sec x + \tan x - x + c}{\sec x + \tan x} \)
\( \implies y = 1 - \frac{x}{\sec x + \tan x} + \frac{c}{\sec x + \tan x} \)
This is the required solution. This problem demonstrates the importance of using trigonometric identities to simplify integrals.
In simple words: We assume the question meant \( \frac { dy }{ dx } \) for the solution to work. We find a special multiplier (integrating factor) using \( \sec x \). Then, we multiply the equation by this factor and integrate. We use a trig identity to simplify the integration, which helps us find 'y'.

🎯 Exam Tip: Be vigilant about using correct trigonometric integral formulas and identities, especially \( \tan^2 x = \sec^2 x - 1 \), which is frequently useful in these types of problems.

 

Question 15. Solve the differential equation:
\( (x + \tan y) dy = \sin 2y dx \)
Answer:
Given differential equation: \( (x + \tan y) dy = \sin 2y dx \)
We need to rearrange this into a linear differential equation. It looks like it will be linear in \(x\).
\( \frac{dx}{dy} = \frac{x + \tan y}{\sin 2y} \)
\( \implies \frac{dx}{dy} = \frac{x}{\sin 2y} + \frac{\tan y}{\sin 2y} \)
Rearrange to the standard form \( \frac{dx}{dy} + Px = Q \):
\( \frac{dx}{dy} - \frac{1}{\sin 2y} x = \frac{\tan y}{\sin 2y} \)
Here, \( P = -\frac{1}{\sin 2y} = -\csc 2y \) and \( Q = \frac{\tan y}{\sin 2y} \).
We can simplify \( Q \):
\( Q = \frac{\frac{\sin y}{\cos y}}{2 \sin y \cos y} = \frac{\sin y}{2 \sin y \cos^2 y} = \frac{1}{2 \cos^2 y} = \frac{1}{2} \sec^2 y \)
So, \( Q = \frac{1}{2} \sec^2 y \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dy} = e^{\int -\csc 2y dy} \)
We know \( \int \csc u du = \log |\tan \frac{u}{2}| \). So, \( \int -\csc 2y dy = -\frac{1}{2} \log |\tan y| \).
\( \implies I.F. = e^{-\frac{1}{2} \log |\tan y|} = e^{\log |(\tan y)^{-1/2}|} = (\tan y)^{-1/2} = \frac{1}{\sqrt{\tan y}} \)
The solution is given by \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + c \):
\( x \cdot \frac{1}{\sqrt{\tan y}} = \int \frac{1}{2} \sec^2 y \cdot \frac{1}{\sqrt{\tan y}} dy + c \)
To solve \( \int \frac{1}{2} \sec^2 y \cdot \frac{1}{\sqrt{\tan y}} dy \), let \( t = \tan y \). Then \( dt = \sec^2 y dy \).
The integral becomes \( \int \frac{1}{2} \cdot \frac{1}{\sqrt{t}} dt = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \frac{t^{1/2}}{1/2} = t^{1/2} = \sqrt{t} \).
Substitute back \( t = \tan y \):
\( x \cdot \frac{1}{\sqrt{\tan y}} = \sqrt{\tan y} + c \)
Now, multiply by \( \sqrt{\tan y} \) to find \( x \):
\( x = \tan y + c \sqrt{\tan y} \)
This is the required solution. Transforming a non-linear equation into a linear one by choosing the right dependent variable is a key skill.
In simple words: We rewrite the equation so that 'x' is the variable we want to find and 'y' is the other variable. We then find a special multiplier (integrating factor) related to \( \tan y \). Using this factor, we solve the equation, and after integrating and simplifying, we find the formula for 'x'.

🎯 Exam Tip: When faced with an equation involving both \( \frac{dy}{dx} \) and \( \frac{dx}{dy} \) type forms, try to rearrange it into a linear differential equation either in \(x\) or in \(y\). Look for terms like \( \frac{x}{f(y)} \) or \( \frac{y}{f(x)} \).

 

Question 16. Solve the following equations:
(i) \( \frac { dy }{ dx } – \frac { y }{ x } = 2x² \)
(ii) \( \frac { dy }{ dx } + \frac { y }{ x } = \sin x \), giving the general solution and also the solution for which \( y = 0 \) and \( x = \frac { \pi }{ 2 } \)
Answer:
(i) Given differential equation: \( \frac { dy }{ dx } – \frac { y }{ x } = 2x² \)
This equation is already in the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = -\frac{1}{x} \) and \( Q = 2x^2 \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int -\frac{1}{x} dx} \)
\( \implies I.F. = e^{-\log x} = e^{\log x^{-1}} = x^{-1} = \frac{1}{x} \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot \frac{1}{x} = \int 2x^2 \cdot \frac{1}{x} dx + c \)
\( \implies \frac{y}{x} = \int 2x dx + c \)
\( \implies \frac{y}{x} = 2 \frac{x^2}{2} + c \)
\( \implies \frac{y}{x} = x^2 + c \)
Now, multiply by \( x \) to find \( y \):
\( y = x^3 + cx \)
This is the required general solution. The integrating factor method effectively transforms the equation into an exact differential, making it straightforward to integrate.
In simple words: The equation is already in a standard form. We find a special multiplier (integrating factor) which is \( \frac{1}{x} \). We then multiply the whole equation by this factor and integrate both sides. This leads us to the final answer for 'y'.

🎯 Exam Tip: Be careful with the negative sign when integrating \( -\frac{1}{x} \); it results in \( -\log x \), which then becomes \( x^{-1} \) for the integrating factor.

 

(ii) Given differential equation: \( \frac { dy }{ dx } + \frac { y }{ x } = \sin x \)
This equation is already in the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = \frac{1}{x} \) and \( Q = \sin x \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int \frac{1}{x} dx} \)
\( \implies I.F. = e^{\log x} = x \)
The general solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot x = \int \sin x \cdot x dx + c \)
\( \implies xy = \int x \sin x dx + c \)
For \( \int x \sin x dx \), use integration by parts with \( u=x \) and \( dv=\sin x dx \), so \( du=dx \) and \( v=-\cos x \).
\( \int x \sin x dx = x(-\cos x) - \int (-\cos x) dx = -x \cos x + \int \cos x dx = -x \cos x + \sin x \)
Substitute this back:
\( xy = -x \cos x + \sin x + c \) (General Solution)
Now, we use the initial condition \( y = 0 \) when \( x = \frac{\pi}{2} \) to find the particular solution:
\( (\frac{\pi}{2})(0) = -(\frac{\pi}{2}) \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) + c \)
\( \implies 0 = -(\frac{\pi}{2})(0) + 1 + c \)
\( \implies 0 = 0 + 1 + c \)
\( \implies c = -1 \)
Substitute \( c = -1 \) back into the general solution to get the particular solution:
\( xy = -x \cos x + \sin x - 1 \)
This is the particular solution for the given initial condition. Initial conditions allow us to find a unique solution from a family of general solutions.
In simple words: First, we find the general answer using the integrating factor method. We find the special multiplier is 'x'. We then use integration by parts to solve for 'y'. Next, we use the given starting values for 'x' and 'y' to find the value of 'c'. Finally, we put 'c' back into the general answer to get the specific answer for this problem.

🎯 Exam Tip: When initial conditions are given, always substitute them into the general solution *after* integration and before solving for \( y \) explicitly to find the constant \( c \). Then, write down the particular solution.

 

Question 17. Solve the following equations:
(i) \( \frac { dy }{ dx } = y \tan x – 2\sin x \)
(ii) \( y' + y = \sin x \)
(iii) \( y' + y = \cos x \)
(iv) \( y' + 2y = \sin x \)
(v) \( 2 \frac { dy }{ dx } + 4y = \sin 2x \)
Answer:
(i) Given differential equation: \( \frac { dy }{ dx } = y \tan x – 2\sin x \)
Rearrange this to the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \):
\( \frac { dy }{ dx } - y \tan x = – 2\sin x \)
Here, \( P = -\tan x \) and \( Q = -2\sin x \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int -\tan x dx} \)
We know \( \int \tan x dx = \log |\sec x| \) or \( -\log |\cos x| \). So, \( \int -\tan x dx = \log |\cos x| \).
\( \implies I.F. = e^{\log (\cos x)} = \cos x \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cos x = \int (-2\sin x) \cos x dx + c \)
\( \implies y \cos x = - \int 2\sin x \cos x dx + c \)
We know that \( \sin 2x = 2 \sin x \cos x \).
\( \implies y \cos x = - \int \sin 2x dx + c \)
\( \implies y \cos x = - (-\frac{\cos 2x}{2}) + c \)
\( \implies y \cos x = \frac{\cos 2x}{2} + c \)
Now, divide by \( \cos x \) to find \( y \):
\( y = \frac{\cos 2x}{2 \cos x} + \frac{c}{\cos x} \)
\( \implies y = \frac{\cos 2x}{2 \cos x} + c \sec x \)
This is the required general solution. The use of double angle formulas like \( \sin 2x \) simplifies the integration greatly.
In simple words: We first put the equation in a standard form. Then, we find a special multiplier (integrating factor), which is \( \cos x \). We multiply the equation by this factor and integrate. We use the double angle formula for sine to make integration easier, which leads to the final answer for 'y'.

🎯 Exam Tip: Always remember fundamental trigonometric identities like \( \sin 2x = 2 \sin x \cos x \) and \( \cos 2x \) identities; they are crucial for simplifying integrals in differential equations.

 

(ii) Given differential equation: \( y' + y = \sin x \)
Rewrite \( y' \) as \( \frac{dy}{dx} \):
\( \frac{dy}{dx} + y = \sin x \)
This equation is already in the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = 1 \) and \( Q = \sin x \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int 1 dx} \)
\( \implies I.F. = e^x \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot e^x = \int e^x \sin x dx + c \)
To solve \( \int e^x \sin x dx \), we use integration by parts twice. Let \( I = \int e^x \sin x dx \).
First application: Let \( u=\sin x \) and \( dv=e^x dx \), so \( du=\cos x dx \) and \( v=e^x \).
\( I = e^x \sin x - \int e^x \cos x dx \)
Second application: Let \( u=\cos x \) and \( dv=e^x dx \), so \( du=-\sin x dx \) and \( v=e^x \).
\( I = e^x \sin x - (e^x \cos x - \int e^x (-\sin x) dx) \)
\( \implies I = e^x \sin x - e^x \cos x - \int e^x \sin x dx \)
\( \implies I = e^x \sin x - e^x \cos x - I \)
Move \( -I \) to the left side:
\( 2I = e^x (\sin x - \cos x) \)
\( \implies I = \frac{e^x}{2} (\sin x - \cos x) \)
Substitute this back into the solution equation:
\( y e^x = \frac{e^x}{2} (\sin x - \cos x) + c \)
Now, divide by \( e^x \) to find \( y \):
\( y = \frac{1}{2} (\sin x - \cos x) + c e^{-x} \)
This is the required general solution. Solving integrals of the form \( \int e^{ax} \sin(bx) dx \) or \( \int e^{ax} \cos(bx) dx \) usually involves two rounds of integration by parts.
In simple words: We start with the equation in a standard form. We find the special multiplier (integrating factor), which is \( e^x \). Then, we solve the integral \( \int e^x \sin x dx \) by doing integration by parts two times. This brings back the original integral, allowing us to solve for it. Finally, we get the answer for 'y'.

🎯 Exam Tip: For integrals of the form \( \int e^{ax} \sin(bx) dx \) or \( \int e^{ax} \cos(bx) dx \), remember that you will apply integration by parts twice, and the original integral will reappear, allowing you to solve for it algebraically.

 

(iii) Given differential equation: \( y' + y = \cos x \)
Rewrite \( y' \) as \( \frac{dy}{dx} \):
\( \frac{dy}{dx} + y = \cos x \)
This equation is already in the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = 1 \) and \( Q = \cos x \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int 1 dx} \)
\( \implies I.F. = e^x \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot e^x = \int e^x \cos x dx + c \)
To solve \( \int e^x \cos x dx \), we use integration by parts twice. Let \( I = \int e^x \cos x dx \).
First application: Let \( u=\cos x \) and \( dv=e^x dx \), so \( du=-\sin x dx \) and \( v=e^x \).
\( I = e^x \cos x - \int e^x (-\sin x) dx \)
\( \implies I = e^x \cos x + \int e^x \sin x dx \)
Second application: Let \( u=\sin x \) and \( dv=e^x dx \), so \( du=\cos x dx \) and \( v=e^x \).
\( I = e^x \cos x + (e^x \sin x - \int e^x \cos x dx) \)
\( \implies I = e^x \cos x + e^x \sin x - I \)
Move \( -I \) to the left side:
\( 2I = e^x (\cos x + \sin x) \)
\( \implies I = \frac{e^x}{2} (\cos x + \sin x) \)
Substitute this back into the solution equation:
\( y e^x = \frac{e^x}{2} (\cos x + \sin x) + c \)
Now, divide by \( e^x \) to find \( y \):
\( y = \frac{1}{2} (\cos x + \sin x) + c e^{-x} \)
This is the required general solution. Solving these types of integrals is a standard technique that appears often.
In simple words: We put the equation in a standard form. We find the special multiplier (integrating factor), which is \( e^x \). Then, we solve the integral \( \int e^x \cos x dx \) by doing integration by parts two times. This brings back the original integral, allowing us to solve for it. Finally, we get the answer for 'y'.

🎯 Exam Tip: The integration of \( e^{ax} \cos(bx) \) or \( e^{ax} \sin(bx) \) is a common pattern. Memorizing the general formula \( \int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2+b^2}(a \cos(bx) + b \sin(bx)) + C \) can save time.

 

(iv) Given differential equation: \( y' + 2y = \sin x \)
Rewrite \( y' \) as \( \frac{dy}{dx} \):
\( \frac{dy}{dx} + 2y = \sin x \)
This equation is already in the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = 2 \) and \( Q = \sin x \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int 2 dx} \)
\( \implies I.F. = e^{2x} \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot e^{2x} = \int e^{2x} \sin x dx + c \)
To solve \( \int e^{2x} \sin x dx \), we use integration by parts twice. Let \( I = \int e^{2x} \sin x dx \).
First application: Let \( u=\sin x \) and \( dv=e^{2x} dx \), so \( du=\cos x dx \) and \( v=\frac{e^{2x}}{2} \).
\( I = \frac{e^{2x}}{2} \sin x - \int \frac{e^{2x}}{2} \cos x dx \)
Second application: Let \( u=\cos x \) and \( dv=\frac{e^{2x}}{2} dx \), so \( du=-\sin x dx \) and \( v=\frac{e^{2x}}{4} \).
\( I = \frac{e^{2x}}{2} \sin x - \left(\frac{e^{2x}}{4} \cos x - \int \frac{e^{2x}}{4} (-\sin x) dx\right) \)
\( \implies I = \frac{e^{2x}}{2} \sin x - \frac{e^{2x}}{4} \cos x - \frac{1}{4} \int e^{2x} \sin x dx \)
\( \implies I = \frac{e^{2x}}{2} \sin x - \frac{e^{2x}}{4} \cos x - \frac{1}{4} I \)
Move \( -\frac{1}{4} I \) to the left side:
\( I + \frac{1}{4} I = \frac{e^{2x}}{2} \sin x - \frac{e^{2x}}{4} \cos x \)
\( \implies \frac{5}{4} I = e^{2x} \left(\frac{\sin x}{2} - \frac{\cos x}{4}\right) \)
\( \implies \frac{5}{4} I = \frac{e^{2x}}{4} (2 \sin x - \cos x) \)
\( \implies I = \frac{e^{2x}}{5} (2 \sin x - \cos x) \)
Substitute this back into the solution equation:
\( y e^{2x} = \frac{e^{2x}}{5} (2 \sin x - \cos x) + c \)
Now, divide by \( e^{2x} \) to find \( y \):
\( y = \frac{1}{5} (2 \sin x - \cos x) + c e^{-2x} \)
This is the required general solution. These integrations, while longer, follow a predictable pattern for exponential and trigonometric functions.
In simple words: We get the equation into a standard form. We find a special multiplier, the integrating factor, which is \( e^{2x} \). Then, we need to solve an integral involving \( e^{2x} \sin x \). This requires doing integration by parts two times, which allows us to solve for the integral algebraically. Finally, we get the equation for 'y'.

🎯 Exam Tip: Be very careful with fractions and signs during the two rounds of integration by parts. A common error is miscalculating the coefficient of \( I \) on the left side (here, \( 1 + \frac{1}{4} = \frac{5}{4} \)).

 

(v) Given differential equation: \( 2 \frac { dy }{ dx } + 4y = \sin 2x \)
First, divide the entire equation by 2 to get the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \):
\( \frac { dy }{ dx } + 2y = \frac{1}{2} \sin 2x \)
Here, \( P = 2 \) and \( Q = \frac{1}{2} \sin 2x \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dx} = e^{\int 2 dx} \)
\( \implies I.F. = e^{2x} \)
The solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \):
\( y \cdot e^{2x} = \int \frac{1}{2} \sin 2x \cdot e^{2x} dx + c \)
Let \( I = \int e^{2x} \sin 2x dx \). We use integration by parts twice.
First application: Let \( u=\sin 2x \) and \( dv=e^{2x} dx \), so \( du=2\cos 2x dx \) and \( v=\frac{e^{2x}}{2} \).
\( I = \frac{e^{2x}}{2} \sin 2x - \int \frac{e^{2x}}{2} (2\cos 2x) dx \)
\( \implies I = \frac{e^{2x}}{2} \sin 2x - \int e^{2x} \cos 2x dx \)
Second application: Let \( u=\cos 2x \) and \( dv=e^{2x} dx \), so \( du=-2\sin 2x dx \) and \( v=\frac{e^{2x}}{2} \).
\( I = \frac{e^{2x}}{2} \sin 2x - \left(\frac{e^{2x}}{2} \cos 2x - \int \frac{e^{2x}}{2} (-2\sin 2x) dx\right) \)
\( \implies I = \frac{e^{2x}}{2} \sin 2x - \frac{e^{2x}}{2} \cos 2x - \int e^{2x} \sin 2x dx \)
\( \implies I = \frac{e^{2x}}{2} \sin 2x - \frac{e^{2x}}{2} \cos 2x - I \)
Move \( -I \) to the left side:
\( 2I = \frac{e^{2x}}{2} (\sin 2x - \cos 2x) \)
\( \implies I = \frac{e^{2x}}{4} (\sin 2x - \cos 2x) \)
Substitute this back into the solution equation:
\( y e^{2x} = \frac{1}{2} \left[\frac{e^{2x}}{4} (\sin 2x - \cos 2x)\right] + c \)
\( \implies y e^{2x} = \frac{e^{2x}}{8} (\sin 2x - \cos 2x) + c \)
Now, divide by \( e^{2x} \) to find \( y \):
\( y = \frac{1}{8} (\sin 2x - \cos 2x) + c e^{-2x} \)
This is the required general solution. It's crucial to correctly handle the coefficients and constants during integration.
In simple words: First, we divide the entire equation by 2 to get it into a standard form. We find the special multiplier (integrating factor), which is \( e^{2x} \). Then, we solve the integral involving \( e^{2x} \sin 2x \) by doing integration by parts two times, which allows us to solve for the integral algebraically. Finally, we get the equation for 'y'.

🎯 Exam Tip: Be careful to apply the initial \( \frac{1}{2} \) factor to the entire result of the integral \( I \) after solving it. Double-check all coefficients, especially when working with \( \sin(ax) \) or \( \cos(ax) \) inside the integral.

 

Question 18. Solve the following equations:
(i) \( (x + 2y²) \frac { dy }{ dx } = y, y > 0 \)
(ii) \( (x + 3y²)\frac { dy }{ dx } = y, \) given that when \( x = 2, y = 1 \).
(iii) \( (3y² – x)dy = ydx \)
(iv) \( y² + (x - \frac { 1 }{ y })\frac { dy }{ dx } = 0 \)
(v) \( (x + y + 1)\frac { dy }{ dx } = 1 \)
Answer:
(i) Given differential equation: \( (x + 2y²) \frac { dy }{ dx } = y \)
Since the equation is not easily arranged into \( \frac{dy}{dx} + Py = Q \), let's try to make it linear in \( x \), i.e., of the form \( \frac{dx}{dy} + Px = Q \).
First, invert \( \frac{dy}{dx} \):
\( \frac{dx}{dy} = \frac{x + 2y^2}{y} \)
\( \implies \frac{dx}{dy} = \frac{x}{y} + \frac{2y^2}{y} \)
\( \implies \frac{dx}{dy} = \frac{x}{y} + 2y \)
Rearrange to the standard form \( \frac{dx}{dy} + Px = Q \):
\( \frac{dx}{dy} - \frac{1}{y} x = 2y \)
Here, \( P = -\frac{1}{y} \) and \( Q = 2y \).
Now, calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P dy} = e^{\int -\frac{1}{y} dy} \)
\( \implies I.F. = e^{-\log y} = e^{\log y^{-1}} = y^{-1} = \frac{1}{y} \)
The solution is given by \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + c \):
\( x \cdot \frac{1}{y} = \int 2y \cdot \frac{1}{y} dy + c \)
\( \implies \frac{x}{y} = \int 2 dy + c \)
\( \implies \frac{x}{y} = 2y + c \)
Now, multiply by \( y \) to find \( x \):
\( x = 2y^2 + cy \)
This is the required general solution. When a differential equation is difficult to solve for \( \frac{dy}{dx} \), considering \( \frac{dx}{dy} \) can simplify it significantly.
In simple words: We change the equation so that 'x' is the main variable we want to find and 'y' is the other variable. We find a special multiplier (integrating factor) which is \( \frac{1}{y} \). We then multiply the equation by this factor and integrate both sides. This helps us find the formula for 'x'.

🎯 Exam Tip: If an equation is not linear in \(y\), check if it can be made linear in \(x\) by writing it as \( \frac{dx}{dy} + Px = Q \). This often involves inverting \( \frac{dy}{dx} \).

 

Question 18.
(i) \( (x + 2y^2) \frac { dy }{ dx } = y, y > 0 \)
Answer:
Given the differential equation: \( (x + 2y^2) \frac { dy }{ dx } = y \). We are told that \( y > 0 \).
We can rewrite this equation to make it linear in terms of \( x \).
First, divide by \( y \):
\( (x + 2y^2) \frac{1}{y} \frac { dy }{ dx } = 1 \)
\( \frac{x}{y} + 2y = \frac{dx}{dy} \)
This can be rearranged as:
\( \frac{dx}{dy} - \frac{1}{y}x = 2y \)
This is a linear differential equation in \( x \), in the form \( \frac{dx}{dy} + Px = Q \).
Here, \( P = -\frac{1}{y} \) and \( Q = 2y \).
The integrating factor (I.F.) is \( e^{\int P dy} \).
\( I.F. = e^{\int -\frac{1}{y} dy} \)
\( = e^{-\log y} \)
\( = e^{\log y^{-1}} \)
\( = y^{-1} = \frac{1}{y} \)
The solution is given by \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + C \).
\( x \cdot \frac{1}{y} = \int 2y \cdot \frac{1}{y} dy + C \)
\( \frac{x}{y} = \int 2 dy + C \)
\( \frac{x}{y} = 2y + C \)
Now, multiply by \( y \) to solve for \( x \):
\( x = 2y^2 + Cy \)
This is the required general solution. This type of equation is often easier to solve by treating y as the independent variable when x and y are mixed.
In simple words: We changed the equation so we could solve for \( x \) instead of \( y \). We found a special multiplier called the integrating factor, which helped us solve the equation easily. The final answer tells us how \( x \) and \( y \) are related.

🎯 Exam Tip: When a differential equation is not easily linear in \( y \), check if it can be rearranged to be linear in \( x \). This often involves treating \( y \) as the independent variable and \( x \) as the dependent variable.

 

Question 18.
(ii) \( (x + 3y^2) \frac { dy }{ dx } = y \), given that when \( x = 2, y = 1 \).
Answer:
Given the differential equation: \( (x + 3y^2) \frac { dy }{ dx } = y \).
We can rearrange this to make it a linear differential equation in \( x \).
\( \frac { dx }{ dy } = \frac{x + 3y^2}{y} \)
\( \frac { dx }{ dy } = \frac{x}{y} + 3y \)
Rearranging into the standard form \( \frac{dx}{dy} + Px = Q \):
\( \frac{dx}{dy} - \frac{1}{y}x = 3y \)
Here, \( P = -\frac{1}{y} \) and \( Q = 3y \).
The integrating factor (I.F.) is \( e^{\int P dy} \).
\( I.F. = e^{\int -\frac{1}{y} dy} \)
\( = e^{-\log y} = e^{\log y^{-1}} = \frac{1}{y} \)
The general solution is \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + C \).
\( x \cdot \frac{1}{y} = \int 3y \cdot \frac{1}{y} dy + C \)
\( \frac{x}{y} = \int 3 dy + C \)
\( \frac{x}{y} = 3y + C \)
So, the general solution is \( x = 3y^2 + Cy \). (Equation 1)
Now, we use the given condition \( x = 2 \) when \( y = 1 \) to find \( C \).
Substitute these values into Equation 1:
\( 2 = 3(1)^2 + C(1) \)
\( 2 = 3 + C \)
\( C = 2 - 3 \)
\( C = -1 \)
Now, substitute \( C = -1 \) back into the general solution:
\( x = 3y^2 - y \)
This is the required particular solution. Finding the constant C is an important step for these problems.
In simple words: First, we solved the equation to find a general formula linking \( x \) and \( y \). Then, we used the given numbers (when \( x=2, y=1 \)) to find the exact value of the unknown constant. This gave us a specific solution that fits those starting conditions.

🎯 Exam Tip: For particular solutions, always remember to substitute the given initial conditions into your general solution to find the value of the constant \( C \). Don't forget to write out the final solution with the calculated \( C \) value.

 

Question 18.
(iii) \( (3y^2 - x)dy = ydx \)
Answer:
Given the differential equation: \( (3y^2 - x)dy = ydx \).
We can rearrange this equation to make it linear in \( x \).
Divide by \( dy \):
\( 3y^2 - x = y \frac{dx}{dy} \)
Now, divide by \( y \):
\( \frac{3y^2}{y} - \frac{x}{y} = \frac{dx}{dy} \)
\( 3y - \frac{x}{y} = \frac{dx}{dy} \)
Rearranging into the standard form \( \frac{dx}{dy} + Px = Q \):
\( \frac{dx}{dy} + \frac{1}{y}x = 3y \)
Here, \( P = \frac{1}{y} \) and \( Q = 3y \).
The integrating factor (I.F.) is \( e^{\int P dy} \).
\( I.F. = e^{\int \frac{1}{y} dy} \)
\( = e^{\log y} = y \)
The general solution is \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + C \).
\( x \cdot y = \int (3y) \cdot (y) dy + C \)
\( xy = \int 3y^2 dy + C \)
Now, integrate \( 3y^2 \):
\( xy = 3 \frac{y^3}{3} + C \)
\( xy = y^3 + C \)
This is the required general solution. Solving these equations often requires careful algebraic manipulation.
In simple words: We first changed the equation to a standard form where \( x \) is the main variable we are solving for. Then, we used an "integrating factor" to help us combine parts of the equation. Finally, we integrated to get the answer, which shows how \( x \) and \( y \) are related.

🎯 Exam Tip: Always look for ways to rewrite a non-standard differential equation into a linear form (\( \frac{dy}{dx} + Py = Q \) or \( \frac{dx}{dy} + Px = Q \)). This is the key first step to solving it using integrating factors.

 

Question 18.
(iv) \( y^2 + (x - \frac { 1 }{ y })\frac { dy }{ dx } = 0 \)
Answer:
Given the differential equation: \( y^2 + (x - \frac { 1 }{ y })\frac { dy }{ dx } = 0 \).
We want to rearrange this into a linear differential equation with \( x \) as the dependent variable.
Move \( y^2 \) to the other side:
\( (x - \frac { 1 }{ y })\frac { dy }{ dx } = -y^2 \)
Now, flip the derivative \( \frac{dy}{dx} \) to \( \frac{dx}{dy} \):
\( x - \frac { 1 }{ y } = -y^2 \frac{dx}{dy} \)
Divide by \( -y^2 \):
\( \frac{x - \frac{1}{y}}{-y^2} = \frac{dx}{dy} \)
\( \frac{dx}{dy} = -\frac{x}{y^2} + \frac{1}{y^3} \)
Rearrange into the standard form \( \frac{dx}{dy} + Px = Q \):
\( \frac{dx}{dy} + \frac{1}{y^2}x = \frac{1}{y^3} \)
Here, \( P = \frac{1}{y^2} \) and \( Q = \frac{1}{y^3} \).
The integrating factor (I.F.) is \( e^{\int P dy} \).
\( I.F. = e^{\int \frac{1}{y^2} dy} \)
\( = e^{\int y^{-2} dy} \)
\( = e^{\frac{y^{-1}}{-1}} \)
\( = e^{-\frac{1}{y}} \)
The general solution is \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + C \).
\( x e^{-\frac{1}{y}} = \int \frac{1}{y^3} e^{-\frac{1}{y}} dy + C \)
To solve the integral \( \int \frac{1}{y^3} e^{-\frac{1}{y}} dy \), we use substitution and then integration by parts.
Let \( t = -\frac{1}{y} \). Then \( \frac{1}{y} = -t \) and \( dt = \frac{1}{y^2} dy \).
The integral becomes \( \int (-t) e^t dt \).
Now, apply integration by parts for \( \int (-t)e^t dt \). Let \( u = -t \) and \( dv = e^t dt \). So \( du = -dt \) and \( v = e^t \).
\( \int (-t)e^t dt = -t e^t - \int e^t (-dt) \)
\( = -t e^t + \int e^t dt \)
\( = -t e^t + e^t + C' \)
\( = e^t (1 - t) + C' \)
Substitute back \( t = -\frac{1}{y} \):
\( = e^{-\frac{1}{y}} (1 - (-\frac{1}{y})) + C' \)
\( = e^{-\frac{1}{y}} (1 + \frac{1}{y}) + C' \)
So, the solution for the differential equation is:
\( x e^{-\frac{1}{y}} = e^{-\frac{1}{y}} (1 + \frac{1}{y}) + C \)
Divide by \( e^{-\frac{1}{y}} \):
\( x = (1 + \frac{1}{y}) + C e^{\frac{1}{y}} \)
\( x = \frac{y+1}{y} + C e^{\frac{1}{y}} \)
This is the required general solution. Recognizing the correct substitution is key here.
In simple words: We rewrote the equation so we could solve for \( x \). We found a special multiplier, the integrating factor, which helped us. The hardest part was solving a tricky integral, which we did by replacing part of the expression with a new variable and then using integration by parts. The final answer gives us \( x \) in terms of \( y \).

🎯 Exam Tip: When faced with integrals like \( \int \frac{1}{y^n} e^{-\frac{1}{y}} dy \), consider the substitution \( t = -\frac{1}{y} \) carefully. This often simplifies the integral into a standard form that can be solved using integration by parts.

 

Question 18.
(v) \( (x + y + 1)\frac { dy }{ dx } = 1 \)
Answer:
Given the differential equation: \( (x + y + 1)\frac { dy }{ dx } = 1 \).
We can rewrite this by taking the reciprocal of the derivative:
\( \frac{dx}{dy} = x + y + 1 \)
Rearranging into the standard form \( \frac{dx}{dy} + Px = Q \):
\( \frac{dx}{dy} - x = y + 1 \)
Here, \( P = -1 \) and \( Q = y + 1 \).
The integrating factor (I.F.) is \( e^{\int P dy} \).
\( I.F. = e^{\int -1 dy} \)
\( = e^{-y} \)
The general solution is \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + C \).
\( x e^{-y} = \int (y + 1) e^{-y} dy + C \)
To solve the integral \( \int (y + 1) e^{-y} dy \), we use integration by parts.
Let \( u = y + 1 \) and \( dv = e^{-y} dy \).
Then \( du = dy \) and \( v = -e^{-y} \).
\( \int (y + 1) e^{-y} dy = (y + 1)(-e^{-y}) - \int (-e^{-y}) dy \)
\( = -(y + 1)e^{-y} + \int e^{-y} dy \)
\( = -(y + 1)e^{-y} - e^{-y} + C' \)
\( = -e^{-y} (y + 1 + 1) + C' \)
\( = -e^{-y} (y + 2) + C' \)
So, the solution for the differential equation is:
\( x e^{-y} = -e^{-y} (y + 2) + C \)
Divide by \( e^{-y} \):
\( x = -(y + 2) + C e^y \)
\( x = -y - 2 + C e^y \)
This is the required general solution. Some equations become much simpler when treated as linear in x.
In simple words: We changed the equation to solve for \( x \) and used an integrating factor. Then, we solved an integral using a method called "integration by parts" to find the general formula that shows how \( x \) and \( y \) are connected.

🎯 Exam Tip: If the equation \( (x+y+1)\frac{dy}{dx} = 1 \) were rearranged to \( \frac{dy}{dx} = \frac{1}{x+y+1} \), it would be non-linear and much harder to solve directly. Always consider taking the reciprocal of the derivative to check if it becomes linear in the other variable.

 

Question 18.
(vi) \( \frac { dx }{ dy } + \frac{x-y^3}{y^2} = 0 \)
Answer:
Given the differential equation: \( \frac { dx }{ dy } + \frac{x-y^3}{y^2} = 0 \).
We separate the terms in the fraction:
\( \frac { dx }{ dy } + \frac{x}{y^2} - \frac{y^3}{y^2} = 0 \)
\( \frac { dx }{ dy } + \frac{1}{y^2}x - y = 0 \)
Rearranging to the standard linear form \( \frac{dx}{dy} + Px = Q \):
\( \frac{dx}{dy} + \frac{1}{y^2}x = y \)
Here, \( P = \frac{1}{y^2} \) and \( Q = y \).
The integrating factor (I.F.) is \( e^{\int P dy} \).
\( I.F. = e^{\int \frac{1}{y^2} dy} = e^{\int y^{-2} dy} = e^{-\frac{1}{y}} \)
The general solution is given by \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + C \).
\( x e^{-\frac{1}{y}} = \int y e^{-\frac{1}{y}} dy + C \)
The integral \( \int y e^{-\frac{1}{y}} dy \) cannot be expressed using elementary functions in a simple closed form. Therefore, the general solution remains in this form, with the integral as part of the expression.
In simple words: We first set up the equation in a standard way and found a special multiplier. When we tried to finish the last step (integration), we found that the remaining part cannot be solved into a simple formula using regular math rules. So, the answer includes that unsolved integral part.

🎯 Exam Tip: Not all integrals have a simple closed-form solution using elementary functions. For such cases, it is acceptable to leave the integral as part of the final solution, provided all other steps are correctly carried out.

 

Question 19.
(i) Solve the differential equation \( \frac { dy }{ dx } - 3 y \cot x = \sin 2x \), given that \( y = 2 \) when \( x = \frac { \pi }{ 2 } \).
Answer:
Given the differential equation: \( \frac { dy }{ dx } - 3 y \cot x = \sin 2x \).
This is a linear differential equation in \( y \) of the form \( \frac { dy }{ dx } + Py = Q \).
Here, \( P = -3 \cot x \) and \( Q = \sin 2x \).
The integrating factor (I.F.) is \( e^{\int P dx} \).
\( I.F. = e^{\int -3 \cot x dx} \)
\( = e^{-3 \log |\sin x|} \)
\( = e^{\log (\sin x)^{-3}} \)
\( = (\sin x)^{-3} = \frac{1}{\sin^3 x} \)
The general solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C \).
\( y \cdot \frac{1}{\sin^3 x} = \int (\sin 2x) \cdot \frac{1}{\sin^3 x} dx + C \)
We know \( \sin 2x = 2 \sin x \cos x \). Substitute this into the integral:
\( \frac{y}{\sin^3 x} = \int \frac{2 \sin x \cos x}{\sin^3 x} dx + C \)
\( \frac{y}{\sin^3 x} = \int \frac{2 \cos x}{\sin^2 x} dx + C \)
\( \frac{y}{\sin^3 x} = \int 2 \cot x \csc x dx + C \)
We know that \( \int \cot x \csc x dx = -\csc x \).
\( \frac{y}{\sin^3 x} = -2 \csc x + C \)
Now, express \( \csc x \) in terms of \( \sin x \):
\( \frac{y}{\sin^3 x} = -\frac{2}{\sin x} + C \)
Multiply by \( \sin^3 x \) to solve for \( y \):
\( y = -2 \sin^2 x + C \sin^3 x \) (Equation 1)
We are given that \( y = 2 \) when \( x = \frac{\pi}{2} \). Substitute these values into Equation 1:
\( 2 = -2 \sin^2 (\frac{\pi}{2}) + C \sin^3 (\frac{\pi}{2}) \)
Since \( \sin (\frac{\pi}{2}) = 1 \):
\( 2 = -2(1)^2 + C(1)^3 \)
\( 2 = -2 + C \)
\( C = 4 \)
Substitute \( C = 4 \) back into Equation 1:
\( y = -2 \sin^2 x + 4 \sin^3 x \)
This is the required particular solution. Remember to always simplify trigonometric expressions where possible.
In simple words: We solved the equation by finding a special multiplier called the integrating factor. After integrating, we used the given starting values for \( x \) and \( y \) to find the exact number for the constant \( C \). This gave us a specific formula for \( y \) that fits the given conditions.

🎯 Exam Tip: Always simplify trigonometric integrals after substituting. Recognizing \( \frac{\cos x}{\sin^2 x} = \cot x \csc x \) is crucial for quickly integrating this type of expression. Pay attention to initial conditions for finding the constant \( C \).

 

Question 19.
(ii) Solve the differential equation \( \frac { dy }{ dx } + 2y \tan x = \sin x \), if \( y = 0 \) for \( x = \frac { \pi }{ 3 } \).
Answer:
Given the differential equation: \( \frac { dy }{ dx } + 2y \tan x = \sin x \).
This is a linear differential equation in \( y \) of the form \( \frac { dy }{ dx } + Py = Q \).
Here, \( P = 2 \tan x \) and \( Q = \sin x \).
The integrating factor (I.F.) is \( e^{\int P dx} \).
\( I.F. = e^{\int 2 \tan x dx} \)
\( = e^{2 \log |\sec x|} \)
\( = e^{\log (\sec x)^2} \)
\( = \sec^2 x \)
The general solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C \).
\( y \cdot \sec^2 x = \int (\sin x) (\sec^2 x) dx + C \)
\( y \sec^2 x = \int \sin x \cdot \frac{1}{\cos^2 x} dx + C \)
\( y \sec^2 x = \int \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} dx + C \)
\( y \sec^2 x = \int \tan x \sec x dx + C \)
We know that \( \int \tan x \sec x dx = \sec x \).
\( y \sec^2 x = \sec x + C \)
To solve for \( y \), divide by \( \sec^2 x \):
\( y = \frac{\sec x}{\sec^2 x} + \frac{C}{\sec^2 x} \)
\( y = \frac{1}{\sec x} + C \cos^2 x \)
\( y = \cos x + C \cos^2 x \) (Equation 1)
We are given that \( y = 0 \) when \( x = \frac{\pi}{3} \). Substitute these values into Equation 1:
\( 0 = \cos (\frac{\pi}{3}) + C \cos^2 (\frac{\pi}{3}) \)
Since \( \cos (\frac{\pi}{3}) = \frac{1}{2} \):
\( 0 = \frac{1}{2} + C (\frac{1}{2})^2 \)
\( 0 = \frac{1}{2} + \frac{C}{4} \)
Multiply by 4 to clear fractions:
\( 0 = 2 + C \)
\( C = -2 \)
Substitute \( C = -2 \) back into Equation 1:
\( y = \cos x - 2 \cos^2 x \)
This is the required particular solution. Proper integration of trigonometric functions is essential.
In simple words: We found a special multiplying factor for the equation. After integrating, we used the given pair of \( x \) and \( y \) values to find the specific constant for this problem. This gave us the final exact solution for \( y \).

🎯 Exam Tip: Remember standard integral forms like \( \int \tan x \sec x dx = \sec x \). When finding \( C \) for particular solutions, be careful with trigonometric values and algebraic manipulation.

 

Question 19.
(iii) Solve the differential equation \( \frac { dx }{ dy } + x \cot y = 2y + y^2 \cot y \), (\( y \ne 0 \)) given that \( x = 0 \) when \( y = \frac { \pi }{ 2 } \).
Answer:
Given the differential equation: \( \frac { dx }{ dy } + x \cot y = 2y + y^2 \cot y \).
This is a linear differential equation in \( x \) of the form \( \frac { dx }{ dy } + Px = Q \).
Here, \( P = \cot y \) and \( Q = 2y + y^2 \cot y \).
The integrating factor (I.F.) is \( e^{\int P dy} \).
\( I.F. = e^{\int \cot y dy} \)
\( = e^{\log |\sin y|} \)
\( = \sin y \)
The general solution is given by \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + C \).
\( x \sin y = \int (2y + y^2 \cot y) \sin y dy + C \)
Now, distribute \( \sin y \) inside the integral:
\( x \sin y = \int (2y \sin y + y^2 \cot y \sin y) dy + C \)
\( x \sin y = \int (2y \sin y + y^2 \frac{\cos y}{\sin y} \sin y) dy + C \)
\( x \sin y = \int (2y \sin y + y^2 \cos y) dy + C \)
The integral \( \int (2y \sin y + y^2 \cos y) dy \) can be solved by recognizing it as the derivative of a product.
We know that \( \frac{d}{dy}(y^2 \sin y) = 2y \sin y + y^2 \cos y \).
So, \( \int (2y \sin y + y^2 \cos y) dy = y^2 \sin y \).
Therefore, the general solution is:
\( x \sin y = y^2 \sin y + C \) (Equation 1)
We are given that \( x = 0 \) when \( y = \frac{\pi}{2} \). Substitute these values into Equation 1:
\( 0 \cdot \sin (\frac{\pi}{2}) = (\frac{\pi}{2})^2 \sin (\frac{\pi}{2}) + C \)
Since \( \sin (\frac{\pi}{2}) = 1 \):
\( 0 \cdot 1 = (\frac{\pi}{2})^2 \cdot 1 + C \)
\( 0 = \frac{\pi^2}{4} + C \)
\( C = -\frac{\pi^2}{4} \)
Substitute \( C = -\frac{\pi^2}{4} \) back into Equation 1:
\( x \sin y = y^2 \sin y - \frac{\pi^2}{4} \)
This is the required particular solution. Recognizing the product rule in reverse is a clever shortcut.
In simple words: We solved this equation by using an integrating factor for \( x \). The tricky part was seeing that a part of the integral was the result of differentiating a product. After that, we used the given starting values to find the exact constant for our solution.

🎯 Exam Tip: Always be on the lookout for integrals that are derivatives of products or quotients. For example, \( \int (f'(x)g(x) + f(x)g'(x)) dx = f(x)g(x) \). This can save a lot of time compared to using integration by parts multiple times.

 

Question 19.
(iv) Solve the differential equation \( x \frac { dy }{ dx } + y = x \cos x + \sin x \), given \( y(\frac { \pi }{ 2 }) = 1 \).
Answer:
Given the differential equation: \( x \frac { dy }{ dx } + y = x \cos x + \sin x \).
First, divide by \( x \) to get the standard linear form \( \frac { dy }{ dx } + Py = Q \):
\( \frac { dy }{ dx } + \frac{1}{x}y = \cos x + \frac{\sin x}{x} \)
Here, \( P = \frac{1}{x} \) and \( Q = \cos x + \frac{\sin x}{x} \).
The integrating factor (I.F.) is \( e^{\int P dx} \).
\( I.F. = e^{\int \frac{1}{x} dx} \)
\( = e^{\log |x|} = x \)
The general solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C \).
\( y \cdot x = \int \left( \cos x + \frac{\sin x}{x} \right) x dx + C \)
\( xy = \int (x \cos x + \sin x) dx + C \)
The integral \( \int (x \cos x + \sin x) dx \) can be solved by recognizing it as the derivative of a product.
We know that \( \frac{d}{dx}(x \sin x) = 1 \cdot \sin x + x \cos x \).
So, \( \int (x \cos x + \sin x) dx = x \sin x \).
Therefore, the general solution is:
\( xy = x \sin x + C \) (Equation 1)
We are given \( y(\frac{\pi}{2}) = 1 \), which means \( y = 1 \) when \( x = \frac{\pi}{2} \). Substitute these values into Equation 1:
\( (\frac{\pi}{2}) \cdot 1 = (\frac{\pi}{2}) \sin (\frac{\pi}{2}) + C \)
Since \( \sin (\frac{\pi}{2}) = 1 \):
\( \frac{\pi}{2} = \frac{\pi}{2} \cdot 1 + C \)
\( \frac{\pi}{2} = \frac{\pi}{2} + C \)
\( C = 0 \)
Substitute \( C = 0 \) back into Equation 1:
\( xy = x \sin x \)
Divide by \( x \) (assuming \( x \neq 0 \)):
\( y = \sin x \)
This is the required particular solution. This problem is a good example of the reverse product rule use.
In simple words: We changed the equation into a standard form and found an integrating factor. After integrating, we noticed that part of the integral was simply the result of multiplying and differentiating two functions. Using the given starting values, we found the constant to be zero, which simplified the final answer for \( y \).

🎯 Exam Tip: Whenever you see an integrand of the form \( (f'(x)g(x) + f(x)g'(x)) \), recognize it as \( \frac{d}{dx}(f(x)g(x)) \). Here, \( x \cos x + \sin x \) is \( \frac{d}{dx}(x \sin x) \). This makes integration much faster.

 

Question 20.
Solve the differential equation \( (1 + x^2)\frac { dy }{ dx } + 2xy - 4x^2 = 0 \) subject to the initial condition \( y(0) = 0 \).
Answer:
Given the differential equation: \( (1 + x^2)\frac { dy }{ dx } + 2xy - 4x^2 = 0 \).
First, rearrange it into the standard linear form \( \frac { dy }{ dx } + Py = Q \).
Move \( -4x^2 \) to the right side:
\( (1 + x^2)\frac { dy }{ dx } + 2xy = 4x^2 \)
Now, divide the entire equation by \( (1 + x^2) \):
\( \frac { dy }{ dx } + \frac{2x}{1 + x^2}y = \frac{4x^2}{1 + x^2} \)
Here, \( P = \frac{2x}{1 + x^2} \) and \( Q = \frac{4x^2}{1 + x^2} \).
The integrating factor (I.F.) is \( e^{\int P dx} \).
\( I.F. = e^{\int \frac{2x}{1 + x^2} dx} \)
Let \( u = 1 + x^2 \), then \( du = 2x dx \).
\( I.F. = e^{\int \frac{1}{u} du} = e^{\log |u|} = u \)
\( I.F. = 1 + x^2 \)
The general solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C \).
\( y (1 + x^2) = \int \left( \frac{4x^2}{1 + x^2} \right) (1 + x^2) dx + C \)
\( y (1 + x^2) = \int 4x^2 dx + C \)
Now, integrate \( 4x^2 \):
\( y (1 + x^2) = 4 \frac{x^3}{3} + C \) (Equation 1)
We are given the initial condition \( y(0) = 0 \), which means \( y = 0 \) when \( x = 0 \). Substitute these values into Equation 1:
\( 0 (1 + 0^2) = 4 \frac{0^3}{3} + C \)
\( 0 = 0 + C \)
\( C = 0 \)
Substitute \( C = 0 \) back into Equation 1:
\( y (1 + x^2) = \frac{4x^3}{3} \)
\( y = \frac{4x^3}{3(1 + x^2)} \)
This is the required particular solution. This type of equation often appears in physics and engineering.
In simple words: We rearranged the given equation into a standard form. We then found a special multiplier called the integrating factor, which helped us solve the equation. Finally, using the given starting condition that \( y \) is zero when \( x \) is zero, we found the exact constant for our final specific solution.

🎯 Exam Tip: When the integrand for the integrating factor is \( \frac{f'(x)}{f(x)} \), its integral is \( \log |f(x)| \), making the I.F. simply \( f(x) \). This is a common pattern in linear differential equations.

 

Question 21.
Solve \( \frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{1}{\left(1+x^2\right)^2} \), if \( y = 0 \) when \( x = 1 \).
Answer:
Given the differential equation: \( \frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{1}{\left(1+x^2\right)^2} \).
This is already in the standard linear form \( \frac { dy }{ dx } + Py = Q \).
Here, \( P = \frac{2x}{1+x^2} \) and \( Q = \frac{1}{(1+x^2)^2} \).
The integrating factor (I.F.) is \( e^{\int P dx} \).
\( I.F. = e^{\int \frac{2x}{1+x^2} dx} \)
As seen in the previous problem, if \( u = 1+x^2 \), then \( du = 2x dx \).
\( I.F. = e^{\int \frac{1}{u} du} = e^{\log |u|} = u \)
\( I.F. = 1+x^2 \)
The general solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C \).
\( y (1+x^2) = \int \frac{1}{(1+x^2)^2} \cdot (1+x^2) dx + C \)
\( y (1+x^2) = \int \frac{1}{1+x^2} dx + C \)
We know that \( \int \frac{1}{1+x^2} dx = \tan^{-1} x \).
\( y (1+x^2) = \tan^{-1} x + C \) (Equation 1)
We are given the condition \( y = 0 \) when \( x = 1 \). Substitute these values into Equation 1:
\( 0 (1+1^2) = \tan^{-1} (1) + C \)
\( 0 (2) = \frac{\pi}{4} + C \)
\( 0 = \frac{\pi}{4} + C \)
\( C = -\frac{\pi}{4} \)
Substitute \( C = -\frac{\pi}{4} \) back into Equation 1:
\( y (1+x^2) = \tan^{-1} x - \frac{\pi}{4} \)
\( y = \frac{\tan^{-1} x - \frac{\pi}{4}}{1+x^2} \)
This is the required particular solution. Remember to correctly evaluate the inverse tangent function.
In simple words: We solved the given equation by finding an integrating factor, which helped us simplify it. After integrating, we used the specific starting values for \( y \) and \( x \) to find the exact value of the constant. This gave us the final answer, which is a specific formula for \( y \).

🎯 Exam Tip: Be familiar with common integrals such as \( \int \frac{1}{1+x^2} dx = \tan^{-1} x \). Also, accurately evaluate inverse trigonometric functions for standard angles when applying initial conditions.

 

Question 22.
Solve the differential equation \( x\frac { dy }{ dx } + y = x^3 \), given that \( y = 1 \) when \( x = 2 \).
Answer:
Given the differential equation: \( x\frac { dy }{ dx } + y = x^3 \).
First, divide by \( x \) to get the standard linear form \( \frac { dy }{ dx } + Py = Q \):
\( \frac { dy }{ dx } + \frac{1}{x}y = x^2 \)
Here, \( P = \frac{1}{x} \) and \( Q = x^2 \).
The integrating factor (I.F.) is \( e^{\int P dx} \).
\( I.F. = e^{\int \frac{1}{x} dx} \)
\( = e^{\log |x|} = x \)
The general solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C \).
\( y \cdot x = \int (x^2) \cdot (x) dx + C \)
\( xy = \int x^3 dx + C \)
Now, integrate \( x^3 \):
\( xy = \frac{x^4}{4} + C \) (Equation 1)
We are given that \( y = 1 \) when \( x = 2 \). Substitute these values into Equation 1:
\( (2)(1) = \frac{(2)^4}{4} + C \)
\( 2 = \frac{16}{4} + C \)
\( 2 = 4 + C \)
\( C = 2 - 4 \)
\( C = -2 \)
Substitute \( C = -2 \) back into Equation 1:
\( xy = \frac{x^4}{4} - 2 \)
To express \( y \) explicitly:
\( y = \frac{x^3}{4} - \frac{2}{x} \)
This is the required particular solution. This method applies broadly to linear differential equations.
In simple words: We started by making the equation into a standard form and then found a special multiplier. After integrating, we used the given starting values for \( y \) and \( x \) to figure out the exact number for the constant. This gave us the final specific formula for \( y \).

🎯 Exam Tip: Always simplify the differential equation to the standard linear form before identifying \( P \) and \( Q \). For definite integrals, be careful with the arithmetic when substituting values to find \( C \).

 

Question 23.
Find the particular solution of the differential equation \( \frac{d y}{d x}=-\frac{x+y \cos x}{1+\sin x} \), given that \( y = 1 \) when \( x = 0 \).
Answer:
Given the differential equation: \( \frac{d y}{d x}=-\frac{x+y \cos x}{1+\sin x} \).
First, rearrange the equation into a linear differential form.
\( \frac{d y}{d x} = -\frac{x}{1+\sin x} - \frac{y \cos x}{1+\sin x} \)
Move the term with \( y \) to the left side:
\( \frac{d y}{d x} + \frac{\cos x}{1+\sin x} y = -\frac{x}{1+\sin x} \)
This is in the standard linear form \( \frac { dy }{ dx } + Py = Q \).
Here, \( P = \frac{\cos x}{1+\sin x} \) and \( Q = -\frac{x}{1+\sin x} \).
The integrating factor (I.F.) is \( e^{\int P dx} \).
\( I.F. = e^{\int \frac{\cos x}{1+\sin x} dx} \)
Let \( u = 1+\sin x \), then \( du = \cos x dx \).
\( I.F. = e^{\int \frac{1}{u} du} = e^{\log |u|} = u \)
\( I.F. = 1+\sin x \)
The general solution is given by \( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C \).
\( y (1+\sin x) = \int \left( -\frac{x}{1+\sin x} \right) (1+\sin x) dx + C \)
\( y (1+\sin x) = \int -x dx + C \)
Now, integrate \( -x \):
\( y (1+\sin x) = -\frac{x^2}{2} + C \) (Equation 1)
We are given that \( y = 1 \) when \( x = 0 \). Substitute these values into Equation 1:
\( 1 (1+\sin 0) = -\frac{0^2}{2} + C \)
Since \( \sin 0 = 0 \):
\( 1 (1+0) = 0 + C \)
\( 1 = C \)
\( C = 1 \)
Substitute \( C = 1 \) back into Equation 1:
\( y (1+\sin x) = -\frac{x^2}{2} + 1 \)
This is the required particular solution. Recognizing the integral for I.F. is key here.
In simple words: We first put the equation into a standard format and found a special multiplier. After that, we integrated the right side and used the given starting values of \( y \) and \( x \) to find the exact constant for our final specific solution.

🎯 Exam Tip: Always simplify the integral \( \int Q \cdot (I.F.) dx \) before evaluating it. Here, the \( (1+\sin x) \) terms canceled out, leading to a much simpler integral.

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