GSEB Class 12 Maths Solutions Chapter 9 Differential Equations Exercise 9.6

Question 1. For each of the differential equations given in questions 1 to 12, find the general solution:
\( \frac{dy}{dx} + 2y = \sin x \)
Answer: The given differential equation is \( \frac{dy}{dx} + 2y = \sin x \). This equation is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = 2 \) and \( Q = \sin x \).
First, we find the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int 2 dx} = e^{2x} \)
Now, the solution to the differential equation is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y e^{2x} = \int (\sin x) e^{2x} dx + C \)
Let \( I_1 = \int e^{2x} \sin x dx \). We will integrate by parts, choosing \( e^{2x} \) as the first function.
\( I_1 = e^{2x} (-\cos x) - \int (2e^{2x}) (-\cos x) dx \)
\( = -e^{2x} \cos x + 2 \int e^{2x} \cos x dx \)
Now, we integrate \( \int e^{2x} \cos x dx \) by parts again, choosing \( e^{2x} \) as the first function.
\( \int e^{2x} \cos x dx = e^{2x} (\sin x) - \int (2e^{2x}) (\sin x) dx \)
\( = e^{2x} \sin x - 2 \int e^{2x} \sin x dx \)
Substitute this back into the expression for \( I_1 \):
\( I_1 = -e^{2x} \cos x + 2 (e^{2x} \sin x - 2 \int e^{2x} \sin x dx) \)
\( I_1 = -e^{2x} \cos x + 2e^{2x} \sin x - 4 I_1 \)
Rearrange to solve for \( I_1 \):
\( I_1 + 4 I_1 = 2e^{2x} \sin x - e^{2x} \cos x \)
\( 5 I_1 = e^{2x} (2 \sin x - \cos x) \)
\( I_1 = \frac{e^{2x}}{5} (2 \sin x - \cos x) \)
Substituting \( I_1 \) back into the general solution equation:
\( y e^{2x} = \frac{e^{2x}}{5} (2 \sin x - \cos x) + C \)
To simplify, we can divide by \( e^{2x} \) (assuming \( e^{2x} \ne 0 \)):
\( y = \frac{1}{5} (2 \sin x - \cos x) + C e^{-2x} \)
Alternatively, multiplying by 5:
\( 5y = 2 \sin x - \cos x + 5Ce^{-2x} \)
In simple words: First, identify the type of equation and find the special multiplying factor. Then, use this factor to solve the equation, often requiring integration by parts. Finally, simplify the result to get the general solution.

Exam Tip: Remember to correctly identify \(P\) and \(Q\) in the linear differential equation. Pay close attention to integration by parts, especially when it needs to be applied multiple times, as common mistakes occur with signs and function choices.

Question 2.
\( \frac{dy}{dx} + 3y = e^{-2x} \)
Answer: The given differential equation is \( \frac{dy}{dx} + 3y = e^{-2x} \). This equation is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = 3 \) and \( Q = e^{-2x} \).
First, we determine the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int 3 dx} = e^{3x} \)
Next, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y e^{3x} = \int (e^{-2x} \cdot e^{3x}) dx + C \)
\( y e^{3x} = \int e^{x} dx + C \)
\( y e^{3x} = e^{x} + C \)
To express \( y \) explicitly:
\( y = \frac{e^{x}}{e^{3x}} + \frac{C}{e^{3x}} \)
\( y = e^{x-3x} + C e^{-3x} \)
\( y = e^{-2x} + C e^{-3x} \)
In simple words: This equation is linear. Find the integrating factor by raising 'e' to the integral of P. Then, multiply by this factor and integrate Q times the factor to get the solution. Finally, solve for y.

Exam Tip: Always simplify the product of exponentials \( e^{-2x} \cdot e^{3x} \) before integrating to avoid unnecessary complications. Ensure the constant of integration \( C \) is correctly handled in the final step.

Question 3.
\( \frac{dy}{dx} + \frac{y}{x} = x^2 \)
Answer: The given differential equation is \( \frac{dy}{dx} + \frac{y}{x} = x^2 \). This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{x} \) and \( Q = x^2 \).
First, we find the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log |x|} \). Since \( x^2 \) is involved, we can assume \( x > 0 \) for \( \log x \). So, \( e^{\log x} = x \).
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y \cdot x = \int (x^2 \cdot x) dx + C \)
\( yx = \int x^3 dx + C \)
\( yx = \frac{x^4}{4} + C \)
To solve for \( y \):
\( y = \frac{x^4}{4x} + \frac{C}{x} \)
\( y = \frac{x^3}{4} + \frac{C}{x} \)
In simple words: Recognize this as a linear equation. Calculate the integrating factor by integrating \(1/x\). Multiply the equation by this factor and then integrate the right side to find the solution for y.

Exam Tip: When \( \int \frac{1}{x} dx \) is calculated, it results in \( \log |x| \). For positive \( x \), this simplifies to \( \log x \). Always be careful with the constant of integration \( C \) and ensure it's added at the correct step.

Question 4.
\( \frac{dy}{dx} + \sec x \cdot y = \tan x \), \( (0 \le x < \frac{\pi}{2}) \)
Answer: The given differential equation is \( \frac{dy}{dx} + \sec x \cdot y = \tan x \). This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \sec x \) and \( Q = \tan x \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int \sec x dx} = e^{\log |\sec x + \tan x|} \). Since \( 0 \le x < \frac{\pi}{2} \), \( \sec x + \tan x \) is positive, so we can write it as \( \sec x + \tan x \).
I.F. \( = \sec x + \tan x \)
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y (\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx + C \)
\( y (\sec x + \tan x) = \int (\sec x \tan x + \tan^2 x) dx + C \)
We know that \( \tan^2 x = \sec^2 x - 1 \). So, we substitute this:
\( y (\sec x + \tan x) = \int (\sec x \tan x + \sec^2 x - 1) dx + C \)
Now, we integrate each term:
\( \int \sec x \tan x dx = \sec x \)
\( \int \sec^2 x dx = \tan x \)
\( \int -1 dx = -x \)
So, the solution becomes:
\( y (\sec x + \tan x) = \sec x + \tan x - x + C \)
In simple words: This is a linear equation. Find the integrating factor by integrating \( \sec x \). Then, use the general solution formula, and you'll need to integrate \( \tan x (\sec x + \tan x) \), which simplifies using trigonometric identities before integration.

Exam Tip: Be familiar with standard integral formulas like \( \int \sec x dx = \log |\sec x + \tan x| \) and trigonometric identities such as \( \tan^2 x = \sec^2 x - 1 \). These are essential for solving such problems efficiently.

Question 5.
\( \cos^2 x \frac{dy}{dx} + y = \tan x \), \( (0 \le x < \frac{\pi}{2}) \)
Answer: The given differential equation is \( \cos^2 x \frac{dy}{dx} + y = \tan x \). To convert it into the standard linear form \( \frac{dy}{dx} + Py = Q \), we divide by \( \cos^2 x \):
\( \frac{dy}{dx} + \frac{1}{\cos^2 x} y = \frac{\tan x}{\cos^2 x} \)
\( \frac{dy}{dx} + \sec^2 x \cdot y = \tan x \sec^2 x \)
Here, \( P = \sec^2 x \) and \( Q = \tan x \sec^2 x \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int \sec^2 x dx} = e^{\tan x} \)
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y e^{\tan x} = \int (\tan x \sec^2 x \cdot e^{\tan x}) dx + C \)
To solve this integral, we use substitution. Let \( t = \tan x \). Then \( dt = \sec^2 x dx \).
The integral becomes \( \int t e^t dt \). We now integrate by parts, with \( t \) as the first function and \( e^t \) as the second.
\( \int t e^t dt = t e^t - \int (1 \cdot e^t) dt \)
\( = t e^t - e^t + C \)
Substitute back \( t = \tan x \):
\( y e^{\tan x} = \tan x e^{\tan x} - e^{\tan x} + C \)
To solve for \( y \), divide by \( e^{\tan x} \):
\( y = \frac{\tan x e^{\tan x}}{e^{\tan x}} - \frac{e^{\tan x}}{e^{\tan x}} + \frac{C}{e^{\tan x}} \)
\( y = \tan x - 1 + C e^{-\tan x} \)
In simple words: First, rearrange the equation into a standard linear form by dividing by \( \cos^2 x \). Then, find the integrating factor using the new P value. Apply the solution formula and integrate using substitution and integration by parts. Finally, express y explicitly.

Exam Tip: For equations like this, always start by converting to the standard \( \frac{dy}{dx} + Py = Q \) form. Recognizing the substitution \( t = \tan x \) and using integration by parts correctly for \( \int t e^t dt \) are critical steps.

Question 6.
\( x \frac{dy}{dx} + 2y = x^2 \log x \)
Answer: The given differential equation is \( x \frac{dy}{dx} + 2y = x^2 \log x \). To convert it into the standard linear form \( \frac{dy}{dx} + Py = Q \), we divide by \( x \):
\( \frac{dy}{dx} + \frac{2}{x} y = x \log x \)
Here, \( P = \frac{2}{x} \) and \( Q = x \log x \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \log |x|} = e^{\log x^2} = x^2 \). (Assuming \( x \ne 0 \)).
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y x^2 = \int (x \log x \cdot x^2) dx + C \)
\( y x^2 = \int x^3 \log x dx + C \)
We integrate by parts, choosing \( \log x \) as the first function and \( x^3 \) as the second function:
\( \int u dv = uv - \int v du \)
Let \( u = \log x \implies du = \frac{1}{x} dx \)
Let \( dv = x^3 dx \implies v = \int x^3 dx = \frac{x^4}{4} \)
So, \( \int x^3 \log x dx = (\log x) \frac{x^4}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x} dx \)
\( = \frac{x^4}{4} \log x - \int \frac{x^3}{4} dx \)
\( = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 dx \)
\( = \frac{x^4}{4} \log x - \frac{1}{4} \left( \frac{x^4}{4} \right) + C \)
\( = \frac{x^4}{4} \log x - \frac{x^4}{16} + C \)
Substituting this back into the solution equation:
\( y x^2 = \frac{x^4}{4} \log x - \frac{x^4}{16} + C \)
To solve for \( y \), divide by \( x^2 \):
\( y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{C}{x^2} \)
We can also write it as:
\( 16y = 4x^2 \log x - x^2 + 16 C x^{-2} \)
\( 16y = x^2 (4 \log x - 1) + 16 C x^{-2} \)
In simple words: First, divide the whole equation by x to get the standard form. Then find the integrating factor, which will be \( x^2 \). After that, multiply and integrate using integration by parts, taking \( \log x \) as the first function. Finally, divide by \( x^2 \) to solve for y.

Exam Tip: When using integration by parts, carefully choose which function is \(u\) and which is \(dv\). For terms like \(x^n \log x\), it's generally best to pick \(u = \log x\). Don't forget to divide by \(x^2\) at the end to get \(y\) alone.

Question 7.
\( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \)
Answer: The given differential equation is \( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \). To convert it into the standard linear form \( \frac{dy}{dx} + Py = Q \), we divide by \( x \log x \):
\( \frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x^2} \)
Here, \( P = \frac{1}{x \log x} \) and \( Q = \frac{2}{x^2} \).
First, we determine the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int \frac{1}{x \log x} dx} \)
To integrate \( \int \frac{1}{x \log x} dx \), let \( t = \log x \). Then \( dt = \frac{1}{x} dx \).
So, \( \int \frac{1}{x \log x} dx = \int \frac{1}{t} dt = \log |t| = \log |\log x| \).
Therefore, I.F. \( = e^{\log |\log x|} = |\log x| \). Assuming \( \log x > 0 \), we use \( \log x \).
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y \log x = \int \left( \frac{2}{x^2} \cdot \log x \right) dx + C \)
\( y \log x = 2 \int x^{-2} \log x dx + C \)
We integrate by parts, choosing \( \log x \) as the first function and \( x^{-2} \) as the second function:
Let \( u = \log x \implies du = \frac{1}{x} dx \)
Let \( dv = x^{-2} dx \implies v = \int x^{-2} dx = \frac{x^{-1}}{-1} = -\frac{1}{x} \)
So, \( \int x^{-2} \log x dx = (\log x) \left( -\frac{1}{x} \right) - \int \left( -\frac{1}{x} \right) \cdot \frac{1}{x} dx \)
\( = -\frac{\log x}{x} + \int x^{-2} dx \)
\( = -\frac{\log x}{x} + \left( -\frac{1}{x} \right) = -\frac{\log x}{x} - \frac{1}{x} \)
Substitute this back into the solution equation:
\( y \log x = 2 \left( -\frac{\log x}{x} - \frac{1}{x} \right) + C \)
\( y \log x = -\frac{2 \log x}{x} - \frac{2}{x} + C \)
To solve for \( y \):
\( y = -\frac{2}{x} - \frac{2}{x \log x} + \frac{C}{\log x} \)
\( y = -\frac{2}{x} \left( 1 + \frac{1}{\log x} \right) + \frac{C}{\log x} \)
The required solution is \( y \log x = -\frac{2}{x} (1 + \log x) + C \).
In simple words: Begin by making the equation linear by dividing by \( x \log x \). Find the integrating factor, which will be \( \log x \). Then, solve the equation using integration by parts, letting \( \log x \) be the first function. Finally, simplify to find y.

Exam Tip: Be cautious when integrating \( \frac{1}{x \log x} \) for the I.F., as it requires a substitution of \( t = \log x \). Similarly, for integration by parts, choose \( \log x \) as the first function. Remember to account for absolute values in logarithms unless the domain specifies positive values.

Question 8.
\( (1 + x^2)dy + 2xy dx = \cot x dx \), \( (x \ne 0) \)
Answer: The given differential equation is \( (1 + x^2)dy + 2xy dx = \cot x dx \). First, we divide by \( dx \) to get \( \frac{dy}{dx} \) and then rearrange it into the standard linear form \( \frac{dy}{dx} + Py = Q \):
\( (1 + x^2) \frac{dy}{dx} + 2xy = \cot x \)
\( \frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{\cot x}{1 + x^2} \)
Here, \( P = \frac{2x}{1 + x^2} \) and \( Q = \frac{\cot x}{1 + x^2} \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int \frac{2x}{1 + x^2} dx} \)
To integrate \( \int \frac{2x}{1 + x^2} dx \), let \( t = 1 + x^2 \). Then \( dt = 2x dx \).
So, \( \int \frac{2x}{1 + x^2} dx = \int \frac{1}{t} dt = \log |t| = \log (1 + x^2) \) (since \( 1 + x^2 \) is always positive).
Therefore, I.F. \( = e^{\log (1 + x^2)} = 1 + x^2 \).
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y (1 + x^2) = \int \left( \frac{\cot x}{1 + x^2} \cdot (1 + x^2) \right) dx + C \)
\( y (1 + x^2) = \int \cot x dx + C \)
\( y (1 + x^2) = \log |\sin x| + C \)
In simple words: Convert the equation to the standard linear form. Find the integrating factor by integrating \( \frac{2x}{1+x^2} \), which gives \( 1+x^2 \). Then, use the solution formula and integrate \( \cot x \) to get the final solution.

Exam Tip: Remember to simplify the differential equation into the standard form before finding the integrating factor. The integral \( \int \frac{2x}{1+x^2} dx \) is a direct application of the \( \int \frac{f'(x)}{f(x)} dx = \log |f(x)| \) rule.

Question 9.
\( x \frac{dy}{dx} + y - x + xy \cot x = 0 \), \( (x \ne 0) \)
Answer: The given differential equation is \( x \frac{dy}{dx} + y - x + xy \cot x = 0 \). First, we rearrange the terms to get it into the standard linear form \( \frac{dy}{dx} + Py = Q \):
\( x \frac{dy}{dx} + y (1 + x \cot x) = x \)
Now, divide by \( x \) to make the coefficient of \( \frac{dy}{dx} \) equal to 1:
\( \frac{dy}{dx} + \frac{1 + x \cot x}{x} y = 1 \)
\( \frac{dy}{dx} + \left( \frac{1}{x} + \cot x \right) y = 1 \)
Here, \( P = \frac{1}{x} + \cot x \) and \( Q = 1 \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int (\frac{1}{x} + \cot x) dx} \)
\( = e^{\int \frac{1}{x} dx + \int \cot x dx} \)
\( = e^{\log |x| + \log |\sin x|} \)
\( = e^{\log |x \sin x|} \)
\( = |x \sin x| \). Assuming \( x \sin x > 0 \), we use \( x \sin x \).
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y (x \sin x) = \int (1 \cdot x \sin x) dx + C \)
\( yx \sin x = \int x \sin x dx + C \)
We integrate \( \int x \sin x dx \) by parts, choosing \( x \) as the first function and \( \sin x \) as the second function:
Let \( u = x \implies du = dx \)
Let \( dv = \sin x dx \implies v = \int \sin x dx = -\cos x \)
So, \( \int x \sin x dx = x(-\cos x) - \int (-\cos x) dx \)
\( = -x \cos x + \int \cos x dx \)
\( = -x \cos x + \sin x \)
Substituting this back into the solution equation:
\( yx \sin x = -x \cos x + \sin x + C \)
To solve for \( y \), divide by \( x \sin x \):
\( y = \frac{-x \cos x}{x \sin x} + \frac{\sin x}{x \sin x} + \frac{C}{x \sin x} \)
\( y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x} \)
In simple words: First, rewrite the equation into the standard linear form by isolating \( \frac{dy}{dx} \) and gathering terms with y. Calculate the integrating factor, which will be \( x \sin x \). Then, use the solution formula and apply integration by parts to solve for y.

Exam Tip: Pay close attention to rearranging the equation into the standard linear form \( \frac{dy}{dx} + Py = Q \). The integral for the integrating factor \( \int (\frac{1}{x} + \cot x) dx \) is straightforward, but the integration by parts for \( \int x \sin x dx \) needs to be done carefully.

Question 10.
\( (x + y)\frac{dy}{dx} = 1 \)
Answer: The given differential equation is \( (x + y)\frac{dy}{dx} = 1 \). This equation is not easily solvable as a linear equation in \( y \). However, if we rewrite it as \( \frac{dx}{dy} \), it becomes a linear equation in \( x \):
\( \frac{dx}{dy} = x + y \)
Rearrange this into the standard linear form \( \frac{dx}{dy} + Px = Q \):
\( \frac{dx}{dy} - x = y \)
Here, \( P = -1 \) and \( Q = y \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dy} = e^{\int -1 dy} = e^{-y} \)
Now, the general solution is given by:
\( x \times \text{I.F.} = \int (Q \times \text{I.F.}) dy + C \)
\( x e^{-y} = \int (y \cdot e^{-y}) dy + C \)
We integrate \( \int y e^{-y} dy \) by parts, choosing \( y \) as the first function and \( e^{-y} \) as the second function:
Let \( u = y \implies du = dy \)
Let \( dv = e^{-y} dy \implies v = \int e^{-y} dy = -e^{-y} \)
So, \( \int y e^{-y} dy = y(-e^{-y}) - \int (-e^{-y}) dy \)
\( = -y e^{-y} + \int e^{-y} dy \)
\( = -y e^{-y} - e^{-y} \)
Substituting this back into the solution equation:
\( x e^{-y} = -y e^{-y} - e^{-y} + C \)
To solve for \( x \), divide by \( e^{-y} \):
\( x = \frac{-y e^{-y}}{e^{-y}} - \frac{e^{-y}}{e^{-y}} + \frac{C}{e^{-y}} \)
\( x = -y - 1 + C e^y \)
The required solution is \( x + y + 1 = C e^y \).
In simple words: When \( \frac{dy}{dx} \) is complex, try flipping it to \( \frac{dx}{dy} \). This makes it a linear equation in x. Find the integrating factor with respect to y, then use integration by parts to solve for x.

Exam Tip: If an equation is not linear in \(y\), consider if it can be made linear in \(x\) by using \( \frac{dx}{dy} \). This often simplifies the problem significantly. Pay attention to the variable of integration when calculating the I.F. and in the main integral.

Question 11.
\( y dx + (x - y^2) dy = 0 \)
Answer: The given differential equation is \( y dx + (x - y^2) dy = 0 \). This equation is not easily solvable as a linear equation in \( y \). Let's rearrange it to express \( \frac{dx}{dy} \) and see if it becomes linear in \( x \):
\( y \frac{dx}{dy} + (x - y^2) = 0 \)
\( y \frac{dx}{dy} + x = y^2 \)
Now, divide by \( y \) to get it into the standard linear form \( \frac{dx}{dy} + Px = Q \):
\( \frac{dx}{dy} + \frac{1}{y} x = y \)
Here, \( P = \frac{1}{y} \) and \( Q = y \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dy} = e^{\int \frac{1}{y} dy} = e^{\log |y|} = |y| \). Assuming \( y > 0 \), we use \( y \).
Now, the general solution is given by:
\( x \times \text{I.F.} = \int (Q \times \text{I.F.}) dy + C \)
\( x \cdot y = \int (y \cdot y) dy + C \)
\( xy = \int y^2 dy + C \)
\( xy = \frac{y^3}{3} + C \)
To solve for \( x \):
\( x = \frac{y^3}{3y} + \frac{C}{y} \)
\( x = \frac{y^2}{3} + \frac{C}{y} \)
In simple words: When the equation is not linear in \( y \), try to rewrite it as a linear equation in \( x \) by finding \( \frac{dx}{dy} \). Then, find the integrating factor, which will be \( y \). Finally, integrate the right-hand side to get the solution for x.

Exam Tip: If an equation isn't linear in \(y\), always check if it can be transformed into a linear equation in \(x\) by using \( \frac{dx}{dy} \). This approach is often the key to solving such problems efficiently. Make sure to choose \(P\) and \(Q\) based on the new variable of integration.

Question 12.
\( (x + 3y^2)\frac{dy}{dx} = y \), \( (y > 0) \)
Answer: The given differential equation is \( (x + 3y^2)\frac{dy}{dx} = y \). This equation is not linear in \( y \). We rearrange it to find \( \frac{dx}{dy} \) to see if it becomes linear in \( x \):
\( \frac{dx}{dy} = \frac{x + 3y^2}{y} \)
\( \frac{dx}{dy} = \frac{x}{y} + 3y \)
Rearrange this into the standard linear form \( \frac{dx}{dy} + Px = Q \):
\( \frac{dx}{dy} - \frac{1}{y} x = 3y \)
Here, \( P = -\frac{1}{y} \) and \( Q = 3y \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\log |y|} = e^{\log |y^{-1}|} = |y^{-1}| = \frac{1}{y} \) (since \( y > 0 \)).
Now, the general solution is given by:
\( x \times \text{I.F.} = \int (Q \times \text{I.F.}) dy + C \)
\( x \cdot \frac{1}{y} = \int \left( 3y \cdot \frac{1}{y} \right) dy + C \)
\( \frac{x}{y} = \int 3 dy + C \)
\( \frac{x}{y} = 3y + C \)
To solve for \( x \):
\( x = y (3y + C) \)
\( x = 3y^2 + Cy \)
In simple words: First, transform the given equation into a linear form with \( x \) as the dependent variable. Then, calculate the integrating factor, which will be \( \frac{1}{y} \). Finally, integrate and solve for \( x \) to get the required general solution.

Exam Tip: Recognizing that the equation is linear in \(x\) and not in \(y\) is the most important step. When calculating the integrating factor for \(P = -\frac{1}{y}\), remember the negative sign in the exponent and how it affects the logarithm.

Question 13. For each of the differential equations given in questions 13 to 15, find particular solution satisfying the given conditions:
\( \frac{dy}{dx} + 2y \tan x = \sin x \); \( y = 0 \), when \( x = \frac{\pi}{3} \)
Answer: The given differential equation is \( \frac{dy}{dx} + 2y \tan x = \sin x \). This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = 2 \tan x \) and \( Q = \sin x \).
First, we determine the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int 2 \tan x dx} \)
\( = e^{2 \log |\sec x|} = e^{\log (\sec^2 x)} = \sec^2 x \).
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y \sec^2 x = \int (\sin x \cdot \sec^2 x) dx + C \)
\( y \sec^2 x = \int \frac{\sin x}{\cos^2 x} dx + C \)
\( y \sec^2 x = \int \left( \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \right) dx + C \)
\( y \sec^2 x = \int (\tan x \sec x) dx + C \)
\( y \sec^2 x = \sec x + C \)
To find the particular solution, we use the given condition: \( y = 0 \) when \( x = \frac{\pi}{3} \).
Substitute these values into the general solution:
\( 0 \cdot \sec^2 \left( \frac{\pi}{3} \right) = \sec \left( \frac{\pi}{3} \right) + C \)
We know that \( \sec \left( \frac{\pi}{3} \right) = \frac{1}{\cos (\pi/3)} = \frac{1}{1/2} = 2 \).
\( 0 = 2 + C \)
\( C = -2 \)
Now, substitute the value of \( C \) back into the general solution to get the particular solution:
\( y \sec^2 x = \sec x - 2 \)
\( y = \frac{\sec x}{\sec^2 x} - \frac{2}{\sec^2 x} \)
\( y = \cos x - 2 \cos^2 x \)
In simple words: This is a linear equation. Calculate the integrating factor, which will be \( \sec^2 x \). Then, integrate \( \sin x \sec^2 x \) and simplify. Use the given starting conditions (y=0 at x=\(\pi/3\)) to find the value of the constant C, and then write the specific solution.

Exam Tip: For particular solutions, always find the general solution first, then use the given initial conditions to determine the value of the constant \(C\). Be careful with trigonometric values for common angles like \( \pi/3 \).

Question 14.
\( (1 + x^2)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2} \); \( y = 0 \), when \( x = 1 \)
Answer: The given differential equation is \( (1 + x^2)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2} \). To convert it into the standard linear form \( \frac{dy}{dx} + Py = Q \), we divide by \( (1 + x^2) \):
\( \frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{1}{(1 + x^2)^2} \)
Here, \( P = \frac{2x}{1 + x^2} \) and \( Q = \frac{1}{(1 + x^2)^2} \).
First, we determine the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int \frac{2x}{1 + x^2} dx} \)
To integrate \( \int \frac{2x}{1 + x^2} dx \), let \( t = 1 + x^2 \). Then \( dt = 2x dx \).
So, \( \int \frac{2x}{1 + x^2} dx = \int \frac{1}{t} dt = \log |t| = \log (1 + x^2) \) (since \( 1 + x^2 \) is always positive).
Therefore, I.F. \( = e^{\log (1 + x^2)} = 1 + x^2 \).
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y (1 + x^2) = \int \left( \frac{1}{(1 + x^2)^2} \cdot (1 + x^2) \right) dx + C \)
\( y (1 + x^2) = \int \frac{1}{1 + x^2} dx + C \)
\( y (1 + x^2) = \tan^{-1} x + C \)
To find the particular solution, we use the given condition: \( y = 0 \) when \( x = 1 \).
Substitute these values into the general solution:
\( 0 \cdot (1 + 1^2) = \tan^{-1} (1) + C \)
\( 0 = \frac{\pi}{4} + C \)
\( C = -\frac{\pi}{4} \)
Now, substitute the value of \( C \) back into the general solution to get the particular solution:
\( y (1 + x^2) = \tan^{-1} x - \frac{\pi}{4} \)
\( y = \frac{\tan^{-1} x}{1 + x^2} - \frac{\pi}{4(1 + x^2)} \)
In simple words: First, rewrite the equation into the standard linear form. Find the integrating factor, which will be \( 1+x^2 \). Then, integrate \( \frac{1}{1+x^2} \) to get \( \tan^{-1} x \). Use the initial conditions \( y=0 \) when \( x=1 \) to find the constant C, and finally, state the particular solution.

Exam Tip: The integral for the integrating factor is a common one involving substitution. Also, remember the standard integral \( \int \frac{1}{1+x^2} dx = \tan^{-1} x \). Accurately applying the initial conditions to find \(C\) is crucial for particular solutions.

Question 15.
\( \frac{dy}{dx} - 3y \cot x = \sin 2x \); \( y = 2 \), when \( x = \frac{\pi}{2} \)
Answer: The given differential equation is \( \frac{dy}{dx} - 3y \cot x = \sin 2x \). This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = -3 \cot x \) and \( Q = \sin 2x \).
First, we determine the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int -3 \cot x dx} \)
\( = e^{-3 \int \cot x dx} = e^{-3 \log |\sin x|} \)
\( = e^{\log |\sin x|^{-3}} = |\sin x|^{-3} = \frac{1}{\sin^3 x} = \operatorname{cosec}^3 x \).
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y \operatorname{cosec}^3 x = \int (\sin 2x \cdot \operatorname{cosec}^3 x) dx + C \)
We use the identity \( \sin 2x = 2 \sin x \cos x \):
\( y \operatorname{cosec}^3 x = \int (2 \sin x \cos x \cdot \frac{1}{\sin^3 x}) dx + C \)
\( y \operatorname{cosec}^3 x = \int \frac{2 \cos x}{\sin^2 x} dx + C \)
\( y \operatorname{cosec}^3 x = 2 \int \cot x \operatorname{cosec} x dx + C \)
\( y \operatorname{cosec}^3 x = 2 (-\operatorname{cosec} x) + C \)
\( y \operatorname{cosec}^3 x = -2 \operatorname{cosec} x + C \)
To find the particular solution, we use the given condition: \( y = 2 \) when \( x = \frac{\pi}{2} \).
Substitute these values into the general solution:
\( 2 \cdot \operatorname{cosec}^3 \left( \frac{\pi}{2} \right) = -2 \operatorname{cosec} \left( \frac{\pi}{2} \right) + C \)
We know that \( \operatorname{cosec} \left( \frac{\pi}{2} \right) = \frac{1}{\sin (\pi/2)} = \frac{1}{1} = 1 \).
\( 2 \cdot (1)^3 = -2 \cdot (1) + C \)
\( 2 = -2 + C \)
\( C = 4 \)
Now, substitute the value of \( C \) back into the general solution to get the particular solution:
\( y \operatorname{cosec}^3 x = -2 \operatorname{cosec} x + 4 \)
To solve for \( y \), multiply by \( \sin^3 x \):
\( y = -2 \frac{\operatorname{cosec} x}{\operatorname{cosec}^3 x} + 4 \frac{1}{\operatorname{cosec}^3 x} \)
\( y = -2 \sin^2 x + 4 \sin^3 x \)
Alternatively, \( y = -2 \sin^2 x (1 - 2 \sin x) \).
In simple words: This is a linear equation. Calculate the integrating factor, which will be \( \operatorname{cosec}^3 x \). Integrate \( \sin 2x \operatorname{cosec}^3 x \) using a trigonometric identity and standard integrals. Use the given initial conditions \( y=2 \) at \( x=\pi/2 \) to find C and then write the specific solution.

Exam Tip: Carefully handle the negative sign in \( P = -3 \cot x \) when calculating the integrating factor. The simplification of \( \sin 2x \operatorname{cosec}^3 x \) using trigonometric identities is key to making the integral solvable.

Question 16. Find the equation of the curve passing through the origin, given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.
Answer: Let the equation of the curve be \( y = f(x) \). The slope of the tangent to the curve at any point \( (x, y) \) is given by \( \frac{dy}{dx} \).
According to the problem, the slope of the tangent is equal to the sum of the coordinates of the point, so:
\( \frac{dy}{dx} = x + y \)
Rearrange this into the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \):
\( \frac{dy}{dx} - y = x \)
Here, \( P = -1 \) and \( Q = x \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int -1 dx} = e^{-x} \)
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y e^{-x} = \int (x \cdot e^{-x}) dx + C \)
We integrate \( \int x e^{-x} dx \) by parts, choosing \( x \) as the first function and \( e^{-x} \) as the second function:
Let \( u = x \implies du = dx \)
Let \( dv = e^{-x} dx \implies v = \int e^{-x} dx = -e^{-x} \)
So, \( \int x e^{-x} dx = x(-e^{-x}) - \int (-e^{-x}) dx \)
\( = -x e^{-x} + \int e^{-x} dx \)
\( = -x e^{-x} - e^{-x} \)
Substituting this back into the solution equation:
\( y e^{-x} = -x e^{-x} - e^{-x} + C \)
To solve for \( y \), divide by \( e^{-x} \):
\( y = \frac{-x e^{-x}}{e^{-x}} - \frac{e^{-x}}{e^{-x}} + \frac{C}{e^{-x}} \)
\( y = -x - 1 + C e^x \)
To find the particular solution, we use the given condition: the curve passes through the origin \( (0, 0) \).
Substitute these values into the general solution:
\( 0 = -0 - 1 + C e^0 \)
\( 0 = -1 + C \cdot 1 \)
\( C = 1 \)
Now, substitute the value of \( C \) back into the general solution to get the particular solution:
\( y = -x - 1 + 1 \cdot e^x \)
\( y = e^x - x - 1 \)
The required equation of the curve is \( y + x + 1 = e^x \).
In simple words: Translate the problem statement into a differential equation, \( \frac{dy}{dx} = x+y \). Convert it to the linear form. Find the integrating factor \( e^{-x} \). Integrate using parts, then apply the condition that the curve passes through the origin \( (0,0) \) to find the constant C.

Exam Tip: Carefully translating the word problem into a differential equation is the first step. For integration by parts, apply the LIATE rule (Logarithms, Inverse trig, Algebraic, Trigonometric, Exponential) to choose \(u\). Don't forget to use \( e^0 = 1 \) when applying initial conditions at the origin.

Question 17. Find the equation of the curve passing through the point (0, 2), given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
Answer: Let the coordinates of a point on the curve be \( (x, y) \). The slope of the tangent to the curve at that point is \( \frac{dy}{dx} \).
According to the problem statement, the sum of the coordinates \( (x+y) \) exceeds the magnitude of the slope \( |\frac{dy}{dx}| \) by 5. This can be written as:
\( x + y - |\frac{dy}{dx}| = 5 \)
\( |\frac{dy}{dx}| = x + y - 5 \)
This means \( \frac{dy}{dx} = \pm (x + y - 5) \). We need to consider two cases:

(i) Taking the positive sign: \( \frac{dy}{dx} = x + y - 5 \)
Rearrange into the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \):
\( \frac{dy}{dx} - y = x - 5 \)
Here, \( P = -1 \) and \( Q = x - 5 \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int -1 dx} = e^{-x} \)
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y e^{-x} = \int (x - 5) e^{-x} dx + C \)
We integrate \( \int (x - 5) e^{-x} dx \) by parts, choosing \( (x - 5) \) as the first function and \( e^{-x} \) as the second function:
Let \( u = x - 5 \implies du = dx \)
Let \( dv = e^{-x} dx \implies v = \int e^{-x} dx = -e^{-x} \)
So, \( \int (x - 5) e^{-x} dx = (x - 5)(-e^{-x}) - \int (-e^{-x}) dx \)
\( = -(x - 5)e^{-x} + \int e^{-x} dx \)
\( = -(x - 5)e^{-x} - e^{-x} \)
Substituting this back into the solution equation:
\( y e^{-x} = -(x - 5)e^{-x} - e^{-x} + C \)
To solve for \( y \), divide by \( e^{-x} \):
\( y = -(x - 5) - 1 + C e^x \)
\( y = -x + 5 - 1 + C e^x \)
\( y = 4 - x + C e^x \)
To find the particular solution, we use the given condition: the curve passes through the point \( (0, 2) \).
Substitute these values into the general solution:
\( 2 = 4 - 0 + C e^0 \)
\( 2 = 4 + C \cdot 1 \)
\( C = -2 \)
Now, substitute the value of \( C \) back into the general solution to get the particular solution for this case:
\( y = 4 - x - 2e^x \)

(ii) Taking the negative sign: \( \frac{dy}{dx} = -(x + y - 5) \)
\( \frac{dy}{dx} = -x - y + 5 \)
Rearrange into the standard linear differential equation form \( \frac{dy}{dx} + Py = Q \):
\( \frac{dy}{dx} + y = -x + 5 \)
Here, \( P = 1 \) and \( Q = -x + 5 \).
First, we calculate the integrating factor (I.F.):
I.F. \( = e^{\int P dx} = e^{\int 1 dx} = e^x \)
Now, the general solution is given by:
\( y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C \)
\( y e^x = \int (-x + 5) e^x dx + C \)
We integrate \( \int (-x + 5) e^x dx \) by parts, choosing \( (-x + 5) \) as the first function and \( e^x \) as the second function:
Let \( u = -x + 5 \implies du = -dx \)
Let \( dv = e^x dx \implies v = \int e^x dx = e^x \)
So, \( \int (-x + 5) e^x dx = (-x + 5)e^x - \int (e^x)(-dx) \)
\( = (-x + 5)e^x + \int e^x dx \)
\( = (-x + 5)e^x + e^x \)
Substituting this back into the solution equation:
\( y e^x = (-x + 5)e^x + e^x + C \)
To solve for \( y \), divide by \( e^x \):
\( y = (-x + 5) + 1 + C e^{-x} \)
\( y = 6 - x + C e^{-x} \)
To find the particular solution, we use the given condition: the curve passes through the point \( (0, 2) \).
Substitute these values into the general solution:
\( 2 = 6 - 0 + C e^0 \)
\( 2 = 6 + C \cdot 1 \)
\( C = -4 \)
Now, substitute the value of \( C \) back into the general solution to get the particular solution for this case:
\( y = 6 - x - 4e^{-x} \)
Thus, there are two possible equations for the curve based on the magnitude condition.
In simple words: This problem involves magnitude, so break it into two cases: plus and minus. For each case, set up a linear differential equation and find its integrating factor. Then, integrate using parts. Finally, use the point \( (0,2) \) to find the constant C for both solutions.

Exam Tip: When a problem involves the "magnitude" of a slope, always consider both positive and negative cases for the differential equation. Each case will lead to a distinct general solution and, subsequently, a distinct particular solution after applying the initial conditions.

Question 18. The integrating factor of the differential equation \( x \frac{dy}{dx} - y = 2x^2 \) is
(a) \( e^{-x} \)
(b) \( e^{-y} \)
(c) \( \frac{1}{x} \)
(d) \( x \)
Answer: (c) \( \frac{1}{x} \)
In simple words: First, rewrite the equation into the standard linear form by dividing by x. Then, identify P and calculate the integrating factor \( e^{\int P dx} \). This will give \( 1/x \).

Exam Tip: To find the integrating factor of a linear differential equation, always ensure the equation is in the standard form \( \frac{dy}{dx} + Py = Q \). Then, the integrating factor is \( e^{\int P dx} \).

Question 19. The Integrating Factor of the differential equation \( (1 - y^2) \frac{dx}{dy} + yx = 2y \) \( (-1 < y < 1) \) is
(a) \( \frac{1}{y^2 - 1} \)
(b) \( \frac{1}{\sqrt{y^2 - 1}} \)
(c) \( \frac{1}{1 - y^2} \)
(d) \( \frac{1}{\sqrt{1 - y^2}} \)
Answer: (d) \( \frac{1}{\sqrt{1 - y^2}} \)
In simple words: First, rewrite the given equation into the standard linear form by dividing by \( (1-y^2) \). Then, identify P and calculate the integrating factor \( e^{\int P dy} \), which involves a substitution to solve the integral. This results in \( \frac{1}{\sqrt{1-y^2}} \).

Exam Tip: For equations linear in \(x\) (i.e., involving \( \frac{dx}{dy} \)), the integrating factor is \( e^{\int P dy} \). Remember to correctly handle the integral \( \int \frac{y}{1-y^2} dy \), which typically requires a substitution.