GSEB Class 12 Maths Solutions Chapter 9 Differential Equations Exercise 9.4

For each of the differential equations in questions 1 to 10, find the general solution:

Question 1. \( \frac{dy}{dx} = \frac{1-\cos x}{1+\cos x} \)
Answer: We have:
\( \frac{dy}{dx} = \frac{1-\cos x}{1+\cos x} = \frac{2\sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}} = \tan^2 \frac{x}{2} \)
\( \implies dy = \tan^2 \frac{x}{2} dx \)
Integrating both sides, we obtain:
\( \int dy = \int \tan^2 \frac{x}{2} dx \)
\( \int dy = \int (\sec^2 \frac{x}{2} - 1) dx \)
\( \implies y = 2 \tan \frac{x}{2} - x + C \)
This represents the necessary solution.
In simple words: We changed the given equation using trigonometric rules. Then, we integrated both sides to find a general solution. This result shows the relationship between y and x.

Exam Tip: Remember to use trigonometric identities to simplify expressions before integrating, especially when dealing with \( \tan^2 x \) or similar terms.

Question 2. \( \frac{dy}{dx} = \sqrt{4-y^{2}} \quad (-2 < y < 2) \)
Answer: Given the differential equation:
\( \frac{dy}{dx} = \sqrt{4-y^{2}} \)
\( \implies \frac{dy}{\sqrt{4-y^{2}}} = dx \)
Integrating both sides, we obtain:
\( \int \frac{dy}{\sqrt{4-y^{2}}} = \int dx \)
\( \implies \sin^{-1} \frac{y}{2} = x + C \)
\( \implies y = 2 \sin(x + C) \)
This equation provides the necessary solution.
In simple words: We separated the variables (y with dy, x with dx) and then integrated both parts. The \( \sin^{-1} \) function appeared, giving us y in terms of x and an integration constant.

Exam Tip: Recognize standard integral forms like \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \frac{x}{a} + C \) to quickly solve such problems.

Question 3. \( \frac{dy}{dx} + y = 1 \quad (y \neq 1) \)
Answer: The given differential equation is \( \frac{dy}{dx} + y = 1 \).
\( \implies \frac{dy}{dx} = 1 - y \)
\( \implies \frac{dy}{1-y} = dx \)
Integrating both sides, we get:
\( \int \frac{dy}{1-y} = \int dx \)
\( \implies -\log |1 - y| = x + \log C \)
\( \implies x = -\log C - \log |1 - y| \)
\( \implies x = -\log (C|1 - y|) \)
\( \implies C(1 - y) = e^{-x} \)
\( \implies 1 - y = \frac{1}{C}e^{-x} \)
\( \implies y = 1 - \frac{1}{C}e^{-x} \)
Let \( -\frac{1}{C} = A \). So, \( y = 1 + Ae^{-x} \) is the necessary solution.
In simple words: We rearranged the equation to separate y and x terms. After integrating, we used logarithm properties to simplify the expression and found y as a function of x with a constant A.

Exam Tip: When integrating \( \frac{dy}{a-y} \), remember the negative sign: \( -\log |a-y| \). Also, combining constants like \( \log C_1 + \log C_2 = \log (C_1C_2) \) or \( \log C_1 - \log C_2 = \log (C_1/C_2) \) can simplify the final form of the solution.

Question 4. \( \sec^2x \tan y dx + \sec^2y \tan x dy = 0 \)
Answer: We have the equation:
\( \sec^2x \tan y dx + \sec^2y \tan x dy = 0 \)
Divide by \( \tan x \tan y \) to separate variables:
\( \implies \frac{\sec^2x}{\tan x} dx + \frac{\sec^2y}{\tan y} dy = 0 \)
Integrating both sides, we get:
\( \int \frac{\sec^2x}{\tan x} dx + \int \frac{\sec^2y}{\tan y} dy = \int 0 \)
\( \implies \log |\tan x| + \log |\tan y| = \log C \)
\( \implies \log |\tan x \tan y| = \log C \)
\( \implies \tan x \tan y = C \)
This is the necessary solution, where \( x \neq \text{odd multiple of } \frac{\pi}{2} \) and \( x \in R \).
In simple words: We organized the equation so x terms are with dx and y terms are with dy. After integrating, we used logarithm rules to combine terms and found a simple equation showing the product of tan x and tan y equals a constant.

Exam Tip: When integrating expressions like \( \frac{f'(x)}{f(x)} \), remember the result is \( \log |f(x)| \). This technique is often useful in variable separable differential equations.

Question 5. \( (e^x + e^{-x})dy – (e^x – e^{-x})dx = 0 \)
Answer: We have:
\( (e^x + e^{-x})dy = (e^x – e^{-x})dx \)
\( \implies dy = \frac{e^x – e^{-x}}{e^x + e^{-x}}dx \)
Integrating both sides, we get:
\( \int dy = \int \frac{e^x – e^{-x}}{e^x + e^{-x}}dx \)
Let \( t = e^x + e^{-x} \).
Then \( dt = (e^x – e^{-x})dx \).
So, the integral becomes:
\( \int dy = \int \frac{dt}{t} \)
\( \implies y = \log |t| + C \)
Substitute \( t = e^x + e^{-x} \) back:
\( \implies y = \log |e^x + e^{-x}| + C \)
Since \( e^x + e^{-x} > 0 \), we can write:
\( y = \log (e^x + e^{-x}) + C \)
This is the necessary solution.
In simple words: We rearranged the equation to put y with dy and x with dx. We then used a substitution method to easily integrate the x part, which gave us the natural logarithm of \( e^x + e^{-x} \) plus a constant.

Exam Tip: For integrals of the form \( \int \frac{f'(x)}{f(x)} dx \), always consider the substitution \( t = f(x) \). This simplifies the integral to \( \int \frac{1}{t} dt = \log|t| + C \).

Question 6. \( \frac{dy}{dx} = (1 + x)^2(1 + y^2) \)
Answer: The given differential equation is:
\( \frac{dy}{dx} = (1 + x)^2(1 + y^2) \)
Separate the variables:
\( \implies \frac{dy}{1+y^2} = (1 + x)^2 dx \)
Integrating both sides, we obtain:
\( \int \frac{dy}{1+y^2} = \int (1 + x)^2 dx \)
\( \implies \tan^{-1}y = \int (1 + 2x + x^2) dx \)
\( \implies \tan^{-1}y = x + x^2 + \frac{x^3}{3} + C \)
This is the necessary equation.
In simple words: We moved the y terms to one side with dy and x terms to the other with dx. Then, we integrated each side separately. The left side gave an inverse tangent, and the right side gave a polynomial in x.

Exam Tip: Remember the standard integral \( \int \frac{1}{1+y^2} dy = \tan^{-1}y + C \). Also, expand \( (1+x)^2 \) before integrating to handle it easily.

Question 7. \( y \log y dx – x dy = 0 \)
Answer: The given differential equation is:
\( y \log y dx - x dy = 0 \)
Rearrange the terms:
\( \implies y \log y dx = x dy \)
Separate the variables:
\( \implies \frac{dy}{y \log y} = \frac{dx}{x} \)
Integrating both sides, we obtain:
\( \int \frac{dy}{y \log y} = \int \frac{dx}{x} \)
For the left side, let \( t = \log y \). Then \( dt = \frac{1}{y} dy \).
So, \( \int \frac{dt}{t} = \int \frac{dx}{x} \)
\( \implies \log |t| = \log |x| + \log C \)
\( \implies \log |\log y| = \log |Cx| \)
\( \implies \log y = Cx \)
\( \implies y = e^{Cx} \) is the necessary solution.
In simple words: We separated the y terms with dy and x terms with dx. We then used a substitution for the y integral. After integrating both sides, we applied logarithm properties to simplify and found y as an exponential function of x.

Exam Tip: When you see \( y \log y \) in the denominator, consider substituting \( t = \log y \). This is a common trick for such integrals. Remember to include the absolute value for the argument of the logarithm, \( \log|x| \), unless the context explicitly shows it's always positive.

Question 8. \( x^5\frac{dy}{dx} = -y^5 \)
Answer: The given differential equation is:
\( x^5 \frac{dy}{dx} = -y^5 \)
Separate the variables:
\( \implies \frac{dy}{y^5} = -\frac{dx}{x^5} \)
\( \implies y^{-5} dy = -x^{-5} dx \)
Integrating both sides, we obtain:
\( \int y^{-5} dy = \int -x^{-5} dx \)
\( \implies \frac{y^{-5+1}}{-5+1} = - \frac{x^{-5+1}}{-5+1} + C' \)
\( \implies \frac{y^{-4}}{-4} = - \frac{x^{-4}}{-4} + C' \)
\( \implies -\frac{1}{4y^4} = \frac{1}{4x^4} + C' \)
Multiply by 4:
\( \implies -\frac{1}{y^4} = \frac{1}{x^4} + 4C' \)
Let \( -4C' = C \):
\( \implies \frac{1}{y^4} + \frac{1}{x^4} = C \)
This is the necessary solution.
In simple words: We moved y terms to one side and x terms to the other. Then, we integrated each side using the power rule for integration. After simplification, we found a relationship between \( \frac{1}{y^4} \) and \( \frac{1}{x^4} \) equal to a constant.

Exam Tip: When integrating negative powers, be careful with the exponent: \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \). Also, remember to combine arbitrary constants at the end for a cleaner solution.

Question 9. \( \frac{dy}{dx} = \sin^{-1}x \)
Answer: The given differential equation is:
\( \frac{dy}{dx} = \sin^{-1}x \)
\( \implies dy = \sin^{-1}x \, dx \)
Integrating both sides, we get:
\( \int dy = \int \sin^{-1}x \, dx + C \)
\( \implies y = \int \sin^{-1}x \cdot 1 \, dx + C \)
We use integration by parts, taking \( \sin^{-1}x \) as the first function and 1 as the second function:
\( \int u \, dv = uv - \int v \, du \)
Here, \( u = \sin^{-1}x \implies du = \frac{1}{\sqrt{1-x^2}} dx \)
And \( dv = 1 \, dx \implies v = x \)
So, \( \int \sin^{-1}x \, dx = x \sin^{-1}x - \int x \frac{1}{\sqrt{1-x^2}} dx \)
\( \int \sin^{-1}x \, dx = x \sin^{-1}x - \int \frac{x}{\sqrt{1-x^2}} dx \)
For the integral \( \int \frac{x}{\sqrt{1-x^2}} dx \), let \( t = 1-x^2 \). Then \( dt = -2x \, dx \implies x \, dx = -\frac{1}{2} dt \).
\( \int \frac{-\frac{1}{2} dt}{\sqrt{t}} = -\frac{1}{2} \int t^{-1/2} dt = -\frac{1}{2} \frac{t^{1/2}}{1/2} = -\sqrt{t} = -\sqrt{1-x^2} \)
Substituting this back:
\( y = x \sin^{-1}x - (-\sqrt{1-x^2}) + C \)
\( \implies y = x \sin^{-1}x + \sqrt{1-x^2} + C \)
This is the necessary solution.
In simple words: We needed to integrate inverse sine. We used a method called integration by parts for this. Then, we solved another small integral by substitution and combined all the parts to get our final answer with a constant.

Exam Tip: For integrals involving inverse trigonometric functions, integration by parts is often the key. Remember the formula \( \int u \, dv = uv - \int v \, du \) and choose u as the inverse function and dv as dx.

Question 10. \( e^x \tan y dx + (1 - e^x) \sec^2y dy = 0 \)
Answer: The given differential equation is:
\( e^x \tan y dx + (1 - e^x) \sec^2y dy = 0 \)
Rearrange the terms:
\( \implies (1 - e^x) \sec^2y dy = -e^x \tan y dx \)
Separate the variables by dividing by \( (1 - e^x) \tan y \):
\( \implies \frac{\sec^2y}{\tan y} dy = \frac{-e^x}{1 - e^x} dx \)
Integrating both sides, we obtain:
\( \int \frac{\sec^2y}{\tan y} dy = \int \frac{-e^x}{1 - e^x} dx \)
For the left side, let \( u = \tan y \). Then \( du = \sec^2y dy \). So \( \int \frac{du}{u} = \log |\tan y| \).
For the right side, let \( v = 1 - e^x \). Then \( dv = -e^x dx \). So \( \int \frac{dv}{v} = \log |1 - e^x| \).
\( \implies \log |\tan y| = \log |1 - e^x| + \log C \)
\( \implies \log |\tan y| = \log |C(1 - e^x)| \)
\( \implies \tan y = C(1 - e^x) \)
This is the necessary solution.
In simple words: We moved all y terms to one side and x terms to the other. Then, we integrated each side, noting that both integrals were of the form \( \int \frac{f'(x)}{f(x)} dx \). Finally, we used logarithm properties to combine the terms into a single, neat equation.

Exam Tip: Always look for opportunities to use the \( \int \frac{f'(x)}{f(x)} dx = \log |f(x)| + C \) integral form. It simplifies many variable separable differential equations, especially those with products and quotients of functions.

For each of the differential equations in questions from 11 to 14, find a particular solution satisfying the given condition:

Question 11. \( (x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x; y = 1, \text{ when } x = 0 \)
Answer: The given differential equation is:
\( (x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x \)
Separate variables:
\( \implies dy = \frac{2x^2 + x}{x^3 + x^2 + x + 1} dx \)
Factor the denominator: \( x^3 + x^2 + x + 1 = x^2(x+1) + 1(x+1) = (x+1)(x^2+1) \)
\( \implies dy = \frac{2x^2 + x}{(x+1)(x^2+1)} dx \)
Integrating both sides:
\( \int dy = \int \frac{2x^2 + x}{(x+1)(x^2+1)} dx \)
Let's use partial fractions for the right side:
\( \frac{2x^2 + x}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1} \)
\( 2x^2 + x = A(x^2+1) + (Bx+C)(x+1) \)
If \( x = -1 \): \( 2(-1)^2 + (-1) = A((-1)^2+1) \implies 2 - 1 = A(2) \implies 1 = 2A \implies A = \frac{1}{2} \)
Comparing coefficients of \( x^2 \): \( 2 = A + B \implies 2 = \frac{1}{2} + B \implies B = \frac{3}{2} \)
Comparing constant terms: \( 0 = A + C \implies 0 = \frac{1}{2} + C \implies C = -\frac{1}{2} \)
So, \( \frac{2x^2 + x}{(x+1)(x^2+1)} = \frac{1}{2(x+1)} + \frac{\frac{3}{2}x - \frac{1}{2}}{x^2+1} = \frac{1}{2(x+1)} + \frac{3x - 1}{2(x^2+1)} \)
\( y = \int \left( \frac{1}{2(x+1)} + \frac{3x - 1}{2(x^2+1)} \right) dx \)
\( y = \frac{1}{2} \int \frac{1}{x+1} dx + \frac{1}{2} \int \frac{3x - 1}{x^2+1} dx \)
\( y = \frac{1}{2} \log |x+1| + \frac{1}{2} \left( \int \frac{3x}{x^2+1} dx - \int \frac{1}{x^2+1} dx \right) \)
For \( \int \frac{3x}{x^2+1} dx \), let \( u = x^2+1 \implies du = 2x \, dx \). So \( \frac{3}{2} \int \frac{du}{u} = \frac{3}{2} \log |x^2+1| \).
\( y = \frac{1}{2} \log |x+1| + \frac{1}{2} \left( \frac{3}{2} \log |x^2+1| - \tan^{-1}x \right) + C \)
\( y = \frac{1}{2} \log |x+1| + \frac{3}{4} \log |x^2+1| - \frac{1}{2} \tan^{-1}x + C \)
Now, use the initial condition \( y=1 \) when \( x=0 \):
\( 1 = \frac{1}{2} \log (0+1) + \frac{3}{4} \log (0^2+1) - \frac{1}{2} \tan^{-1}(0) + C \)
\( 1 = \frac{1}{2} \log 1 + \frac{3}{4} \log 1 - \frac{1}{2} (0) + C \)
\( 1 = 0 + 0 - 0 + C \)
\( \implies C = 1 \)
Substitute C back into the general solution:
\( y = \frac{1}{2} \log |x+1| + \frac{3}{4} \log |x^2+1| - \frac{1}{2} \tan^{-1}x + 1 \)
This can be written as:
\( y = \frac{1}{4} [2\log |x+1| + 3\log |x^2+1|] - \frac{1}{2} \tan^{-1}x + 1 \)
\( y = \frac{1}{4} [\log (x+1)^2 + \log (x^2+1)^3] - \frac{1}{2} \tan^{-1}x + 1 \)
\( y = \frac{1}{4} \log [(x+1)^2 (x^2+1)^3] - \frac{1}{2} \tan^{-1}x + 1 \)
This is the necessary particular solution.
In simple words: First, we separated the variables and used partial fractions to break down the complex fraction into simpler parts. Then, we integrated each part separately, which included a logarithm and an inverse tangent. Finally, we used the given initial values to find the specific constant of integration, giving us the unique solution for the curve.

Exam Tip: Partial fraction decomposition is crucial for integrating rational functions. Remember to handle linear and irreducible quadratic factors correctly. When solving for the constant of integration, be very careful with calculations involving log(1) which is 0 and tan inverse of 0 which is 0.

Question 12. \( x(x^2 - 1)\frac{dy}{dx} = 1; y = 0, \text{ when } x = 2 \)
Answer: The given differential equation is:
\( x(x^2 - 1)\frac{dy}{dx} = 1 \)
Separate the variables:
\( \implies dy = \frac{1}{x(x^2 - 1)} dx \)
\( \implies dy = \frac{1}{x(x-1)(x+1)} dx \)
Integrating both sides:
\( \int dy = \int \frac{1}{x(x-1)(x+1)} dx \)
For the right side, use partial fractions:
\( \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} \)
\( 1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1) \)
If \( x = 0 \): \( 1 = A(-1)(1) \implies A = -1 \)
If \( x = 1 \): \( 1 = B(1)(2) \implies B = \frac{1}{2} \)
If \( x = -1 \): \( 1 = C(-1)(-2) \implies C = \frac{1}{2} \)
So, \( \int dy = \int \left( -\frac{1}{x} + \frac{1}{2(x-1)} + \frac{1}{2(x+1)} \right) dx \)
\( y = -\log |x| + \frac{1}{2} \log |x-1| + \frac{1}{2} \log |x+1| + C \)
\( y = -\log |x| + \frac{1}{2} (\log |x-1| + \log |x+1|) + C \)
\( y = -\log |x| + \frac{1}{2} \log |(x-1)(x+1)| + C \)
\( y = -\log |x| + \frac{1}{2} \log |x^2-1| + C \)
\( y = \log |x^{-1}| + \log |(x^2-1)^{1/2}| + C \)
\( y = \log \left| \frac{\sqrt{x^2-1}}{x} \right| + C \)
Now, use the initial condition \( y=0 \) when \( x=2 \):
\( 0 = \log \left| \frac{\sqrt{2^2-1}}{2} \right| + C \)
\( 0 = \log \left| \frac{\sqrt{3}}{2} \right| + C \)
\( \implies C = -\log \left( \frac{\sqrt{3}}{2} \right) \)
Substitute C back into the general solution:
\( y = \log \left| \frac{\sqrt{x^2-1}}{x} \right| - \log \left( \frac{\sqrt{3}}{2} \right) \)
\( y = \log \left| \frac{\sqrt{x^2-1}}{x} \cdot \frac{2}{\sqrt{3}} \right| \)
This is the necessary particular solution.
In simple words: We separated the variables and used partial fractions to simplify the fraction on the x-side. After integrating, we used logarithm properties to combine the terms. Then, we used the given starting values to find the specific constant, which completed our unique solution.

Exam Tip: Factorizing the denominator completely is the first step in partial fraction decomposition. Always double-check your arithmetic when solving for the constants A, B, and C. Remember the properties of logarithms like \( \log A - \log B = \log (A/B) \).

Question 13. \( \cos\left(\frac{dy}{dx}\right) = a; (a \in R), y = 2, \text{ when } x = 0 \)
Answer: The given differential equation is:
\( \cos\left(\frac{dy}{dx}\right) = a \)
Take \( \cos^{-1} \) on both sides:
\( \implies \frac{dy}{dx} = \cos^{-1}a \)
Note that \( \cos^{-1}a \) is a constant. Let \( k = \cos^{-1}a \).
\( \implies \frac{dy}{dx} = k \)
\( \implies dy = k \, dx \)
Integrating both sides:
\( \int dy = \int k \, dx \)
\( \implies y = kx + C \)
Substitute \( k = \cos^{-1}a \):
\( y = (\cos^{-1}a)x + C \)
Now, use the initial condition \( y=2 \) when \( x=0 \):
\( 2 = (\cos^{-1}a)(0) + C \)
\( \implies C = 2 \)
Substitute C back into the general solution:
\( y = (\cos^{-1}a)x + 2 \)
This is the necessary particular solution. This can also be expressed as \( \cos^{-1}a = \frac{y-2}{x} \) or \( a = \cos \left( \frac{y-2}{x} \right) \).
In simple words: Since the cosine of the derivative equals a constant, the derivative itself must be another constant. We integrated this simple derivative to get a linear equation for y. Then, we used the initial values to find the exact constant for our specific solution.

Exam Tip: If \( \frac{dy}{dx} \) is equal to a constant, the general solution will always be a linear equation \( y = mx + C \), where m is the constant value of \( \frac{dy}{dx} \). Applying initial conditions will help you determine the specific value of C.

Question 14. \( \frac{dy}{dx} = y \tan x; y = 1, \text{ when } x = 0 \)
Answer: The given differential equation is:
\( \frac{dy}{dx} = y \tan x \)
Separate the variables:
\( \implies \frac{dy}{y} = \tan x \, dx \)
Integrating both sides:
\( \int \frac{dy}{y} = \int \tan x \, dx \)
\( \implies \log |y| = \log |\sec x| + \log C \)
(Alternatively, \( \int \tan x \, dx = -\log |\cos x| \). Let's use this for consistency with some textbooks.)
\( \implies \log |y| = -\log |\cos x| + \log C \)
\( \implies \log |y| + \log |\cos x| = \log C \)
\( \implies \log |y \cos x| = \log C \)
\( \implies y \cos x = C \)
Now, use the initial condition \( y=1 \) when \( x=0 \):
\( (1)(\cos 0) = C \)
\( (1)(1) = C \)
\( \implies C = 1 \)
Substitute C back into the general solution:
\( y \cos x = 1 \)
\( \implies y = \frac{1}{\cos x} = \sec x \)
This is the necessary particular solution.
In simple words: We organized the equation to put y terms with dy and x terms with dx. After integrating, we used logarithm rules to combine them into one equation. Then, using the given starting values, we found the specific constant, resulting in the unique function y = sec x.

Exam Tip: Remember the integral of \( \tan x \) is \( \log |\sec x| + C \) or \( -\log |\cos x| + C \). Either form is correct, but the latter often leads to easier simplification when combining logarithms. Always apply the initial conditions carefully to find the particular solution.

Question 15. Find the equation of the curve passing through the point (0, 0) and whose differential equation is \( y' = e^x \sin x \).
Answer: The given differential equation is \( y' = e^x \sin x \).
This means \( \frac{dy}{dx} = e^x \sin x \).
\( \implies dy = e^x \sin x \, dx \)
Integrating both sides, we get:
\( \int dy = \int e^x \sin x \, dx \)
\( \implies y = \int e^x \sin x \, dx \)
We use integration by parts for \( \int e^x \sin x \, dx \). Let \( I = \int e^x \sin x \, dx \).
Using \( \int u \, dv = uv - \int v \, du \):
Let \( u = \sin x \implies du = \cos x \, dx \)
Let \( dv = e^x \, dx \implies v = e^x \)
\( I = e^x \sin x - \int e^x \cos x \, dx \)
Now, apply integration by parts again to \( \int e^x \cos x \, dx \):
Let \( u = \cos x \implies du = -\sin x \, dx \)
Let \( dv = e^x \, dx \implies v = e^x \)
\( \int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx \)
So, \( I = e^x \sin x - (e^x \cos x + I) \)
\( I = e^x \sin x - e^x \cos x - I \)
\( \implies 2I = e^x (\sin x - \cos x) \)
\( \implies I = \frac{e^x}{2} (\sin x - \cos x) \)
So, the general solution is:
\( y = \frac{e^x}{2} (\sin x - \cos x) + C \)
The curve passes through the point \( (0, 0) \). Substitute \( x=0, y=0 \):
\( 0 = \frac{e^0}{2} (\sin 0 - \cos 0) + C \)
\( 0 = \frac{1}{2} (0 - 1) + C \)
\( 0 = -\frac{1}{2} + C \)
\( \implies C = \frac{1}{2} \)
Substitute C back into the general solution:
\( y = \frac{e^x}{2} (\sin x - \cos x) + \frac{1}{2} \)
This is the necessary equation of the curve.
In simple words: To find the curve's equation, we integrated \( e^x \sin x \). This required using integration by parts twice. After finding the general solution, we used the given point (0, 0) to find the specific constant of integration, giving us the unique curve equation.

Exam Tip: Integrals of the form \( \int e^{ax} \sin(bx) dx \) or \( \int e^{ax} \cos(bx) dx \) always require integration by parts twice, leading to an equation that can be solved for the integral itself. Be careful with signs and constants during the process.

Question 16. If for the differential equation \( xy\frac{dy}{dx} = (x + 2)(y + 2) \), find the solution curve passing through the point \( (1, – 1) \).
Answer: The given differential equation is:
\( xy\frac{dy}{dx} = (x + 2)(y + 2) \)
Rearrange the terms:
\( \implies xy \, dy = (x + 2)(y + 2) \, dx \)
Separate the variables:
\( \implies \frac{y}{y+2} dy = \frac{x+2}{x} dx \)
Integrate both sides:
\( \int \frac{y}{y+2} dy = \int \frac{x+2}{x} dx \)
For the left side: \( \int \frac{y+2-2}{y+2} dy = \int \left( 1 - \frac{2}{y+2} \right) dy = y - 2\log |y+2| \)
For the right side: \( \int \left( 1 + \frac{2}{x} \right) dx = x + 2\log |x| \)
So, the general solution is:
\( y - 2\log |y+2| = x + 2\log |x| + C \)
The curve passes through \( (1, -1) \). Substitute \( x=1, y=-1 \):
\( -1 - 2\log |-1+2| = 1 + 2\log |1| + C \)
\( -1 - 2\log |1| = 1 + 2(0) + C \)
\( -1 - 0 = 1 + 0 + C \)
\( \implies -1 = 1 + C \)
\( \implies C = -2 \)
Substitute C back into the general solution:
\( y - 2\log |y+2| = x + 2\log |x| - 2 \)
This can be rearranged as:
\( y - x = 2\log |y+2| + 2\log |x| - 2 \)
\( \implies y - x = 2(\log |x| + \log |y+2|) - 2 \)
\( \implies y - x = 2\log |x(y+2)| - 2 \)
This is the necessary solution.
In simple words: We rearranged the given equation to separate y and x terms. After integrating both sides, which involved splitting fractions into simpler parts, we obtained a general solution with logarithms. Finally, we used the given point (1, -1) to find the exact constant of integration, giving us the unique equation for the curve.

Exam Tip: When dealing with fractions like \( \frac{y}{y+a} \) or \( \frac{x}{x+a} \), rewrite them as \( 1 - \frac{a}{y+a} \) or \( 1 - \frac{a}{x+a} \) to simplify integration. This approach helps in handling the logarithmic terms more easily.

Question 17. Find the equation of the curve passing through the point (0, – 2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
Answer: Let the point on the curve be \( (x, y) \).
The slope of the tangent to the curve at \( (x, y) \) is \( \frac{dy}{dx} \).
According to the problem statement, the product of the slope of the tangent and the y-coordinate is equal to the x-coordinate:
\( y \left(\frac{dy}{dx}\right) = x \)
Separate the variables:
\( \implies y \, dy = x \, dx \)
Integrating both sides, we get:
\( \int y \, dy = \int x \, dx \)
\( \implies \frac{y^2}{2} = \frac{x^2}{2} + C' \)
\( \implies y^2 = x^2 + 2C' \)
Let \( 2C' = C \):
\( \implies y^2 = x^2 + C \)
The curve passes through the point \( (0, -2) \). Substitute \( x=0, y=-2 \):
\( (-2)^2 = (0)^2 + C \)
\( 4 = 0 + C \)
\( \implies C = 4 \)
Substitute C back into the general equation:
\( y^2 = x^2 + 4 \)
This can also be written as \( y^2 - x^2 = 4 \). This is the necessary equation of the curve.
In simple words: We wrote the problem as a differential equation, linking the slope, y-coordinate, and x-coordinate. We then separated y and x terms and integrated them. Using the given point (0, -2), we found the specific constant, which completed the unique equation of the curve.

Exam Tip: Translate word problems into mathematical equations carefully. "Slope of the tangent" always means \( \frac{dy}{dx} \). Once you have a differential equation, check if it's variable separable for easier integration. Remember to consolidate constants for a cleaner final solution.

Question 18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (- 4, – 3). Find the equation of the curve given that it passes through (- 2, 1).
Answer: Let the point of contact be \( (x, y) \).
The slope of the tangent to the curve at \( (x, y) \) is \( \frac{dy}{dx} \).
The slope of the line segment joining \( (x, y) \) and \( (-4, -3) \) is \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{y - (-3)}{x - (-4)} = \frac{y+3}{x+4} \).
According to the problem statement, the slope of the tangent is twice the slope of this line segment:
\( \frac{dy}{dx} = 2 \left( \frac{y+3}{x+4} \right) \)
Separate the variables:
\( \implies \frac{dy}{y+3} = \frac{2}{x+4} dx \)
Integrating both sides:
\( \int \frac{dy}{y+3} = \int \frac{2}{x+4} dx \)
\( \implies \log |y+3| = 2\log |x+4| + \log C' \)
\( \implies \log |y+3| = \log (x+4)^2 + \log C' \)
\( \implies \log |y+3| = \log [C'(x+4)^2] \)
\( \implies y+3 = C(x+4)^2 \) (where \( C = \pm C' \) is a new arbitrary constant)
The curve passes through \( (-2, 1) \). Substitute \( x=-2, y=1 \):
\( 1+3 = C(-2+4)^2 \)
\( 4 = C(2)^2 \)
\( 4 = 4C \)
\( \implies C = 1 \)
Substitute C back into the general equation:
\( y+3 = 1 \cdot (x+4)^2 \)
\( \implies y = (x+4)^2 - 3 \)
This is the necessary equation of the curve.
In simple words: We translated the problem's conditions into slopes and set up a differential equation. After separating the variables and integrating, we used logarithm properties to simplify. Then, we applied the given point (-2, 1) to find the precise constant, which gave us the unique curve equation.

Exam Tip: Clearly define the slopes involved: the slope of the tangent \( \frac{dy}{dx} \) and the slope of the line segment between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) as \( \frac{y_2 - y_1}{x_2 - x_1} \). Pay attention to logarithm properties when combining terms and simplifying the constant of integration.

Question 19. The volume of a spherical balloon, being inlated, changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units, find the radius of balloon after t seconds.
Answer: Let V be the volume of the balloon and r be its radius.
The volume of a sphere is given by \( V = \frac{4}{3}\pi r^3 \).
The problem states that the volume changes at a constant rate, so \( \frac{dV}{dt} = k \), where k is a constant.
Differentiate V with respect to t:
\( \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) = \frac{4}{3}\pi \cdot 3r^2 \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt} \)
So, we have: \( 4\pi r^2 \frac{dr}{dt} = k \)
Separate variables:
\( \implies 4\pi r^2 dr = k \, dt \)
Integrate both sides:
\( \int 4\pi r^2 dr = \int k \, dt \)
\( \implies 4\pi \frac{r^3}{3} = kt + C \) (Equation 1)
Now, use the given conditions to find k and C:
Condition 1: When \( t=0, r=3 \).
\( 4\pi \frac{(3)^3}{3} = k(0) + C \)
\( 4\pi \frac{27}{3} = C \)
\( 4\pi (9) = C \)
\( \implies C = 36\pi \)
Condition 2: When \( t=3, r=6 \).
\( 4\pi \frac{(6)^3}{3} = k(3) + 36\pi \) (using C = 36π)
\( 4\pi \frac{216}{3} = 3k + 36\pi \)
\( 4\pi (72) = 3k + 36\pi \)
\( 288\pi = 3k + 36\pi \)
\( 3k = 288\pi - 36\pi \)
\( 3k = 252\pi \)
\( \implies k = 84\pi \)
Substitute k and C back into Equation 1:
\( \frac{4}{3}\pi r^3 = 84\pi t + 36\pi \)
Divide the entire equation by \( 4\pi \):
\( \frac{r^3}{3} = 21t + 9 \)
Multiply by 3:
\( r^3 = 3(21t + 9) \)
\( r^3 = 63t + 27 \)
\( \implies r = (63t + 27)^{1/3} \)
This is the radius of the balloon after t seconds.
In simple words: We started with the formula for a sphere's volume and its given constant growth rate. By differentiating and integrating, we found a general equation relating radius, time, and two constants. We then used the two given conditions (initial radius and radius after 3 seconds) to solve for these constants, providing a final formula for the radius at any time t.

Exam Tip: Remember to differentiate the volume formula with respect to time to get the rate of change of volume. Use the given initial conditions carefully to set up a system of equations to find the values of the constants of integration.

Question 20. Principal increases at the rate of r% per year. Find the value of r, if Rs.100 doubles itself in 10 years (log 2 = 0.6931).
Answer: Let P be the principal amount. The rate of increase is r% per year.
So, the differential equation is:
\( \frac{dP}{dt} = \frac{r}{100} P \)
Separate variables:
\( \implies \frac{dP}{P} = \frac{r}{100} dt \)
Integrate both sides:
\( \int \frac{dP}{P} = \int \frac{r}{100} dt \)
\( \implies \log P = \frac{r}{100} t + \log C' \)
\( \implies \log P - \log C' = \frac{r}{100} t \)
\( \implies \log \left( \frac{P}{C'} \right) = \frac{r}{100} t \)
\( \implies \frac{P}{C'} = e^{\frac{r}{100} t} \)
\( \implies P = C' e^{\frac{r}{100} t} \) (Equation 1)
Given initial condition: Initially, \( t=0, P=100 \).
\( 100 = C' e^{\frac{r}{100} (0)} \)
\( 100 = C' e^0 \)
\( \implies C' = 100 \)
Substitute C' back into Equation 1:
\( P = 100 e^{\frac{r}{100} t} \)
Given that the amount doubles in 10 years, so when \( t=10, P=200 \).
\( 200 = 100 e^{\frac{r}{100} (10)} \)
\( 200 = 100 e^{\frac{r}{10}} \)
Divide by 100:
\( 2 = e^{\frac{r}{10}} \)
Take natural logarithm on both sides:
\( \log 2 = \frac{r}{10} \)
\( \implies r = 10 \log 2 \)
Given \( \log 2 = 0.6931 \):
\( r = 10 \times 0.6931 \)
\( r = 6.931 \)
Thus, the value of r is \( 6.931\% \).
In simple words: We set up a differential equation for how money grows with a percentage rate. After integrating, we found a formula for the principal amount over time. We used the initial Rs.100 and the fact it doubled in 10 years to solve for the rate 'r', using the given logarithm value.

Exam Tip: Problems involving continuous growth (like principal increasing at a rate proportional to itself) always lead to exponential functions. The formula \( P = P_0 e^{kt} \) (where \( P_0 \) is initial principal, k is rate, t is time) is very useful. Remember to use natural logarithms for base e calculations.

Question 21. In a bank, principal increases at the rate of 5% per year. An amount of Rs.1000 is deposited with this bank. How much will it worth after 10 years (e0.5 = 1.648)?
Answer: Let P be the principal amount. The rate of interest is 5% per year.
So, the differential equation is:
\( \frac{dP}{dt} = \frac{5}{100} P \)
\( \implies \frac{dP}{dt} = 0.05 P \)
Separate variables:
\( \implies \frac{dP}{P} = 0.05 dt \)
Integrate both sides:
\( \int \frac{dP}{P} = \int 0.05 dt \)
\( \implies \log P = 0.05t + \log C' \)
\( \implies \log \left( \frac{P}{C'} \right) = 0.05t \)
\( \implies P = C' e^{0.05t} \) (Equation 1)
Initial condition: An amount of Rs.1000 is deposited, so when \( t=0, P=1000 \).
\( 1000 = C' e^{0.05(0)} \)
\( 1000 = C' e^0 \)
\( \implies C' = 1000 \)
Substitute C' back into Equation 1:
\( P = 1000 e^{0.05t} \)
We need to find the worth after 10 years, so substitute \( t=10 \):
\( P = 1000 e^{0.05 \times 10} \)
\( P = 1000 e^{0.5} \)
Given \( e^{0.5} = 1.648 \):
\( P = 1000 \times 1.648 \)
\( P = 1648 \)
Therefore, after 10 years, the amount will be Rs.1648.
In simple words: We set up an equation showing how the principal grows with a 5% rate. After integrating, we got a formula for the principal over time. We used the initial deposit of Rs.1000 to find the constant, then calculated the amount after 10 years using the given value for \( e^{0.5} \).

Exam Tip: This is a standard continuous compounding problem. The general formula \( P(t) = P_0 e^{rt} \) (where P is principal, \( P_0 \) is initial principal, r is rate, t is time) can be directly applied if you remember it, or derived using the differential equation. Always double-check your exponent calculation.

Question 22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours, will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Answer: Let y denote the number of bacteria at any instant t.
The rate of growth of bacteria is proportional to the number present:
\( \frac{dy}{dt} \propto y \)
\( \implies \frac{dy}{dt} = ky \), where k is the constant of proportionality.
Separate variables:
\( \implies \frac{dy}{y} = k \, dt \)
Integrate both sides:
\( \int \frac{dy}{y} = \int k \, dt \)
\( \implies \log y = kt + C' \) (Equation 1)
Let \( y_0 \) be the initial number of bacteria.
When \( t=0, y=y_0 \):
\( \log y_0 = k(0) + C' \)
\( \implies C' = \log y_0 \)
Substitute C' back into Equation 1:
\( \log y = kt + \log y_0 \)
\( \implies \log y - \log y_0 = kt \)
\( \implies \log \left( \frac{y}{y_0} \right) = kt \) (Equation 2)
Given conditions:
Initial count \( y_0 = 1,00,000 \).
The number increases by 10% in 2 hours. So, when \( t=2 \), \( y = y_0 + 0.10y_0 = 1.10y_0 \).
Using Equation 2:
\( \log \left( \frac{1.10y_0}{y_0} \right) = k(2) \)
\( \implies \log(1.10) = 2k \)
\( \implies k = \frac{1}{2} \log(1.10) \)
Now, we need to find the time \( t_1 \) when the count reaches 2,00,000 (i.e., \( y = 2y_0 \)).
Using Equation 2 again:
\( \log \left( \frac{2y_0}{y_0} \right) = k t_1 \)
\( \implies \log 2 = k t_1 \)
Substitute the value of k:
\( \log 2 = \left( \frac{1}{2} \log(1.10) \right) t_1 \)
\( \implies t_1 = \frac{2 \log 2}{\log(1.10)} \)
Hence, the necessary number of hours for the count to reach 2,00,000 is \( \frac{2 \log 2}{\log(1.10)} \).
In simple words: We modeled the bacteria growth with a differential equation based on proportional growth. We integrated this to get a general formula for the population over time, which included a growth constant 'k'. We used the information about the 10% increase in 2 hours to find 'k'. Finally, we used 'k' to calculate how long it would take for the bacteria count to double.

Exam Tip: Problems involving population growth (or decay) that are proportional to the current amount are solved using exponential growth/decay models. Always formulate the differential equation \( \frac{dy}{dt} = ky \) first, then solve for k and apply it to find the required time or quantity.

Question 23. The general solution of the differential equation \( \frac{dy}{dx} = e^{x+y} \) is
(A) \( e^x + e^{-y} = c \)
(B) \( e^x + e^x = c \)
(C) \( e^{-x} + e^y = c \)
(D) \( e^x + e^{-y} = c \)
Answer: (A) \( e^x + e^{-y} = c \)
We first separate the variables in the given differential equation. This allows us to write it as \( e^{-y} dy = e^x dx \). Next, we integrate both sides of this revised equation. The integration leads to \( -e^{-y} = e^x + C \), where C is the integration constant. By moving terms around, we obtain \( e^x + e^{-y} = -C \). If we consider \( -C \) as a new constant \( c \), the final general solution becomes \( e^x + e^{-y} = c \).
In simple words: To find the solution, we separate the parts with 'y' and 'x'. Then we integrate both sides. This gives us an equation showing the relationship between 'e to the power x' and 'e to the power negative y', plus a constant.

Exam Tip: When solving differential equations, always remember to separate variables first if possible, then integrate each side, and include the constant of integration.