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

Solve The Following Differential Equations:

Question 1. \( \frac { dy }{ dx } = 5x + 7 \)
Answer: Given the differential equation: \[ \frac { dy }{ dx } = 5x + 7 \] First, we separate the variables by moving \( dx \) to the right side:
\( \implies dy = (5x + 7) dx \) Now, we integrate both sides to find the solution. Integrating is like finding the original function before it was differentiated.
\( \implies \int dy = \int (5x + 7) dx \)
\( \implies y = \frac { 5x^2 }{ 2 } + 7x + C \) This is the final solution to the differential equation.
In simple words: We separate \( dy \) and \( dx \), then integrate both sides to get an equation for \( y \) with a constant \( C \).

🎯 Exam Tip: Remember to always include the constant of integration \( C \) when solving indefinite integrals, as it accounts for any constant term that would vanish upon differentiation.

Question 2. \( \frac { dy }{ dx } = \sin x - x \)
Answer: Given the differential equation: \[ \frac { dy }{ dx } = \sin x - x \] First, we separate the variables by moving \( dx \) to the right side:
\( \implies dy = (\sin x - x)dx \) Next, we integrate both sides to find the solution. Integration helps us reverse the differentiation process.
\( \implies \int dy = \int (\sin x - x) dx \)
\( \implies y = -\cos x - \frac { x^2 }{ 2 } + C \) This is the required solution.
In simple words: We separate the \( dy \) and \( dx \) terms, then integrate each side. The integral of \( \sin x \) is \( -\cos x \), and for \( x \) it's \( \frac{x^2}{2} \).

🎯 Exam Tip: Be careful with the signs when integrating trigonometric functions like \( \sin x \) and \( \cos x \). A common mistake is to forget the negative sign for \( \int \sin x dx \).

Question 3. \( \frac { dy }{ dx } = x \log x \)
Answer: Given the differential equation: \[ \frac { dy }{ dx } = x \log x \] First, we separate the variables by moving \( dx \) to the right side:
\( \implies dy = x \log x dx \) Next, we integrate both sides. For integrating \( x \log x \), we use integration by parts, which is a method for integrating products of functions.
\( \implies \int dy = \int x \log x dx \) Using integration by parts \( \int u dv = uv - \int v du \): Let \( u = \log x \) and \( dv = x dx \). Then \( du = \frac{1}{x} dx \) and \( v = \frac{x^2}{2} \).
\( \implies y = (\log x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} dx \)
\( \implies y = \frac{x^2}{2} \log x - \int \frac{x}{2} dx \)
\( \implies y = \frac{x^2}{2} \log x - \frac{x^2}{4} + C \) This is the required solution.
In simple words: We rearrange the equation, then integrate both sides. To integrate \( x \log x \), we use a special method called integration by parts.

🎯 Exam Tip: When using integration by parts, carefully choose \( u \) and \( dv \). The "LIATE" rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) helps: choose \( u \) as the function that comes first in this order.

Question 4. \( \frac { dy }{ dx } + 2x = e^{3x} \)
Answer: Given the differential equation: \[ \frac { dy }{ dx } + 2x = e^{3x} \] First, we rearrange the equation to separate the variables, getting all \( x \) terms with \( dx \) and \( dy \) by itself:
\( \implies \frac { dy }{ dx } = e^{3x} - 2x \)
\( \implies dy = (e^{3x} - 2x) dx \) Next, we integrate both sides to find the solution. Integrating exponential functions and power functions is straightforward.
\( \implies \int dy = \int (e^{3x} - 2x) dx \)
\( \implies y = \frac{e^{3x}}{3} - \frac{2x^2}{2} + C \)
\( \implies y = \frac{e^{3x}}{3} - x^2 + C \) This is the required solution.
In simple words: We first move all \( x \) terms to the right side with \( dx \), then integrate both sides to solve for \( y \).

🎯 Exam Tip: Remember that \( \int e^{ax} dx = \frac{e^{ax}}{a} + C \). Also, always simplify fractions like \( \frac{2x^2}{2} \) to \( x^2 \).

Question 5. \( (x + 1)\frac { dy }{ dx } = x^2 \)
Answer: Given the differential equation: \[ (x + 1)\frac { dy }{ dx } = x^2 \] First, we rearrange the equation to separate the variables, getting all \( x \) terms with \( dx \):
\( \implies \frac { dy }{ dx } = \frac{x^2}{x + 1} \)
\( \implies dy = \frac{x^2}{x + 1} dx \) Next, we integrate both sides. To integrate the right side, we perform polynomial division or add and subtract 1 in the numerator to simplify.
\( \implies \int dy = \int \frac{x^2}{x + 1} dx \) We can rewrite \( x^2 \) as \( x^2 - 1 + 1 \) in the numerator:
\( \implies y = \int \frac{x^2 - 1 + 1}{x + 1} dx \)
\( \implies y = \int \left( \frac{x^2 - 1}{x + 1} + \frac{1}{x + 1} \right) dx \) Since \( x^2 - 1 = (x - 1)(x + 1) \):
\( \implies y = \int \left( (x - 1) + \frac{1}{x + 1} \right) dx \)
\( \implies y = \frac{x^2}{2} - x + \log |x + 1| + C \) This is the required solution.
In simple words: We move \( (x+1) \) to the right side, then break the fraction into simpler parts by adding and subtracting 1. Then we integrate each part to find \( y \).

🎯 Exam Tip: For rational functions where the degree of the numerator is greater than or equal to the degree of the denominator, always perform polynomial division or algebraic manipulation before integrating to simplify the expression.

Question 6. \( (x + 1)^2 \frac { dy }{ dx } = x e^x \)
Answer: Given the differential equation: \[ (x + 1)^2 \frac { dy }{ dx } = x e^x \] First, we rearrange the equation to separate the variables:
\( \implies \frac { dy }{ dx } = \frac{x e^x}{(x + 1)^2} \)
\( \implies dy = \frac{x e^x}{(x + 1)^2} dx \) Next, we integrate both sides. To integrate the right side, we can add and subtract 1 in the numerator to create terms that simplify the expression, often used with parts where \( e^x \) is involved.
\( \implies \int dy = \int \frac{x e^x}{(x + 1)^2} dx \) We rewrite \( x \) as \( (x+1) - 1 \) in the numerator:
\( \implies y = \int \frac{(x + 1 - 1) e^x}{(x + 1)^2} dx \)
\( \implies y = \int \left( \frac{(x + 1) e^x}{(x + 1)^2} - \frac{e^x}{(x + 1)^2} \right) dx \)
\( \implies y = \int \left( \frac{e^x}{x + 1} - \frac{e^x}{(x + 1)^2} \right) dx \) This integral is of the form \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \). Here, let \( f(x) = \frac{1}{x + 1} \). Then \( f'(x) = -\frac{1}{(x + 1)^2} \). So the integral becomes:
\( \implies y = \frac{e^x}{x + 1} + C \) This is the required solution.
In simple words: We separate the variables and then integrate. The integral on the right side looks like a special form involving \( e^x \) and a function plus its derivative, which simplifies the integration.

🎯 Exam Tip: Recognize the integral form \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \). This can save a lot of time and effort compared to using integration by parts multiple times.

Question 7. \( \frac { dy }{ dx } = \sin^3 x \cos^2 x + x e^x \)
Answer: Given the differential equation: \[ \frac { dy }{ dx } = \sin^3 x \cos^2 x + x e^x \] First, we separate the variables by moving \( dx \) to the right side:
\( \implies dy = (\sin^3 x \cos^2 x + x e^x) dx \) Next, we integrate both sides:
\( \implies \int dy = \int (\sin^3 x \cos^2 x + x e^x) dx \) We can split this into two integrals:
\( \implies y = \int \sin^3 x \cos^2 x dx + \int x e^x dx \) For the first integral \( \int \sin^3 x \cos^2 x dx \), we can write \( \sin^3 x = \sin^2 x \sin x = (1 - \cos^2 x) \sin x \).
\( \implies \int (1 - \cos^2 x) \cos^2 x \sin x dx \) Let \( t = \cos x \), then \( dt = -\sin x dx \). So \( \sin x dx = -dt \). The integral becomes \( \int (1 - t^2) t^2 (-dt) = \int (t^4 - t^2) dt = \frac{t^5}{5} - \frac{t^3}{3} \). Substituting back \( t = \cos x \): \( \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} \). For the second integral \( \int x e^x dx \), we use integration by parts \( \int u dv = uv - \int v du \). Let \( u = x \) and \( dv = e^x dx \). Then \( du = dx \) and \( v = e^x \). So \( \int x e^x dx = x e^x - \int e^x dx = x e^x - e^x = e^x (x - 1) \). Combining both results:
\( \implies y = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + e^x (x - 1) + C \) This is the required solution.
In simple words: We split the integral into two parts. For the first part, we change \( \sin^3 x \) to use \( \cos x \) and substitute. For the second part, we use integration by parts.

🎯 Exam Tip: When integrating powers of sine and cosine, look for opportunities to use trigonometric identities (like \( \sin^2 x = 1 - \cos^2 x \)) and substitution. For product functions like \( x e^x \), integration by parts is often the key.

Question 8. \( \frac{d y}{d x}=\frac{1}{\sin ^4 x+\cos ^4 x} \)
Answer: Given the differential equation: \[ \frac{d y}{d x}=\frac{1}{\sin ^4 x+\cos ^4 x} \] First, we separate the variables:
\( \implies dy = \frac{1}{\sin ^4 x+\cos ^4 x} dx \) Next, we integrate both sides:
\( \implies \int dy = \int \frac{1}{\sin ^4 x+\cos ^4 x} dx \) To simplify the denominator, divide the numerator and denominator by \( \cos^4 x \):
\( \implies y = \int \frac{\frac{1}{\cos^4 x}}{\frac{\sin^4 x}{\cos^4 x}+\frac{\cos^4 x}{\cos^4 x}} dx \)
\( \implies y = \int \frac{\sec^4 x}{\tan^4 x+1} dx \) We can write \( \sec^4 x = \sec^2 x \cdot \sec^2 x = (1 + \tan^2 x) \sec^2 x \).
\( \implies y = \int \frac{(1+\tan^2 x)\sec^2 x}{\tan^4 x+1} dx \) Let \( t = \tan x \). Then \( dt = \sec^2 x dx \). The integral becomes:
\( \implies y = \int \frac{1+t^2}{t^4+1} dt \) To integrate this, divide the numerator and denominator by \( t^2 \):
\( \implies y = \int \frac{\frac{1}{t^2}+1}{t^2+\frac{1}{t^2}} dt \) The denominator \( t^2+\frac{1}{t^2} \) can be written as \( \left( t - \frac{1}{t} \right)^2 + 2 \). Let \( u = t - \frac{1}{t} \). Then \( du = \left( 1 + \frac{1}{t^2} \right) dt \).
\( \implies y = \int \frac{du}{u^2+2} \) This is a standard integral form \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \). Here \( a^2 = 2 \), so \( a = \sqrt{2} \).
\( \implies y = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{u}{\sqrt{2}} \right) + C \) Substitute back \( u = t - \frac{1}{t} = \frac{t^2-1}{t} \):
\( \implies y = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\frac{t^2-1}{t}}{\sqrt{2}} \right) + C \)
\( \implies y = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t^2-1}{\sqrt{2}t} \right) + C \) Substitute back \( t = \tan x \):
\( \implies y = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\tan^2 x-1}{\sqrt{2} \tan x} \right) + C \) This is the required solution.
In simple words: We first move \( dx \) to the right side. To integrate the fraction, we divide by \( \cos^4 x \), use a substitution \( t = \tan x \), then further simplify the new fraction by dividing by \( t^2 \) and use another substitution to solve it.

🎯 Exam Tip: Integrals with \( \sin^4 x+\cos^4 x \) in the denominator often require dividing by the highest power of \( \cos x \) to convert to \( \tan x \) and \( \sec x \), followed by appropriate substitutions and algebraic manipulations.

Question 9. \( \frac{d y}{d x}=x \sin ^2 x+\frac{1}{x \log x} \)
Answer: Given the differential equation: \[ \frac{d y}{d x}=x \sin ^2 x+\frac{1}{x \log x} \] First, we separate the variables:
\( \implies dy = \left( x \sin ^2 x+\frac{1}{x \log x} \right) dx \) Next, we integrate both sides:
\( \implies \int dy = \int \left( x \sin ^2 x+\frac{1}{x \log x} \right) dx \) We can split this into two integrals:
\( \implies y = \int x \sin ^2 x dx + \int \frac{1}{x \log x} dx \) For the first integral \( \int x \sin ^2 x dx \), use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \):
\( \implies \int x \left( \frac{1 - \cos 2x}{2} \right) dx = \frac{1}{2} \int (x - x \cos 2x) dx \)
\( \implies \frac{1}{2} \left( \int x dx - \int x \cos 2x dx \right) \) \( \int x dx = \frac{x^2}{2} \). For \( \int x \cos 2x dx \), use integration by parts (\( u=x, dv=\cos 2x dx \)): \( v = \frac{\sin 2x}{2} \). So, \( x \frac{\sin 2x}{2} - \int \frac{\sin 2x}{2} dx = \frac{x \sin 2x}{2} - \frac{1}{2} \left( -\frac{\cos 2x}{2} \right) = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} \). Thus, the first integral part is \( \frac{1}{2} \left( \frac{x^2}{2} - \left( \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} \right) \right) = \frac{x^2}{4} - \frac{x \sin 2x}{4} - \frac{\cos 2x}{8} \). For the second integral \( \int \frac{1}{x \log x} dx \), let \( t = \log x \). Then \( dt = \frac{1}{x} dx \).
\( \implies \int \frac{1}{t} dt = \log |t| + C = \log |\log x| + C \). Combining both results:
\( \implies y = \frac{x^2}{4} - \frac{x \sin 2x}{4} - \frac{\cos 2x}{8} + \log |\log x| + C \) This is the required solution.
In simple words: We split the problem into two parts. For the first part with \( \sin^2 x \), we use a trigonometric identity and then integration by parts. For the second part, we use a simple substitution.

🎯 Exam Tip: When integrating terms like \( \sin^2 x \) or \( \cos^2 x \) multiplied by \( x \) (or any algebraic term), convert them to \( \cos 2x \) terms using identities \( \sin^2 x = \frac{1 - \cos 2x}{2} \) or \( \cos^2 x = \frac{1 + \cos 2x}{2} \). Also, recognize forms like \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C \), which applies to \( \frac{1}{x \log x} \).

Question 10. \( \sqrt{a+x}dy + xdx = 0 \)
Answer: Given the differential equation: \[ \sqrt{a+x}dy + xdx = 0 \] First, we rearrange the equation to separate the variables:
\( \implies \sqrt{a+x}dy = -xdx \)
\( \implies dy = -\frac{x}{\sqrt{a+x}} dx \) Next, we integrate both sides:
\( \implies \int dy = \int -\frac{x}{\sqrt{a+x}} dx \)
\( \implies y = -\int \frac{x}{\sqrt{a+x}} dx \) To solve the integral on the right, let \( u = a+x \). Then \( x = u-a \) and \( du = dx \).
\( \implies y = -\int \frac{u-a}{\sqrt{u}} du \)
\( \implies y = -\int \left( \frac{u}{\sqrt{u}} - \frac{a}{\sqrt{u}} \right) du \)
\( \implies y = -\int \left( u^{1/2} - a u^{-1/2} \right) du \)
\( \implies y = -\left( \frac{u^{3/2}}{3/2} - a \frac{u^{1/2}}{1/2} \right) + C \)
\( \implies y = -\frac{2}{3} u^{3/2} + 2a u^{1/2} + C \) Substitute back \( u = a+x \):
\( \implies y = -\frac{2}{3} (a+x)^{3/2} + 2a (a+x)^{1/2} + C \) This is the required solution.
In simple words: We separate \( dy \) and \( dx \), then use a substitution \( u = a+x \) to simplify the integral. After integrating the powers of \( u \), we substitute back \( a+x \) to get the final answer.

🎯 Exam Tip: When integrating expressions involving \( \sqrt{a+x} \) or similar forms, a substitution like \( u = a+x \) often simplifies the integral significantly, turning it into easily integrable power functions.

Question 11. \( \frac{d y}{d x}=\sqrt{4-y^2} \)
Answer: Given the differential equation: \[ \frac{d y}{d x}=\sqrt{4-y^2} \] First, we separate the variables, bringing all \( y \) terms to the left side with \( dy \) and \( x \) terms to the right with \( dx \):
\( \implies \frac{dy}{\sqrt{4-y^2}} = dx \) Next, we integrate both sides:
\( \implies \int \frac{dy}{\sqrt{4-y^2}} = \int dx \) The integral on the left side is a standard inverse sine form: \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1} \left( \frac{x}{a} \right) + C \). Here, \( a^2 = 4 \), so \( a = 2 \).
\( \implies \sin^{-1} \left( \frac{y}{2} \right) = x + C \) This is the required solution.
In simple words: We move the \( \sqrt{4-y^2} \) term to the left with \( dy \), then integrate both sides. The left side becomes an inverse sine function, and the right side becomes \( x \).

🎯 Exam Tip: Recognize common integral forms. The integral of \( \frac{1}{\sqrt{a^2-x^2}} \) is a direct formula leading to \( \sin^{-1}(\frac{x}{a}) \), which is frequently tested.

Question 12. \( \frac { dy }{ dx } = \sec y \)
Answer: Given the differential equation: \[ \frac { dy }{ dx } = \sec y \] First, we separate the variables, bringing all \( y \) terms to the left side with \( dy \) and \( x \) terms to the right with \( dx \): Since \( \sec y = \frac{1}{\cos y} \):
\( \implies \frac{dy}{\sec y} = dx \)
\( \implies \cos y dy = dx \) Next, we integrate both sides:
\( \implies \int \cos y dy = \int dx \) The integral of \( \cos y \) is \( \sin y \).
\( \implies \sin y = x + C \) This is the required solution.
In simple words: We move \( \sec y \) to the left side by writing it as \( \cos y \), then integrate both sides. The integral of \( \cos y \) is \( \sin y \).

🎯 Exam Tip: When dealing with trigonometric functions in differential equations, it's often helpful to rewrite them in terms of sine and cosine before separating variables, as their integrals are more commonly known.

Question 13. \( \frac{d y}{d x}=2^{-y} \)
Answer: Given the differential equation: \[ \frac{d y}{d x}=2^{-y} \] First, we separate the variables, bringing all \( y \) terms to the left side with \( dy \) and \( x \) terms to the right with \( dx \):
\( \implies \frac{1}{2^{-y}} dy = dx \)
\( \implies 2^y dy = dx \) Next, we integrate both sides:
\( \implies \int 2^y dy = \int dx \) The integral of \( a^x \) is \( \frac{a^x}{\log a} \). So, \( \int 2^y dy = \frac{2^y}{\log 2} \).
\( \implies \frac{2^y}{\log 2} = x + C_1 \) We can write \( C_1 = \frac{C}{\log 2} \) for a simplified constant \( C \):
\( \implies \frac{2^y}{\log 2} = x + \frac{C}{\log 2} \)
\( \implies 2^y = x \log 2 + C \) This is the required solution.
In simple words: We move the \( 2^{-y} \) term to the left side as \( 2^y \), then integrate both sides. The integral of \( 2^y \) is \( \frac{2^y}{\log 2} \).

🎯 Exam Tip: Remember the general integration formula for exponential functions: \( \int a^x dx = \frac{a^x}{\log a} + C \). Be careful with constants like \( \log 2 \) in the denominator.

Question 14. Find the particular solution of \( e^{\frac{dy}{dx}} = x + 1 \), given that \( x = 0, y = 3 \).
Answer: Given the differential equation: \[ e^{\frac{dy}{dx}} = x + 1 \] To solve this, we first take the natural logarithm on both sides:
\( \implies \log(e^{\frac{dy}{dx}}) = \log(x + 1) \)
\( \implies \frac{dy}{dx} = \log(x + 1) \) Next, we separate the variables:
\( \implies dy = \log(x + 1) dx \) Now, we integrate both sides. To integrate \( \log(x+1) \), we use integration by parts, considering \( \log(x+1) \cdot 1 \).
\( \implies \int dy = \int \log(x + 1) \cdot 1 dx \) Using integration by parts \( \int u dv = uv - \int v du \): Let \( u = \log(x + 1) \) and \( dv = 1 dx \). Then \( du = \frac{1}{x + 1} dx \) and \( v = x \).
\( \implies y = x \log(x + 1) - \int x \cdot \frac{1}{x + 1} dx \)
\( \implies y = x \log(x + 1) - \int \frac{x}{x + 1} dx \) To integrate \( \frac{x}{x + 1} \), we write \( \frac{x}{x + 1} = \frac{x + 1 - 1}{x + 1} = 1 - \frac{1}{x + 1} \).
\( \implies y = x \log(x + 1) - \int \left( 1 - \frac{1}{x + 1} \right) dx \)
\( \implies y = x \log(x + 1) - (x - \log|x + 1|) + C \)
\( \implies y = x \log(x + 1) - x + \log|x + 1| + C \) We can combine the log terms:
\( \implies y = (x + 1) \log|x + 1| - x + C \) (Equation 1) Now, we use the given condition \( x = 0, y = 3 \) to find the particular solution. Substitute these values into Equation 1:
\( \implies 3 = (0 + 1) \log|0 + 1| - 0 + C \)
\( \implies 3 = 1 \cdot \log(1) - 0 + C \) Since \( \log(1) = 0 \):
\( \implies 3 = 1 \cdot 0 - 0 + C \)
\( \implies 3 = C \) Substitute \( C = 3 \) back into Equation 1:
\( \implies y = (x + 1) \log|x + 1| - x + 3 \) This is the required particular solution. It is good practice to remember \( \log 1 = 0 \).
In simple words: First, we take the logarithm to remove \( e \), then separate the variables and integrate using integration by parts. After finding the general solution, we use the given \( x \) and \( y \) values to find the specific constant \( C \).

🎯 Exam Tip: For particular solutions, always find the general solution first, then use the given initial conditions (values of \( x \) and \( y \)) to solve for the constant of integration \( C \). Remember that \( \log 1 = 0 \).

Question 15. Find the particular solution of the differential equation \( \log\left(\frac { dy }{ dx }\right) = 3x + 4y \), given that \( y = 0 \) when \( x = 0 \).
Answer: Given the differential equation: \[ \log\left(\frac { dy }{ dx }\right) = 3x + 4y \] First, we convert this logarithmic equation to an exponential form:
\( \implies \frac { dy }{ dx } = e^{3x + 4y} \) We can separate the exponential term:
\( \implies \frac { dy }{ dx } = e^{3x} e^{4y} \) Now, we separate the variables, bringing all \( y \) terms to the left side with \( dy \) and \( x \) terms to the right with \( dx \):
\( \implies \frac{dy}{e^{4y}} = e^{3x} dx \)
\( \implies e^{-4y} dy = e^{3x} dx \) Next, we integrate both sides:
\( \implies \int e^{-4y} dy = \int e^{3x} dx \) Integrating \( e^{ay} \) with respect to \( y \) gives \( \frac{e^{ay}}{a} \). So:
\( \implies \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C \) (Equation 1) Now, we use the given condition \( x = 0, y = 0 \) to find the particular solution. Substitute these values into Equation 1:
\( \implies \frac{e^{-4(0)}}{-4} = \frac{e^{3(0)}}{3} + C \)
\( \implies \frac{e^0}{-4} = \frac{e^0}{3} + C \) Since \( e^0 = 1 \):
\( \implies \frac{1}{-4} = \frac{1}{3} + C \)
\( \implies -\frac{1}{4} = \frac{1}{3} + C \) To find \( C \), subtract \( \frac{1}{3} \) from both sides:
\( \implies C = -\frac{1}{4} - \frac{1}{3} \) To combine these fractions, find a common denominator, which is 12:
\( \implies C = -\frac{3}{12} - \frac{4}{12} \)
\( \implies C = -\frac{7}{12} \) Substitute \( C = -\frac{7}{12} \) back into Equation 1:
\( \implies \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} - \frac{7}{12} \) This is the required particular solution. To make the denominators positive and clearer, multiply the entire equation by -12:
\( \implies 3e^{-4y} = -4e^{3x} + 7 \)
In simple words: First, we convert the logarithm to an exponential form, then separate the terms with \( y \) and \( x \). We integrate both sides and use the starting conditions to find the exact value of the constant \( C \).

🎯 Exam Tip: When given initial conditions, substitute them into the general solution *before* simplifying constants to avoid errors. Also, be careful with negative signs and fractions during algebraic manipulation.