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Linear Differential Equation – Integrating Factor | JEE

JEE Maths question with a full step-by-step solution.

Question

Let

\dfrac{x\,dy}{dx} - y = x^2(xe^x + e^x - 1)

for all

x \in R - \{0\}

with

y(1) = e - 1

. If

y(2) = k\,y(1)\big(y(1)+2\big)

, then the value of

k

is:

1

2

3

4

correct

Solution

Step 1: Write in linear form (divide by

x

\frac{dy}{dx} - \frac{y}{x} = x(xe^x + e^x - 1).

\text{IF} = e^{\int -\frac{1}{x}dx} = e^{-\ln x} = \frac{1}{x}.

Step 2: Solve:

y\cdot\frac{1}{x} = \int \big(xe^x + e^x - 1\big)\,dx = xe^x - x + C \quad\text{(since } \tfrac{d}{dx}(xe^x)=xe^x+e^x\text{)}.

\Rightarrow \frac{y}{x} = xe^x - x + C.

Step 3: Apply

y(1) = e-1

\dfrac{e-1}{1} = e - 1 + C \Rightarrow C = 0

. Hence

y = x^2e^x - x^2 = x^2(e^x - 1)

. Step 4: Evaluate.

y(2) = 4(e^2 - 1) = 4(e-1)(e+1)

. With

y(1)=e-1

and

y(1)+2 = e+1

y(2) = 4\,y(1)\big(y(1)+2\big) \Rightarrow k = 4.

Correct answer: (4)

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