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Linear Differential Equation – Integrating Factor | JEE

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Question
Let x dydx−y=x2(xex+ex−1)\dfrac{x\,dy}{dx} - y = x^2(xe^x + e^x - 1) for all x∈R−{0}x \in R - \{0\} with y(1)=e−1y(1) = e - 1. If y(2)=k y(1)(y(1)+2)y(2) = k\,y(1)\big(y(1)+2\big), then the value of kk is:
A11
B22
C33
D44correct
Solution
Step 1: Write in linear form (divide by xx):
dydx−yx=x(xex+ex−1).\frac{dy}{dx} - \frac{y}{x} = x(xe^x + e^x - 1).
IF=e∫−1xdx=e−ln⁡x=1x.\text{IF} = e^{\int -\frac{1}{x}dx} = e^{-\ln x} = \frac{1}{x}.
 Step 2: Solve:
y⋅1x=∫(xex+ex−1) dx=xex−x+C(since ddx(xex)=xex+ex).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{)}.
⇒yx=xex−x+C.\Rightarrow \frac{y}{x} = xe^x - x + C.
 Step 3: Apply y(1)=e−1y(1) = e-1: e−11=e−1+C⇒C=0\dfrac{e-1}{1} = e - 1 + C \Rightarrow C = 0. Hence y=x2ex−x2=x2(ex−1)y = x^2e^x - x^2 = x^2(e^x - 1). Step 4: Evaluate. y(2)=4(e2−1)=4(e−1)(e+1)y(2) = 4(e^2 - 1) = 4(e-1)(e+1). With y(1)=e−1y(1)=e-1 and y(1)+2=e+1y(1)+2 = e+1,
y(2)=4 y(1)(y(1)+2)⇒k=4.y(2) = 4\,y(1)\big(y(1)+2\big) \Rightarrow k = 4.
 Correct answer: (4)
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