For x∈R,x≠0 if y(x) is a differentiable function such that xx∫1y(t)dt=(x+1)x∫1ty(t)dt, then y(x) equals:
A
Cx3e1x
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B
Cx2e−1x
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C
Cx3e−1x
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D
Cxe−1x
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Solution
The correct option is CCx3e−1x xx∫1y(t)dt=(x+1)x∫1ty(t)dt
Differentiating w.r.t x, we get x∫1y(t)dt+xy(x)=x∫1ty(t)dt+(x+1)xy(x) ⇒x2y(x)=x∫1(y(t)−ty(t))dt
Again differentiating w.r.t x, we get 2xy(x)+x2dydx=y(x)−xy(x) ⇒dydx=1−3xx2y ⇒∫dyy=∫1−3xx2dx ⇒logy=−1x−3logx+logC ⇒logy=−1x−logx3+logC ⇒logyx3C=−1x ⇒y=Cx3e−1x