JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q76

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MathematicsSingle correct+4 / -1Question 76 of 90

For x ∈ R, x ≠ 0, if y(x) is a differentiable function such that $x\displaystyle\int_1^xy(t)\,dt=(x+1)\int_1^xt\,y(t)\,dt$ , then y(x) equals :

(where C is a constant.)

\dfrac Cx e^{-\frac1x}

\dfrac C{x^2}e^{-\frac1x}

\dfrac C{x^3}e^{-\frac1x}

Cx^3e^{\frac1x}

Answer and worked solution

Official answer: C

Differentiating, $\int_1^xy\,dt+xy=\int_1^xty\,dt+(x+1)xy$ , so $\int_1^x(1-t)y\,dt=x^2y$ . Differentiating again, $(1-x)y=2xy+x^2y'$ , i.e. $\frac{y'}y=\frac1{x^2}-\frac3x$ . So $y=\frac C{x^3}e^{-1/x}$ .

Topics:Leibniz rule Variable separable

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JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 76