JEE Main 2014 · 11 Apr · Shift 1 · Mathematics Q77

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If the general solution of the differential equation $y'=\dfrac yx+\Phi\left(\dfrac xy\right)$ , for some function $\Phi$ , is given by $y\ln|cx|=x$ , where $c$ is an arbitrary constant, then $\Phi(2)$ is equal to :

\dfrac14

-4

-\dfrac14

Answer and worked solution

Official answer: D

$\ln|cx|=\frac xy$ . Differentiating, $\dfrac1x=\dfrac{y-xy'}{y^2}$ , so
$y'=\frac yx-\frac{y^2}{x^2}.$
Hence $\Phi(t)=-\dfrac1{t^2}$ with $t=\frac xy$ , and $\Phi(2)=-\dfrac14$ .

Topics:Homogeneous equations

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JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 77