Q35 / 90 · JEE Main 2010 · Mathematics

Solution of the differential equation $\cos x d y=y(\sin x-y) d x, 0 < x < \frac{\pi}{2}$ is

  • A
    $y \sec x=\tan x+c$
  • B
    $y \tan x=\sec x+c$
  • C
    $\tan x=(\sec x+c) y$
  • Correct answer
    D
    $\sec x=(\tan x+c) y$

Solution

$$ \begin{aligned} & \cos x d y=y(\sin x-y) d x \\ & \frac{d y}{d x}=y \tan x-y^2 \sec x \\ & \frac{1}{y^2} \frac{d y}{d x}-\frac{1}{y} \tan x=-\sec x \\ & \text { Let } \frac{1}{y}=t \\ & -\frac{1}{y^2} \frac{d y}{d x}=\frac{d t}{d x} \\ & -\frac{d y}{d x}-t \tan x=-\sec x \Rightarrow \frac{d t}{d x}+(\tan x) t=\sec x . \\ & \text { I.F. }=e^{f \tan x d x}=\sec x \end{aligned} $$ Solution is $t\left(\right.$ I.F) $=\int$ (I.F) $\sec x d x$ $\frac{1}{y} \sec x=\tan x+c$

Hence, the correct answer is option: 'D'.

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