Solution of the differential equation $\cos x d y=y(\sin x-y) d x, 0 < x < \frac{\pi}{2}$ is
Solution of the differential equation $\cos x d y=y(\sin x-y) d x, 0 < x < \frac{\pi}{2}$ is
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$y \sec x=\tan x+c$
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$y \tan x=\sec x+c$
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$\tan x=(\sec x+c) y$
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$\sec x=(\tan x+c) y$
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) $\sec x d x$
$\frac{1}{y} \sec x=\tan x+c$
Asked in: JEE Main 2010
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