The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=y \tan x-y^2 \sec x$ is
The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=y \tan x-y^2 \sec x$ is
- $\tan x=(\mathrm{c}+\sec x) y$, where c is constant of integration.
- $\sec y=(\mathrm{c}+\tan y) x$, where c is constant of integration.
- $\sec x=(\mathrm{c}+\tan x) y$, where c is constant of integration.
- $\cos y=(\mathrm{c}+\tan y)$, where c is constant of integration.
Solution
$\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=y \tan x-y^2 \sec x \\
& \Rightarrow \frac{1}{y^2} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}+\tan x\left(-\frac{1}{y}\right)=-\sec x \ldots (i)\\
& \text { Put } \mathrm{v}=-\frac{1}{y} \Rightarrow \frac{\mathrm{dv}}{\mathrm{~d} x}=\frac{1}{y^2} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}
\end{aligned}$
$\therefore \quad \frac{\mathrm{dv}}{\mathrm{d} x}+(\tan x) \mathrm{v}=-\sec x \quad \ldots[$ From (i) $]$
$\therefore \quad$ I.F. $=\mathrm{e}^{\int \tan x \mathrm{dx}}=\mathrm{e}^{\log (\sec x)}=\sec x$
$\therefore \quad$ solution of the given equation is
$\begin{aligned} & \text { v. } \sec x=\int-\sec x \cdot \sec x \mathrm{~d} x+\mathrm{c}_1 \\ & \Rightarrow \mathrm{v} \sec x=-\tan x+\mathrm{c}_1 \\ & \Rightarrow-\frac{1}{y} \sec x=-\tan x+\mathrm{c}_1 \\ & \Rightarrow \sec x=y(\tan x+\mathrm{c}), \text { where } \mathrm{c}=-\mathrm{c}_1\end{aligned}$
Asked in: MHT CET 2024 (16 May Shift 2)
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