Question

Solve the differntial equation $2\frac{dy}{dx}-y\sec x=y^3\tan x.$

• A. $\frac{1}{y^2}(\sec x+\tan x)=\sec x+\tan x-x+c$
• B. $-\frac{1}{y^2}(\sec x-\tan x)=\sec x+\tan x-x+c$
• C. $-\frac{1}{y^2}(\sec x+\tan x)=\sec x+\tan x-x+c$
• D. None of these.

Answer 1

The correct option is C $-\frac{1}{y^2}(\sec x+\tan x)=\sec x+\tan x-x+c$

Given, $2\frac{dy}{dx}-y\sec x=y^3\tan x$

$\Rightarrow \frac{2}{y^3}\frac{dy}{dx}-\frac{1}{y^2}\sec x=\tan x$

Put $-\frac{1}{y^2}=v\Rightarrow \frac{1}{y}dy=dv$

$\therefore \frac{dv}{dx}+v\sec x=\tan x$ ....(1)

Here $P=\sec x\Rightarrow \int Pdx=\int \sec xdx=\log (\sec x+\tan x)$

$\therefore I.F.=e^{\log (\sec x+\tan x)}=\sec x+\tan x$

Multiplying (1) by $I.F.$ we get

$(\sec x+\tan x)\frac{dv}{dx}+v\sec x(\sec x+\tan x)=\tan x(\sec x+\tan x)$

Integrating both sides, we get

$(\sec x+\tan x)v=\int \tan x(\sec x+\tan x)dx+c$

$\Rightarrow \frac{-1}{y^2}(\sec x+\tan x)=\sec x+\tan x-x+c(\because {\tan }^{2}x={\sec }^{2}x-1)$