Question

For the differential equation $2\frac{dy}{dx}-y\sec x=y^3\tan x$, the solution is $\frac{1}{y^2}=-1+(c+x)\cot (\frac{x}{2}+\theta )$. Find $4\tan \left(\theta \right)$.

Answer 1

$2\frac{dy}{dx}-y\sec x=y^3\tan x\Rightarrow \frac{1}{y^3}\frac{dy}{dx}-\frac{\sec x}{2y^2}=\tan x$
Put $-\frac{1}{2y^2}=v\Rightarrow \frac{1}{y^3}dy=dv$
$\therefore \frac{dv}{dx}+v\sec x=\tan x$ ...(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
$v(\sec x+\tan x)=\int \tan x(\sec x+\tan x)dx+c=-x+(\sec x+\tan x)+c$
$\Rightarrow \frac{1}{y^2}=-1+(c+x)\cot (\frac{x}{2}+\frac{\pi }{4})$
Then $4\tan \theta =4\tan \frac{\pi }{4}=4$