Question

For the differential equation $\frac{dy}{dx}+2y\tan x=y^2.$ Its solution is ${\cos }^{2}x+(\frac{1}{A}x+\frac{1}{B}\sin 2x)y=cy$ find A+B?

Answer 1

$\frac{dy}{dx}+2y\tan x=y^2$ ....(1)
which is a Bernoulli's differential eqn
Dividing (1) by $y^2$, we get
$y^{-2}\frac{dy}{dx}+2y^{-1}\tan x=1$ ....(2)
Put $z=y^{-1}$
$\Rightarrow \frac{dz}{dx}=-\frac{1}{y^2}\frac{dy}{dx}$
So, eqn (2) becomes
$-\frac{dz}{dx}+2z\tan x=1$
$\Rightarrow \frac{dz}{dx}-2z\tan x=-1$ ...(3)
which is a linear differential equation with z as dependent variable.
Integrating factor $I.F.=e^{\int Pdx}$
$=e^{\int -2\tan xdx}$
$=e^{2\log \cos x}$
$\Rightarrow I.F.={\cos }^{2}x$
Solution of (3) is given
$z{\cos }^{2}x=\int {\cos }^{2}dx$
$z{\cos }^{2}x=\int \frac{1+\cos 2x}{2}dx$
$\Rightarrow z{\cos }^{2}x=\frac{1}{2}(x+\frac{\sin 2x}{2})+C$
$\Rightarrow y^{-1}{\cos }^{2}x=(\frac{x}{2}+\frac{\sin 2x}{4})+C$
$\Rightarrow {\cos }^{2}x=(\frac{x}{2}+\frac{\sin 2x}{4})y+Cy$
On comparing with given solution,
$A=2,B=4$
$A+B=6$