Question

Solve the differential equation: $(1+x^2)\frac{dy}{dx}-4x^2{\cos }^{2}y+x\sin 2y=0$

• A. $\tan y(1-x^2)=\frac{4}{3}{x}^3+c.$
• B. $\tan y(1-x^2)=\frac{4}{3}{x}^{-3}+c.$
• C. $\tan y(1+x^2)=\frac{4}{3}{x}^{-3}+c.$
• D. $\tan y(1+x^2)=\frac{4}{3}{x}^3+c.$

Answer 1

The correct option is D $\tan y(1+x^2)=\frac{4}{3}{x}^3+c.$

Given, $(1+x^2)\frac{dy}{dx}-{4x}^2{\cos }^{2}y+x{\sin }^{2}y=0$
$\Rightarrow {\sec }^{2}y\frac{dy}{dx}+\frac{2x}{1+x^2}\tan x=\frac{{4x}^2}{1+x^2}$
Put $\tan y=v\Rightarrow {\sec }^{2}ydy=dv$
$\therefore \frac{dv}{dx}+\frac{2x}{1+x^2}v=\frac{{4x}^2}{1+x^2}$ ...(1)
Here $P=\frac{2x}{1+x^2}\Rightarrow \int Pdx=\int \frac{2x}{1+x^2}=\log (1+x^2)$
$\therefore I.F=e^{\log (1+x^2)}=1+x^2$
Multiplying (1) by $I.F$ we get
$(1+x^2)\frac{dv}{dx}+2xv={4x}^2$
Integrating both sides
$v(1+x^2)=\frac{4}{3}{x}^3+c$

$\Rightarrow \tan y(1+x^2)=\frac{4}{3}{x}^3+c$