Question

Solve the differential equation: $\frac{dy}{dx}-2y\tan x=y^2{\tan }^{2}x.$

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

Answer 1

The correct option is B $\frac{1}{y}{\sec }^{2}x=\frac{1}{3}{\tan }^{3}x+c$

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

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

$\therefore \frac{dv}{dx}+2\tan xv={\tan }^{2}x$ ...(1)

Here $P=2\tan x\Rightarrow \int Pdx=\int 2\tan xdx=2\log \left(\sec x\right)=\log {\sec }^{2}x$

$\therefore I.F.=e^{\log {\sec }^{2}x}={\sec }^{2}x$

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

${\sec }^{2}x\frac{dv}{dx}+2{\sec }^{2}x\tan xv={\sec }^{2}x{\tan }^{2}x$

Integrating both sides we get

$v{\sec }^{2}x=\int {\sec }^{2}x{\tan }^{2}xdx+c=\frac{1}{3}{\tan }^{3}x+c$

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