Question

The solution of the differential equation $\left(1+\frac{2 \tan x}{y}\right) \frac{d y}{d x}=\sec ^2 x$ such that $y\left(\frac{\pi}{4}\right)=1$ is

• A. $2 y+\tan x=y^2+1$
• B. $2 \tan x+y=3 y^2$
• C. $\tan x + y =2 y ^2$
• D. $2 y+\tan x=3 y^2$

Answer 1

Answer: (C)

$ 1+\frac{2 \tan x}{y}=\sec ^2 x \frac{d x}{d y}$
Let $ \tan x=z \Rightarrow \frac{d z}{d y}=\sec ^2 x \frac{d x}{d y}$
$\frac{d z}{d y}-\frac{2 z}{y}=1 \Rightarrow \frac{z}{y^2}=\int \frac{1}{y^2} d y+c $
$z=-y+c y^2 $
$\tan x=c y^2-y $
$y\left(\frac{\pi}{4}\right)=1 \Rightarrow 1=c-1 \Rightarrow c=2 $
$2 y^2=y+\tan x$