Question

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

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

Answer 1

Answer: (C)

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