The solution of dydx+y2 x= x2y, 0 x

Question

The solution of $\dfrac{dy}{dx}+\dfrac{y}{2}\sec x=\dfrac{	an x}{2y}$, $0\le x<\pi/2$, $y(0)=1$, is given by:

Accepted Answer

With $t=y^2$: $\dfrac{dt}{dx}+t\sec x=\tan x$. I.F.$=\sec x+\tan x$.

$$t(\sec x+\tan x) = \sec x+\tan x-x+C.$$

Using $y(0)=1 \Rightarrow t(0)=1$: $C=0$. So $t=1-\dfrac{x}{\sec x+\tan x}$, i.e. $y^2=1-\dfrac{x}{\sec x+\tan x}$.

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