If y=y(x) solves dydx=( x-y)^2x, x(-2,2), y(0)=0, then y(-4) is:

Question

If $y=y(x)$ solves $\dfrac{dy}{dx}=(	an x-y)\sec^2x$, $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}ight)$, $y(0)=0$, then $y\left(-\dfrac{\pi}{4}ight)$ is:

Accepted Answer

$\dfrac{dy}{dx}+y\sec^2x = \sec^2x\tan x$. I.F.$=e^{\tan x}$.

$$ye^{\tan x} = \int e^{\tan x}\sec^2x\tan x\,dx = (\tan x-1)e^{\tan x}+C.$$

Using $y(0)=0$: $C=1$. At $x=-\pi/4$: $y=-2+e$.

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