Let y=y(x) solve dydx+y x=2x+x^2 x, x(-2,2), y(0)=1.

Question

Let $y=y(x)$ solve $\dfrac{dy}{dx}+y	an x=2x+x^2	an x$, $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}ight)$, $y(0)=1$. Then:

Accepted Answer

I.F.$=\sec x$. $y\sec x = x^2\sec x+C \Rightarrow y = x^2+C\cos x$. Using $y(0)=1$: $C=1$, so $y=x^2+\cos x$.

$y'=2x-\sin x$; $y'(\pi/4)=\dfrac{\pi}{2}-\dfrac{1}{\sqrt2}$, $y'(-\pi/4)=-\dfrac{\pi}{2}+\dfrac{1}{\sqrt2}$.

$$y'\!\left(\frac{\pi}{4}\right)-y'\!\left(-\frac{\pi}{4}\right) = \pi-\sqrt2.$$

Solutions · 1