If dydx+3^2xy=1^2x, x(-3,3), y(4)=43, then y(-4) equals:

Question

If $\dfrac{dy}{dx}+\dfrac{3}{\cos^2x}y=\dfrac{1}{\cos^2x}$, $x\in\left(-\dfrac{\pi}{3},\dfrac{\pi}{3}\right)$, $y\left(\dfrac{\pi}{4}\right)=\dfrac43$, then $y\left(-\dfrac{\pi}{4}\right)$ equals:

Accepted Answer

$$\frac{dy}{1-3y} = \sec^2x\,dx \implies -\frac13\ln|1-3y| = \tan x+C.$$

Using $x=\pi/4,y=4/3$: solving gives $C=-1-\dfrac13\ln3$. At $x=-\pi/4$: $\ln|1-3y|=6 \implies y=\dfrac13\pm e^6$; taking the consistent branch, $y=\dfrac13+e^6$.

Solutions · 1