Let y=y(x) solve x(3 x+ x+3),dy = [1+y x(3 x+ x+3)],dx, 0 x/2, y(0)=0.

Question

Let $y=y(x)$ solve $\cos x(3\sin x+\cos x+3)\,dy = [1+y\sin x(3\sin x+\cos x+3)]\,dx$, $0\le x\le\pi/2$, $y(0)=0$. Then $y\left(\dfrac{\pi}{3}\right)$ is:

Accepted Answer

$$\frac{dy}{dx}-(\tan x)y = \frac{1}{\cos x(3\sin x+\cos x+3)}.$$

I.F.$=\cos x$. With Weierstrass substitution $z=\tan(x/2)$, integrating gives
$$y\cos x = \log\left|\frac{z+1}{z+2}\right|+\log2.$$

Using $y(0)=0$ fixes the constant, and evaluating at $x=\pi/3$ gives $y=2\log_e\dfrac{2\sqrt3+10}{11}$.

Solutions · 1