Let y=y(x) solve (2+ x)(y+1)dydx=- x, y0, y(0)=1.

Question

Let $y=y(x)$ solve $(2+\sin x)(y+1)\dfrac{dy}{dx}=-\cos x$, $y>0$, $y(0)=1$. If $y(\pi)=a$ and $y'(\pi)=b$, the ordered pair $(a,b)$ is:

Accepted Answer

$$\int\frac{dy}{y+1} = -\int\frac{\cos x}{2+\sin x}\,dx \implies \ln(y+1)+\ln(2+\sin x)=\ln C.$$

Using $(0,1)$: $C=4$. So $y=\dfrac{2-\sin x}{2+\sin x}$. At $x=\pi$: $y=1 \Rightarrow a=1$; differentiating and evaluating gives $y'(\pi)=1 \Rightarrow b=1$.

Solutions · 1