Let y=y(x) be the solution of the differential equation [(1+ x)(dy)/(dx)+(y+1) x=0, y(0)=0…

Question

Let $y=y(x)$ be the solution of the differential equation
$$(1+\sin x)\frac{dy}{dx}+(y+1)\cos x=0,\qquad y(0)=0.$$
If the curve $y=y(x)$ passes through $\left(\alpha,-\dfrac12\right)$, then a value of $\alpha$ is:

Accepted Answer

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

Using $y(0)=0$: $0=-\ln1+C \Rightarrow C=0$. So
$$y+1 = \frac{1}{1+\sin x}.$$

At $\left(\alpha,-\frac12\right)$: $\dfrac12 = \dfrac{1}{1+\sin\alpha} \Rightarrow 1+\sin\alpha=2 \Rightarrow \sin\alpha=1 \Rightarrow \alpha=\dfrac{\pi}{2}$.

Solutions · 1