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

Question

Let $y=y(x)$ be the solution curve of the differential equation
 $(1+\sin x)\frac{dy}{dx}+(y+1)\cos x=0,y(0)=0.$ If the curve $y=y(x)$ passes through the point $(\alpha ,\frac{-1}{2})$,
 then a value of $\alpha $ is:

Accepted Answer

$\int \frac{dy}{y+1}=-\int \frac{\cos x}{1+\sin x}\, dx$

On the left side, $\int \frac{dy}{y+1}=\ln (y+1)$.

On the right side, take $1+\sin x$ as substitution: let $u=1+\sin x$, then $du=\cos x\, dx$. So, $-\int \frac{\cos x}{1+\sin x}\, dx=-\int \frac{du}{u}=-\ln (1+\sin x).$

So we get $\Rightarrow \ln (y+1)=-\ln (1+\sin x)+C$

Now use the initial condition $y(0)=0$. Put $x=0$ and $y=0$ in the equation:

$	herefore y(0)=0	o \ln (0+1)=-\ln (1+\sin 0)+C$

That is $0=-\ln (1)+C\Rightarrow C=0.$

So $\Rightarrow \ln (y+1)=-\ln (1+\sin x)$

Taking antilog on both sides, $\Rightarrow y+1=\frac{1}{1+\sin x}$

Given that the curve passes through $(\alpha ,-\frac{1}{2})$, substitute $y=-\frac{1}{2}$ and $x=\alpha $:

$-\frac{1}{2}+1=\frac{1}{1+\sin \alpha }$

So $\frac{1}{2}=\frac{1}{1+\sin \alpha }\Rightarrow 1+\sin \alpha =2\Rightarrow \sin \alpha =1$

Now $\sin \alpha =1$ gives $\Rightarrow \alpha =\frac{\pi }{2}$

Solutions · 1