Question

Find the particular solution of differential equation :

$\frac{dy}{dx}=-\frac{x+y\cos x}{1+\sin x}$, given that $y=1$ when $x=0$.

Answer 1

Given:

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

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

Comparing equation (1) with linear differential equation

$\frac{dy}{dx}+Py=Q$

we will get,

$P=\frac{\cos x}{1+\sin x};Q=\frac{-x}{1+\sin x}$

$\therefore$ Intergrating factor (I.F.), $\text{I.F}=e^{\int Pdx}$

$I.F=e^{\int \frac{\cos x}{1+\sin x}dx}=e^{\log (1+\sin x)}=1+\sin x$

For general solution of differential equation,

$\Rightarrow (1+\sin x)y=\int -xdx+C$

$\because \left[y\right(I.F.)=\int Q(I.F.)dx+C]$

$\Rightarrow y(1+\sin x)=\frac{-x^2}{2}+C$ .............(2)

Now, we get $y=1$, when $x=0$

$\Rightarrow 1(1+\sin 0)=\frac{-0}{2}+C\Rightarrow C=1$

Putting $C=1$ in equation (2), we get

$y(1+\sin x)=\frac{-x^2}{2}+1$

Hence the particular solution is,

$2y(1+\sin x)+x^2-2=0$