Question

Find the particular solution of differential equation:

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

Answer 1

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

⇒ ${\displaystyle \frac{dy}{dx}+\frac{\cos x}{1+\sin x}y=\frac{x}{1+\sin x}\text{......i}}$

This is a linear differential equation with

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

${\displaystyle \therefore I.F.=e^{\int }\frac{\cos x}{1+\sin x}dx}$

= ${\displaystyle e^{\log (1+\sin x)}}$

= 1+ sinx

Multiplying both the sides of i by I.F. = 1 + sinx, we get

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

Integrating with respect to x, we get

${\displaystyle y(1+\sin x)=\int -xdx+C}$

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

Given that y = 1 when x = 0

${\displaystyle \therefore 1=\frac{2C}{2(1+0)}}$

⇒ C =1 ................(iii)

Put iii in ii , we get

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