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

$\frac{dy}{dx}=-\frac{x+ycosx}{1+sinx}=\frac{-x}{1+sinx}-y\frac{cosx}{1+sinx}$ , which implies $\frac{dy}{dx}+y\frac{cosx}{1+sinx}=\frac{-x}{1+sinx}$

it is a linear equation which is in the form of $\frac{dy}{dx}+yP\left(x\right)=Q\left(x\right)$ , where $P\left(x\right)=\frac{cosx}{1+sinx}$ and $Q\left(x\right)=\frac{-x}{1+sinx}$

the integral factor is $e^{\int P\left(x\right)dx}=e^{\int \frac{cosx}{1+sinx}dx}=e^{\ln (1+sinx)}=1+sinx$

Therefore the equation is $y(1+sinx)=\int \frac{-x}{1+sinx}(1+sinx)dx+c=\int -xdx+c=-x^2/2+c$

Therefore $y=\frac{-x^2}{2(1+sinx)}+\frac{c}{1+sinx}$ , at $x=0$ , $y=1$ , we have $1=0+c$

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