Question

Find the particular solution of differential equation : $\frac{dy}{dx}=-\frac{x+ycosx}{1+sinx}$ given that y = 1 when x = 0.

Answer 1

Here $\frac{dy}{dx}=-\frac{x+ycosx}{1+sinx}\Rightarrow \frac{dy}{dx}+\left(\begin{array}{c}\frac{cosx}{1+sinx}\end{array}\right)y=-\frac{x}{1+sinx}$
This is in the form of $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)whereP\left(x\right)=\frac{cosx}{1+sinx},Q\left(x\right)=-\frac{x}{1+sinx}$
Now I.F. =$e^{\int \left(\begin{array}{c}\frac{cosx}{1+sinx}\end{array}\right)dx}=e^{log(1+sinx)}=1+sinx$

So the required solution is, $y(1+sinx)=\int -\frac{x}{1+sinx}\times (1+sinx)dx+C$

$\Rightarrow y(1+sinx)=\int -xdx+C\Rightarrow y(1+sinx)=-\frac{1}{2}{x}^2+C$

As it is given that y = 1 when x = 0 so, $1(1+sin0)=-\frac{1}{2}\times 0^2+C\Rightarrow C=1.$

So the particular solution is, $y(1+sinx)=1-\frac{1}{2}{x}^2.$