Question

Find the solution to 2cos(x)y'(x)=2cos^2(x)-sin^2(x)+y^2 if y(x)=sin x denotes a particular solution

Answer 1

$2cosx.y'(x)=2cos^2(x)-sin^2(x)+y^2\\$

Now substituting $y(x)=sinx$ , we get:

$2cosx.y'(x)=2cos^2(x)-sin^2(x)+sin^2(x)\\ \Rightarrow 2cosx.y'(x)=2cos^2x\\ \Rightarrow y'(x)=cosx\\ \Rightarrow dy=cosx.dx$

Integrating both sides, we get:

$y=sinx+c$