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
"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)\\\\\n\\Rightarrow 2cosx.y'(x)=2cos^2x\\\\\n\\Rightarrow y'(x)=cosx\\\\\n\\Rightarrow dy=cosx.dx"
Integrating both sides, we get:
"y=sinx+c"
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