Solve the differential equation
`(dy)/(dx) =(e^(x)(sin^(2)x+sin 2x))/(y (2logy+1)).`

Explanation

We have
`(dy)/(dx) =(e^(x)(sin^(2)x+sin 2x))/(y (2logy+1))`
`rArr y(2log y +1)dy = e^(x) (sin^(2)x + sin 2x)dx`
`rArr 2 int underset(II)(y) underset(I)(log) y dy +int y dy = int e^(x) (sin^(2)x+sin 2x)dx`
`rArr 2[(log y)*(y^(2))/(2)-int(1)/(y)*(y^(2))/(2)dy]+(1)/(2)y^(2)=inte^(x)(sin^(2)x +sin 2x)dx `[integrating by parts]
`rArr y^(2)(logy) - int y dy +(1)/(2) y^(2)=int e^(x)(sin^(2)x+sin 2x)dx`
`rArr y^(2) (log y) = e^(x) sin^(2)x +C [ because inte^(x){f(x) +f'(x)}dx = e^(x) f(x) +C].`
`therefore y^(2)(logy) =e^(x) sin^(2)x +C` is the required solution.