What is the solution for y^{''}+y^'=sin^2(x) ?

The solution for y^{''}+y^'=sin^2(x) is y=c_{1}+c_{2}e^{-x}+1/2 x-1/20 sin(2x)-2/5 cos(2x)-sin^2(x)

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y^{''}+y^'=sin^2(x)

y^{^{''}} + y^{^{'}} = sin^{2} (x)

y = c_{1} + c_{2} e^{- x} + \frac{1}{2} x - \frac{1}{20} sin (2 x) - \frac{2}{5} cos (2 x) - sin^{2} (x)

Solution steps

y^{''} (x) + y^{'} (x) = sin^{2} (x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + \frac{1}{2} x - \frac{1}{20} sin (2 x) - \frac{2}{5} cos (2 x) - sin^{2} (x)

y = c_{1} + c_{2} e^{- x} + \frac{1}{2} x - \frac{1}{20} sin (2 x) - \frac{2}{5} cos (2 x) - sin^{2} (x)

x = - 2, x = 1, y = 11 x, y = x^{2} - 12

\frac{d}{d x} (400 - 5 x)

\int v (u + v^{2})^{4} d u

\int sin (θ) d θ

\frac{d}{d x} (6 x + 7 x)

What is the solution for y^{''}+y^'=sin^2(x) ?
The solution for y^{''}+y^'=sin^2(x) is y=c_{1}+c_{2}e^{-x}+1/2 x-1/20 sin(2x)-2/5 cos(2x)-sin^2(x)