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

The solution for 2y^{''}+y^'=cos^2(x) is y=c_{1}+c_{2}e^{-x/2}+(34x+sin(2x)-4cos(2x))/(68)-1

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

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

y = c_{1} + c_{2} e^{- \frac{x}{2}} + \frac{34 x + sin ( 2 x ) - 4 cos ( 2 x )}{68} - 1

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{x}{2}} + \frac{34 x + sin ( 2 x ) - 4 cos ( 2 x )}{68} - 1

y = c_{1} + c_{2} e^{- \frac{x}{2}} + \frac{34 x + sin ( 2 x ) - 4 cos ( 2 x )}{68} - 1

y^{^{''}} - 5 y^{^{'}} + 6 y = (2 x - 1) e^{4 x}

y^{^{''}} - 2 y^{^{'}} + y = \frac{2 e ^{x}}{x ^{2} - 1}

y^{^{''}} + 5 y^{^{'}} + 6 y = - 19 t e^{(2 t)}

y^{^{''}} + 9 y = g (t), y (0) = 4, y^{^{'}} (0) = 0

y^{^{''}} + 5 y^{^{'}} + 4 y = 2 e^{- 5 x}, y (0) = 4, y^{^{'}} (0) = 3

What is the solution for 2y^{''}+y^'=cos^2(x) ?
The solution for 2y^{''}+y^'=cos^2(x) is y=c_{1}+c_{2}e^{-x/2}+(34x+sin(2x)-4cos(2x))/(68)-1