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

The solution for y^{''}+2y^'=cos(2x) is y=c_{1}+c_{2}e^{-2x}+1/8 sin(2x)-1/8 cos(2x)

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

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

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{8} sin (2 x) - \frac{1}{8} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{8} sin (2 x) - \frac{1}{8} cos (2 x)

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{8} sin (2 x) - \frac{1}{8} cos (2 x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 10 e^{3 t}

y^{^{''}} - 6 y^{^{'}} + 3 y = x e^{x}

3 y^{^{''}} + 8 y^{^{'}} + 4 y = 8 x^{2} - 8 x + 2 + e^{- 2 x}

y^{^{''''}} + 2 y^{^{''}} + y = 3 + cos (3 t)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{t} sec (t)

What is the solution for y^{''}+2y^'=cos(2x) ?
The solution for y^{''}+2y^'=cos(2x) is y=c_{1}+c_{2}e^{-2x}+1/8 sin(2x)-1/8 cos(2x)