What is the solution for y^{''}+4y^'=cos(x) ?

The solution for y^{''}+4y^'=cos(x) is y=c_{1}+c_{2}e^{-4x}+4/17 sin(x)-1/17 cos(x)

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y^{''}+4y^'=cos(x)

y^{^{''}} + 4 y^{^{'}} = cos (x)

y = c_{1} + c_{2} e^{- 4 x} + \frac{4}{17} sin (x) - \frac{1}{17} cos (x)

Solution steps

y^{''} (x) + 4 y^{'} (x) = cos (x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{4}{17} sin (x) - \frac{1}{17} cos (x)

y = c_{1} + c_{2} e^{- 4 x} + \frac{4}{17} sin (x) - \frac{1}{17} cos (x)

y^{^{''}} + 7 y^{^{'}} + 10 y = - 10 + e^{4 t}

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

y^{^{''}} + 2 y^{^{'}} + 2 y = (e^{(- x)}) \cdot cos (x)

y^{^{''}} - 3 y^{^{'}} - 4 y = x e^{x} + 2 cos (x) - sin (x)

D (D - 1)^{3} y = (x^{2} + 2 x) e^{x}

What is the solution for y^{''}+4y^'=cos(x) ?
The solution for y^{''}+4y^'=cos(x) is y=c_{1}+c_{2}e^{-4x}+4/17 sin(x)-1/17 cos(x)