What is the solution for y^{''}+2y^'=cos(4x) ?

The solution for y^{''}+2y^'=cos(4x) is y=c_{1}+c_{2}e^{-2x}+1/40 sin(4x)-1/20 cos(4x)

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y^{''}+2y^'=cos(4x)

y^{^{''}} + 2 y^{^{'}} = cos (4 x)

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{40} sin (4 x) - \frac{1}{20} cos (4 x)

Solution steps

y^{''} (x) + 2 y^{'} (x) = cos (4 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{40} sin (4 x) - \frac{1}{20} cos (4 x)

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{40} sin (4 x) - \frac{1}{20} cos (4 x)

2 x^{^{''}} + 4 x + 20 x = 20 H (t)

2 y^{^{''}} - y^{^{'}} - y = 25 e^{- x}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 5 y = e^{x} sin (2 x)

y^{^{''}} - 4 y = 12 sin (2 t)

y^{^{''}} + 6 y^{^{'}} + 45 y = 6

What is the solution for y^{''}+2y^'=cos(4x) ?
The solution for y^{''}+2y^'=cos(4x) is y=c_{1}+c_{2}e^{-2x}+1/40 sin(4x)-1/20 cos(4x)