What is the solution for y^{''}+4y^'+3y=cos(2x) ?

The solution for y^{''}+4y^'+3y=cos(2x) is y=c_{1}e^{-x}+c_{2}e^{-3x}+8/65 sin(2x)-1/65 cos(2x)

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

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

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{8}{65} sin (2 x) - \frac{1}{65} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{8}{65} sin (2 x) - \frac{1}{65} cos (2 x)

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{8}{65} sin (2 x) - \frac{1}{65} cos (2 x)

y^{^{''}} + π^{2} y = \frac{5}{ln ( 10 )} sin (x ln (2))

\frac{d ^{2} y}{d t ^{2}} + 6 \frac{d y}{d t} + 9 y = e^{- 3 t}

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

y^{^{''}} + x + 4 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = x e^{4 x}

What is the solution for y^{''}+4y^'+3y=cos(2x) ?
The solution for y^{''}+4y^'+3y=cos(2x) is y=c_{1}e^{-x}+c_{2}e^{-3x}+8/65 sin(2x)-1/65 cos(2x)