What is the solution for y^{''}+4y^'+4y=e^{3x} ?

The solution for y^{''}+4y^'+4y=e^{3x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+1/25 e^{3x}

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y^{''}+4y^'+4y=e^{3x}

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{3 x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{1}{25} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{1}{25} e^{3 x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{1}{25} e^{3 x}

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

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 159 = 0

y^{^{''}} + 3 y^{^{'}} + 3 y = 6 sin (2 t)

y^{^{''}} - 14 y^{^{'}} + 49 y = 14 x + 3

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

What is the solution for y^{''}+4y^'+4y=e^{3x} ?
The solution for y^{''}+4y^'+4y=e^{3x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+1/25 e^{3x}