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}+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} + 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} + e^{3 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{3 x}

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

y^{^{''}} - 8 y^{^{'}} - 9 y = 21 e^{2} t

y^{^{''}} + 4 y^{^{'}} = 3 \cdot sin (2 x)

y^{^{''}} - 6 y^{^{'}} + 8 y = x e^{3 x} cos (4 x)

y^{^{''}} - 8 y^{^{'}} - 20 y = 5 e^{- 2 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}+e^{3x}