What is the solution for y^{''}-2y^'+y=4e^{3x} ?

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

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y^{''}-2y^'+y=4e^{3x}

y^{^{''}} - 2 y^{^{'}} + y = 4 e^{3 x}

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 20 y = 2 sin (3 t)

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

2 \frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 3 y = e^{2 x}

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 5 e^{- t} + 2

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 5 e^{- t} + 4

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