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

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

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{4} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{4} e^{3 x}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{4} e^{3 x}

y^{^{''}} + y^{^{'}} - 8 y = 48 t

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

y^{^{''}} + y^{^{'}} - 6 y = 6 e^{2 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{x} - 9 + 24 x + 4

y^{^{''''}} - 2 y^{^{''}} + 1 = 0

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