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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(D^{2} - 4 D + 4) y = sin (3 x)

y^{^{''}} - 2 y^{^{'}} - 15 = 0, y (0) = 2, y^{^{'}} (0) = 6

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{3 x} - x^{2}

\frac{d ^{2} y}{d x ^{2}} + 4 y = - 3 x^{2} + 2 x + 3

y^{^{''}} + 16 y = g (t), y (0) = 5, y^{^{'}} (0) = 0

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