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}

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

y^{^{''}} + 10 y^{^{'}} + 41 y = t e^{- 5 t} sin (4 t)

y^{^{''}} - y^{^{'}} - 6 y = e^{t}, y (0) = 1, y^{^{'}} (0) = 1

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

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{x}, y (0) = 0, y^{^{'}} (0) = 3

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}