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^{^{''}} - 3 y^{^{'}} + 7 y = t^{2} + cos (t)

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

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = 4 cos (x)

(d^{3} + d) y = 6 x (2 x + 1)

y^{^{''}} - 2 y^{^{'}} = - 2

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}