What is the solution for y^{''}-3y^'+2y=-e^t+2e^{2t} ?

The solution for y^{''}-3y^'+2y=-e^t+2e^{2t} is y=c_{1}e^{2t}+c_{2}e^t+e^tt+2te^{2t}

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y^{''}-3y^'+2y=-e^t+2e^{2t}

y^{^{''}} - 3 y^{^{'}} + 2 y = - e^{t} + 2 e^{2 t}

y = c_{1} e^{2 t} + c_{2} e^{t} + e^{t} t + 2 t e^{2 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + e^{t} t + 2 t e^{2 t}

y = c_{1} e^{2 t} + c_{2} e^{t} + e^{t} t + 2 t e^{2 t}

y^{^{''}} + y = 3 t^{2}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y = cos (4 t)

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 = e^{- 5 x}

(d^{2} - 9) y = 30 sin^{3} (x)

What is the solution for y^{''}-3y^'+2y=-e^t+2e^{2t} ?
The solution for y^{''}-3y^'+2y=-e^t+2e^{2t} is y=c_{1}e^{2t}+c_{2}e^t+e^tt+2te^{2t}