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

The solution for y^{''}-2y^'+2y=2e^t is y=e^t(c_{1}cos(t)+c_{2}sin(t))+2e^t

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

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

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) + 2 e^{t}

Solution steps

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

Solve linear ODE:

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) + 2 e^{t}

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) + 2 e^{t}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 9 y = x e^{x}

y^{^{''}} + y^{^{'}} + 2 y = x + 3, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + y^{^{'}} + 4.25 = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- 5 x} (- 55 + 68 x - 30 x^{2})

2 x^{^{''}} + x = 3 t^{2} + 10 t

What is the solution for y^{''}-2y^'+2y=2e^t ?
The solution for y^{''}-2y^'+2y=2e^t is y=e^t(c_{1}cos(t)+c_{2}sin(t))+2e^t