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

The solution for y^{''}-2y^'+3y=3e^{2t} is y=e^t(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))+e^{2t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 t} ln (t)

y^{^{''}} - 7 y^{^{'}} + 6 y = x, y^{1} (0) = e x

y^{^{''}} - 8 y^{^{'}} - 9 y = 10, y (0) = 0, y^{^{'}} (0) = 4

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

y^{^{''}} + 2 y = e^{- 2 x} (x^{2} + 1)

What is the solution for y^{''}-2y^'+3y=3e^{2t} ?
The solution for y^{''}-2y^'+3y=3e^{2t} is y=e^t(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))+e^{2t}