What is the solution for y^{''}-3y^'+2y=4t+e^{3t} ?

The solution for y^{''}-3y^'+2y=4t+e^{3t} is y=c_{1}e^{2t}+c_{2}e^t+2t+3+1/2 e^{3t}

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y^{''}-3y^'+2y=4t+e^{3t}

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 t + e^{3 t}

y = c_{1} e^{2 t} + c_{2} e^{t} + 2 t + 3 + \frac{1}{2} e^{3 t}

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 4 t + e^{3 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + 2 t + 3 + \frac{1}{2} e^{3 t}

y = c_{1} e^{2 t} + c_{2} e^{t} + 2 t + 3 + \frac{1}{2} e^{3 t}

2 y^{^{''}} - 8 y^{^{'}} - 4 y = - 5 x^{2} - 12 x e^{- x}

y^{^{''}} - 5 y^{^{'}} - 24 y = (t + 1) e^{t}

5 y^{^{''}} + y^{^{'}} = - 4 x, y (0) = 0, y^{^{'}} (0) = - 25

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} = 5 t

3 y^{^{''}} + 2 y^{^{'}} - y = 23

What is the solution for y^{''}-3y^'+2y=4t+e^{3t} ?
The solution for y^{''}-3y^'+2y=4t+e^{3t} is y=c_{1}e^{2t}+c_{2}e^t+2t+3+1/2 e^{3t}