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

The solution for y^{''}+4y^'=t^2+2e^t is y=c_{1}+c_{2}e^{-4t}+(t^3)/(12)-(t^2)/(16)+t/(32)+2/5 e^t

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

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

y = c_{1} + c_{2} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{2}{5} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{2}{5} e^{t}

y = c_{1} + c_{2} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{2}{5} e^{t}

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

y^{^{''}} + 2 y^{^{'}} - 15 y = (2 x + 1) e^{3 x} + 10

y^{^{''}} - y^{^{'}} - 2 y = 2 t^{2} - 5

y^{^{''}} - 3 y^{^{'}} - 4 y = 2^{t - 2}, y (2) = 1, y^{^{'}} (2) = - 1

y^{^{''}} + 4 y = cos (2) (x)

What is the solution for y^{''}+4y^'=t^2+2e^t ?
The solution for y^{''}+4y^'=t^2+2e^t is y=c_{1}+c_{2}e^{-4t}+(t^3)/(12)-(t^2)/(16)+t/(32)+2/5 e^t