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

The solution for y^{''}+4y=t^2+4e^t is y=c_{1}cos(2t)+c_{2}sin(2t)+(t^2)/4-1/8+4/5 e^t

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{4}{5} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{4}{5} e^{t}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{4}{5} e^{t}

y^{^{''}} + 2 y = 12 sin (2 t), y (0) = 5

y^{^{''}} + y^{^{'}} - 2 y = 3 e^{- 2 t}, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x, y (0) = 4, y^{^{'}} (0) = 8

y^{^{''}} - 25 y = e^{5 t}, y (0) = 2, y^{^{'}} (0) = 1

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

What is the solution for y^{''}+4y=t^2+4e^t ?
The solution for y^{''}+4y=t^2+4e^t is y=c_{1}cos(2t)+c_{2}sin(2t)+(t^2)/4-1/8+4/5 e^t