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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 9 y = 6 sin (3 x)

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

y^{^{''}} - 2 y^{^{'}} + y = 6

y^{^{''}} - 2 y^{^{'}} + y = 4

y^{^{''}} + 4 y^{^{'}} - 5 y = 35 e^{- 4 t}

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