What is the solution for y^{''}+4y=t^2+8e^t ?

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

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y^{''}+4y=t^2+8e^t

y^{^{''}} + 4 y = t^{2} + 8 e^{t}

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

Solution steps

y^{''} (t) + 4 y = t^{2} + 8 e^{t}

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 (9 x - 2) e^{2 x}

y^{^{''}} + 25 y = 10 cos (5 x) + 15 sin (5 x)

y^{^{''}} - 7 y^{^{'}} + 12 y = x^{2} e^{x}

y^{^{''}} - y^{^{'}} - 20 y = 72 e^{2 t}

y^{^{''}} - y = t, y (0) = 1, y^{^{'}} (0) = 1

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