What is the solution for y^{''}+4y=t^2+10e^t ?

The solution for y^{''}+4y=t^2+10e^t is y=c_{1}cos(2t)+c_{2}sin(2t)+(t^2)/4-1/8+2e^t

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y^{''}+4y=t^2+10e^t

y^{^{''}} + 4 y = t^{2} + 10 e^{t}

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

Solution steps

y^{''} (t) + 4 y = t^{2} + 10 e^{t}

Solve linear ODE:

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

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

What is the solution for y^{''}+4y=t^2+10e^t ?
The solution for y^{''}+4y=t^2+10e^t is y=c_{1}cos(2t)+c_{2}sin(2t)+(t^2)/4-1/8+2e^t