What is the solution for y^{''}+4y=t^2+3e^t ?

The solution for y^{''}+4y=t^2+3e^t is y=c_{1}cos(2t)+c_{2}sin(2t)+(t^2)/4-1/8+3/5 e^t

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y^{''}+4y=t^2+3e^t

y^{^{''}} + 4 y = t^{2} + 3 e^{t}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{3}{5} e^{t}

Solution steps

y^{''} (t) + 4 y = t^{2} + 3 e^{t}

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{3}{5} e^{t}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{3}{5} e^{t}

\int_{0}^{π} cos^{6} (x) d x

y^{^{'}} = 0.285 (1 - y)

100 y^{^{''}} = 0

x \to 3 lim (5 x^{2})

\frac{d}{d x} (x^{- \frac{2}{3}} (x + 5))

What is the solution for y^{''}+4y=t^2+3e^t ?
The solution for y^{''}+4y=t^2+3e^t is y=c_{1}cos(2t)+c_{2}sin(2t)+(t^2)/4-1/8+3/5 e^t