What is the solution for y^{''}+4y=5t^2e^t ?

The solution for y^{''}+4y=5t^2e^t is y=c_{1}cos(2t)+c_{2}sin(2t)+e^tt^2-(4e^tt)/5-2/25 e^t

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y^{''}+4y=5t^2e^t

y^{^{''}} + 4 y = 5 t^{2} e^{t}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + e^{t} t^{2} - \frac{4 e ^{t} t}{5} - \frac{2}{25} e^{t}

Solution steps

y^{''} (t) + 4 y = 5 t^{2} e^{t}

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + e^{t} t^{2} - \frac{4 e ^{t} t}{5} - \frac{2}{25} e^{t}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + e^{t} t^{2} - \frac{4 e ^{t} t}{5} - \frac{2}{25} e^{t}

y^{^{''}} + y = cos (2 t)

D^{2} (D - 2)^{3} y = 48 e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{5 x}

y^{^{''}} - 3 y^{^{'}} - 40 y = 210 e^{2 t}

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = t^{2} - 2 t

What is the solution for y^{''}+4y=5t^2e^t ?
The solution for y^{''}+4y=5t^2e^t is y=c_{1}cos(2t)+c_{2}sin(2t)+e^tt^2-(4e^tt)/5-2/25 e^t