What is the solution for y^{''}-4y^'+5y=125t^2 ?

The solution for y^{''}-4y^'+5y=125t^2 is y=e^{2t}(c_{1}cos(t)+c_{2}sin(t))+25t^2+40t+22

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y^{''}-4y^'+5y=125t^2

y^{^{''}} - 4 y^{^{'}} + 5 y = 125 t^{2}

y = e^{2 t} (c_{1} cos (t) + c_{2} sin (t)) + 25 t^{2} + 40 t + 22

Solution steps

y^{''} (t) - 4 y^{'} (t) + 5 y = 125 t^{2}

Solve linear ODE:

y = e^{2 t} (c_{1} cos (t) + c_{2} sin (t)) + 25 t^{2} + 40 t + 22

y = e^{2 t} (c_{1} cos (t) + c_{2} sin (t)) + 25 t^{2} + 40 t + 22

\frac{d ^{3} y}{d x ^{3}} + \frac{d y}{d x} = tan (x)

3 x^{^{''}} + 25 x^{^{'}} + 48 x = 5 cos (4 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{- x} (2 x + sin (2 x))

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} - 8 y^{^{'}} + 20 y = 100 x^{2} - 39 x e^{x}

What is the solution for y^{''}-4y^'+5y=125t^2 ?
The solution for y^{''}-4y^'+5y=125t^2 is y=e^{2t}(c_{1}cos(t)+c_{2}sin(t))+25t^2+40t+22