What is the solution for y^{''}-4y^'+5y=125t^2,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-4y^'+5y=125t^2,y(0)=0,y^'(0)=0 is y=e^{2t}(-22cos(t)+4sin(t))+25t^2+40t+22

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y^{''}-4y^'+5y=125t^2,y(0)=0,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} + 5 y = 125 t^{2}, y (0) = 0, y^{^{'}} (0) = 0

y = e^{2 t} (- 22 cos (t) + 4 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} (- 22 cos (t) + 4 sin (t)) + 25 t^{2} + 40 t + 22

y = e^{2 t} (- 22 cos (t) + 4 sin (t)) + 25 t^{2} + 40 t + 22

y^{^{''}} + 18 y^{^{'}} + 45 y = e^{2 x} + x

y^{^{''}} + 16 y = 8 sec (4 x)

y^{^{''}} + 3 y = 3 t^{3}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y = 3 sin (2 t) + 2 t cos (2 t)

y^{^{''}} - 16 y = sinh (4 x)

What is the solution for y^{''}-4y^'+5y=125t^2,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y^'+5y=125t^2,y(0)=0,y^'(0)=0 is y=e^{2t}(-22cos(t)+4sin(t))+25t^2+40t+22