What is the solution for y^{''}+4y=8t^2-20t+8 ?

The solution for y^{''}+4y=8t^2-20t+8 is y=c_{1}cos(2t)+c_{2}sin(2t)+2t^2-5t+1

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y^{''}+4y=8t^2-20t+8

y^{^{''}} + 4 y = 8 t^{2} - 20 t + 8

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 2 t^{2} - 5 t + 1

Solution steps

y^{''} (t) + 4 y = 8 t^{2} - 20 t + 8

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 2 t^{2} - 5 t + 1

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 2 t^{2} - 5 t + 1

\frac{d}{d x} (4 x^{3} + \frac{1}{x ^{2}} - x + 3)

\frac{d}{d x} (\frac{5 x}{( x ^{3} - 6 ) ^{2}})

\frac{\partial}{\partial x} (\frac{3}{2 x ^{2}})

y^{2} y^{^{'}} = x^{2}

\int_{0}^{\frac{π}{3}} x sin (12 x) d x

What is the solution for y^{''}+4y=8t^2-20t+8 ?
The solution for y^{''}+4y=8t^2-20t+8 is y=c_{1}cos(2t)+c_{2}sin(2t)+2t^2-5t+1