What is the solution for y^{''}+6y^'+8y=16t^2 ?

The solution for y^{''}+6y^'+8y=16t^2 is y=c_{1}e^{-2t}+c_{2}e^{-4t}+2t^2-3t+7/4

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y^{''}+6y^'+8y=16t^2

y^{^{''}} + 6 y^{^{'}} + 8 y = 16 t^{2}

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + 2 t^{2} - 3 t + \frac{7}{4}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 8 y = 16 t^{2}

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + 2 t^{2} - 3 t + \frac{7}{4}

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + 2 t^{2} - 3 t + \frac{7}{4}

y^{^{''}} + 4 y = u_{π} (t)

y^{^{''}} - y = - 6 sin (x)

y^{^{''}} + 4 y^{^{'}} + 6 y = 1, + e^{- t} y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} = - 1

y^{^{''}} + y^{^{'}} = - 3

What is the solution for y^{''}+6y^'+8y=16t^2 ?
The solution for y^{''}+6y^'+8y=16t^2 is y=c_{1}e^{-2t}+c_{2}e^{-4t}+2t^2-3t+7/4