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

The solution for y^{''}-y^'-2y=-2t+8t^2 is y=c_{1}e^{2t}+c_{2}e^{-t}-4t^2+5t-13/2

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} - 4 t^{2} + 5 t - \frac{13}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - 4 t^{2} + 5 t - \frac{13}{2}

y = c_{1} e^{2 t} + c_{2} e^{- t} - 4 t^{2} + 5 t - \frac{13}{2}

\frac{4}{( s - 1 ) ^{3}}

y^{^{'}} = a - b y^{2}

\frac{\partial}{\partial x} (cos (4 x))

\frac{\partial}{\partial y} (e^{x} + z^{2})

f (x) = \frac{1}{x + 4}

What is the solution for y^{''}-y^'-2y=-2t+8t^2 ?
The solution for y^{''}-y^'-2y=-2t+8t^2 is y=c_{1}e^{2t}+c_{2}e^{-t}-4t^2+5t-13/2