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

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

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} - 4 t^{2} + 6 t - 7

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - 4 t^{2} + 6 t - 7

y = c_{1} e^{2 t} + c_{2} e^{- t} - 4 t^{2} + 6 t - 7

y = x - 2, y = x^{2} - 4, x = 5

\int y cos (z) d z

x y = (- 1 - x^{2}) y^{^{'}}

\int 2 x^{- \frac{1}{2}} d x

7 (4 x^{2} - 7 x + 2)^{- 6}

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