What is the solution for y^{''}-2y^'-8y=-8t-18,y(0)=1,y^'(0)=28 ?

The solution for y^{''}-2y^'-8y=-8t-18,y(0)=1,y^'(0)=28 is y=(31/6-1)e^{4t}-31/6 e^{-2t}+t+2

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y^{''}-2y^'-8y=-8t-18,y(0)=1,y^'(0)=28

y^{^{''}} - 2 y^{^{'}} - 8 y = - 8 t - 18, y (0) = 1, y^{^{'}} (0) = 28

y = (\frac{31}{6} - 1) e^{4 t} - \frac{31}{6} e^{- 2 t} + t + 2

Solution steps

y^{''} (t) - 2 y^{'} (t) - 8 y = - 8 t - 18

Solve linear ODE:

y = (\frac{31}{6} - 1) e^{4 t} - \frac{31}{6} e^{- 2 t} + t + 2

y = (\frac{31}{6} - 1) e^{4 t} - \frac{31}{6} e^{- 2 t} + t + 2

y^{^{''}} - 2 y^{^{'}} + y = e^{x} sec^{2} (x) tan (x)

y^{^{''}} - y = 2 t^{2} - t + 3 e^{- 3 t}

q^{^{''}} + 0.1 q^{^{'}} + q = v sin (0.5 t), q (0) = 0

y^{^{''}} - 2 y^{^{'}} = x + e^{x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 30 e^{2 t}

What is the solution for y^{''}-2y^'-8y=-8t-18,y(0)=1,y^'(0)=28 ?
The solution for y^{''}-2y^'-8y=-8t-18,y(0)=1,y^'(0)=28 is y=(31/6-1)e^{4t}-31/6 e^{-2t}+t+2