What is the solution for y^{''}+6y^'+8y=8t+22,y(0)=-7,y^'(0)=20 ?

The solution for y^{''}+6y^'+8y=8t+22,y(0)=-7,y^'(0)=20 is y=(1/2-9)e^{-2t}-1/2 e^{-4t}+t+2

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y^{''}+6y^'+8y=8t+22,y(0)=-7,y^'(0)=20

y^{^{''}} + 6 y^{^{'}} + 8 y = 8 t + 22, y (0) = - 7, y^{^{'}} (0) = 20

y = (\frac{1}{2} - 9) e^{- 2 t} - \frac{1}{2} e^{- 4 t} + t + 2

Solution steps

y^{''} (t) + 6 y^{'} (t) + 8 y = 8 t + 22

Solve linear ODE:

y = (\frac{1}{2} - 9) e^{- 2 t} - \frac{1}{2} e^{- 4 t} + t + 2

y = (\frac{1}{2} - 9) e^{- 2 t} - \frac{1}{2} e^{- 4 t} + t + 2

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 x \cdot e^{2 x}

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

(d^{2} + 5 d + 6) y = (d + 1) 6 t^{2}

y^{^{''}} + y = 8 e^{- 6 x}

(d^{2} - 1) \cdot y = x \cdot e^{x} + cos^{2} (x) + x^{2}

What is the solution for y^{''}+6y^'+8y=8t+22,y(0)=-7,y^'(0)=20 ?
The solution for y^{''}+6y^'+8y=8t+22,y(0)=-7,y^'(0)=20 is y=(1/2-9)e^{-2t}-1/2 e^{-4t}+t+2