What is the solution for 8y^{''}+2y^'-y=9 ?

The solution for 8y^{''}+2y^'-y=9 is y=c_{1}e^{t/4}+c_{2}e^{-t/2}-9

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8y^{''}+2y^'-y=9

8 y^{^{''}} + 2 y^{^{'}} - y = 9

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

Solution steps

8 y^{''} (t) + 2 y^{'} (t) - y = 9

Solve linear ODE:

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

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

y^{^{''}} - 8 y^{^{'}} + 15 y = 64 e^{t}

d (x) = (3 x - 1) (x - 2)^{- 1}

y^{^{''}} - 2 y + y = x + 2 + cos (2 x)

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 8 y = 3 cos (x)

3 y^{^{''}} + 3 y^{^{'}} - y = x^{2} e^{x}

What is the solution for 8y^{''}+2y^'-y=9 ?
The solution for 8y^{''}+2y^'-y=9 is y=c_{1}e^{t/4}+c_{2}e^{-t/2}-9