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

The solution for 2y^{''}+8y^'-y=8 is y=c_{1}e^{((-4+3sqrt(2))t)/2}+c_{2}e^{-((4+3sqrt(2))t)/2}-8

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

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

y = c_{1} e^{\frac{( - 4 + 3 2 ) t}{2}} + c_{2} e^{- \frac{( 4 + 3 2 ) t}{2}} - 8

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 4 + 3 2 ) t}{2}} + c_{2} e^{- \frac{( 4 + 3 2 ) t}{2}} - 8

y = c_{1} e^{\frac{( - 4 + 3 2 ) t}{2}} + c_{2} e^{- \frac{( 4 + 3 2 ) t}{2}} - 8

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 e^{- 4 t}, y (0) = 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = π x^{2} + 3

y^{^{''}} + y^{^{'}} - 2 y = 9 e^{x} - 2

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 2 x = 2 e^{- 3 t}

y^{^{''}} - y^{^{'}} - 6 y = 19 sin (3 x)

What is the solution for 2y^{''}+8y^'-y=8 ?
The solution for 2y^{''}+8y^'-y=8 is y=c_{1}e^{((-4+3sqrt(2))t)/2}+c_{2}e^{-((4+3sqrt(2))t)/2}-8