What is the solution for 8y^{''}+6y^'-y=11 ?

The solution for 8y^{''}+6y^'-y=11 is y=c_{1}e^{((-3+sqrt(17))t)/8}+c_{2}e^{-((3+sqrt(17))t)/8}-11

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8y^{''}+6y^'-y=11

8 y^{^{''}} + 6 y^{^{'}} - y = 11

y = c_{1} e^{\frac{( - 3 + 17 ) t}{8}} + c_{2} e^{- \frac{( 3 + 17 ) t}{8}} - 11

Solution steps

8 y^{''} (t) + 6 y^{'} (t) - y = 11

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 17 ) t}{8}} + c_{2} e^{- \frac{( 3 + 17 ) t}{8}} - 11

y = c_{1} e^{\frac{( - 3 + 17 ) t}{8}} + c_{2} e^{- \frac{( 3 + 17 ) t}{8}} - 11

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What is the solution for 8y^{''}+6y^'-y=11 ?
The solution for 8y^{''}+6y^'-y=11 is y=c_{1}e^{((-3+sqrt(17))t)/8}+c_{2}e^{-((3+sqrt(17))t)/8}-11