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

h \to 0 lim (cos (\frac{π}{2}))

\int_{0}^{4} e^{x} d x

\frac{d}{d x} (- 25 (\frac{x}{x ^{2} + 1}))

\int cos (9 x) d x

\int cot (\frac{1}{8} x) d x

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