What is the solution for 9y^{''}+4y^'-y=22 ?

The solution for 9y^{''}+4y^'-y=22 is y=c_{1}e^{((-2+sqrt(13))t)/9}+c_{2}e^{-((2+sqrt(13))t)/9}-22

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9y^{''}+4y^'-y=22

9 y^{^{''}} + 4 y^{^{'}} - y = 22

y = c_{1} e^{\frac{( - 2 + 13 ) t}{9}} + c_{2} e^{- \frac{( 2 + 13 ) t}{9}} - 22

Solution steps

9 y^{''} (t) + 4 y^{'} (t) - y = 22

Solve linear ODE:

y = c_{1} e^{\frac{( - 2 + 13 ) t}{9}} + c_{2} e^{- \frac{( 2 + 13 ) t}{9}} - 22

y = c_{1} e^{\frac{( - 2 + 13 ) t}{9}} + c_{2} e^{- \frac{( 2 + 13 ) t}{9}} - 22

y^{^{''}} - y^{^{'}} = 4 e_{- x}

x^{^{''}} + 4 x = 12 \cdot cos (t)

(D^{3} - 2 D^{2} - 5 D + 6) y = e^{4 x}

10 y^{^{''}} + 20 y^{^{'}} + 6260 y = 100

y^{^{''}} + 2 y^{^{'}} - 8 y = 24 e^{2 t}

What is the solution for 9y^{''}+4y^'-y=22 ?
The solution for 9y^{''}+4y^'-y=22 is y=c_{1}e^{((-2+sqrt(13))t)/9}+c_{2}e^{-((2+sqrt(13))t)/9}-22