What is the solution for 9y^{''}+4y^'=-5y ?

The solution for 9y^{''}+4y^'=-5y is y=e^{-(2t)/9}(c_{1}cos((sqrt(41)t)/9)+c_{2}sin((sqrt(41)t)/9))

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9y^{''}+4y^'=-5y

9 y^{^{''}} + 4 y^{^{'}} = - 5 y

y = e^{- \frac{2 t}{9}} (c_{1} cos (\frac{41 t}{9}) + c_{2} sin (\frac{41 t}{9}))

Solution steps

9 y^{''} (t) + 4 y^{'} (t) = - 5 y

Solve linear ODE:

y = e^{- \frac{2 t}{9}} (c_{1} cos (\frac{41 t}{9}) + c_{2} sin (\frac{41 t}{9}))

y = e^{- \frac{2 t}{9}} (c_{1} cos (\frac{41 t}{9}) + c_{2} sin (\frac{41 t}{9}))

y^{^{''}} - \frac{3}{2} y^{^{'}} + y = 0

77 y^{^{''''}} + 39 y^{^{''}} + 4 y = 0

\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} + 10 y = 0

3 \frac{d ^{2} y}{d x ^{2}} + 7 \frac{d y}{d x} - 11 y = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (1) = e^{3}

What is the solution for 9y^{''}+4y^'=-5y ?
The solution for 9y^{''}+4y^'=-5y is y=e^{-(2t)/9}(c_{1}cos((sqrt(41)t)/9)+c_{2}sin((sqrt(41)t)/9))