What is the solution for 100y^{''}+180y^'+145y=0 ?

The solution for 100y^{''}+180y^'+145y=0 is y=e^{-(9t}{10}}(c_{1}cos(\frac{4t)/5)+c_{2}sin((4t)/5))

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100y^{''}+180y^'+145y=0

100 y^{^{''}} + 180 y^{^{'}} + 145 y = 0

y = e^{- \frac{9 t}{10}} (c_{1} cos (\frac{4 t}{5}) + c_{2} sin (\frac{4 t}{5}))

Solution steps

100 y^{''} (t) + 180 y^{'} (t) + 145 y = 0

Solve linear ODE:

y = e^{- \frac{9 t}{10}} (c_{1} cos (\frac{4 t}{5}) + c_{2} sin (\frac{4 t}{5}))

y = e^{- \frac{9 t}{10}} (c_{1} cos (\frac{4 t}{5}) + c_{2} sin (\frac{4 t}{5}))

y^{^{''}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = - 3

(D^{2} + 64)^{2} y = 0

4 \cdot \frac{d ^{2} y}{d x ^{2}} + 12 \cdot \frac{d y}{d x} + 9 y = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = \frac{2}{3}

0.2 q^{^{''}} + 4 q^{^{'}} + 100 q = 0

What is the solution for 100y^{''}+180y^'+145y=0 ?
The solution for 100y^{''}+180y^'+145y=0 is y=e^{-(9t}{10}}(c_{1}cos(\frac{4t)/5)+c_{2}sin((4t)/5))