What is the solution for y^{''}+7y^'+20y=0 ?

The solution for y^{''}+7y^'+20y=0 is y=e^{-(7t)/2}(c_{1}cos((sqrt(31)t)/2)+c_{2}sin((sqrt(31)t)/2))

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y^{''}+7y^'+20y=0

y^{^{''}} + 7 y^{^{'}} + 20 y = 0

y = e^{- \frac{7 t}{2}} (c_{1} cos (\frac{31 t}{2}) + c_{2} sin (\frac{31 t}{2}))

Solution steps

y^{''} (t) + 7 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{- \frac{7 t}{2}} (c_{1} cos (\frac{31 t}{2}) + c_{2} sin (\frac{31 t}{2}))

y = e^{- \frac{7 t}{2}} (c_{1} cos (\frac{31 t}{2}) + c_{2} sin (\frac{31 t}{2}))

2 y^{^{''}} + 14 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} - 8 y = 0

y^{^{''}} + 5 y^{^{'}} = 0, y (0) = 9, y^{^{'}} (0) = - 20

\frac{d ^{2} x}{d t ^{2}} + 6 \cdot \frac{d x}{d t} + 3 \cdot x = 0, x (0) = 0

4 (\frac{d ^{2} y}{d x ^{2}}) + 4 (\frac{d y}{d x}) + y = 0

What is the solution for y^{''}+7y^'+20y=0 ?
The solution for y^{''}+7y^'+20y=0 is y=e^{-(7t)/2}(c_{1}cos((sqrt(31)t)/2)+c_{2}sin((sqrt(31)t)/2))