What is the solution for y^{''}+5y^'+7y=0 ?

The solution for y^{''}+5y^'+7y=0 is y=e^{-(5t)/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))

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y^{''}+5y^'+7y=0

y^{^{''}} + 5 y^{^{'}} + 7 y = 0

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

Solution steps

y^{''} (t) + 5 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} + 6 y = 0

x^{^{''}} + 10 x^{^{'}} + 15 x = 0

y^{^{''}} + (3 - α) y^{^{'}} - 2 (α - 1) y = 0

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 5, y^{^{''}} (0) = 13

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

What is the solution for y^{''}+5y^'+7y=0 ?
The solution for y^{''}+5y^'+7y=0 is y=e^{-(5t)/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))