What is the solution for x^{''}+7x^'+16x=0 ?

The solution for x^{''}+7x^'+16x=0 is x=e^{-(7t)/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))

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x^{''}+7x^'+16x=0

x^{^{''}} + 7 x^{^{'}} + 16 x = 0

x = e^{- \frac{7 t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

Solution steps

x^{''} (t) + 7 x^{'} (t) + 16 x = 0

Solve linear ODE:

x = e^{- \frac{7 t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

x = e^{- \frac{7 t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

2 y^{^{''}} - 4 y^{^{'}} - 2 y = 0

4 y^{^{''}} - 21 y^{^{'}} + 20 y = 0, y^{^{'}} (0) = - 2.75

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = e^{2 x}

x^{^{''}} + 15 x^{^{'}} + 24 x = 0

3 y^{^{''}} - 12 y^{^{'}} + 87 y = 0

What is the solution for x^{''}+7x^'+16x=0 ?
The solution for x^{''}+7x^'+16x=0 is x=e^{-(7t)/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))