What is the solution for y^{''}+16y^'+3y=0 ?

The solution for y^{''}+16y^'+3y=0 is y=c_{1}e^{(-8+sqrt(61))t}+c_{2}e^{(-8-sqrt(61))t}

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y^{''}+16y^'+3y=0

y^{^{''}} + 16 y^{^{'}} + 3 y = 0

y = c_{1} e^{(- 8 + 61) t} + c_{2} e^{(- 8 - 61) t}

Solution steps

y^{''} (t) + 16 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = c_{1} e^{(- 8 + 61) t} + c_{2} e^{(- 8 - 61) t}

y = c_{1} e^{(- 8 + 61) t} + c_{2} e^{(- 8 - 61) t}

y^{^{''}} - 25 y^{^{'}} + 5 y = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (1) = 0, y^{^{'}} (1) = - 1

4 y^{^{''}} + 75 y = 0

x^{^{''}} + 5 x^{^{'}} + 9 x = 0

25 y^{^{''}} - 30 y^{^{'}} + 9 y = 0, y^{1} (0) = e^{\frac{3 x}{5}}

What is the solution for y^{''}+16y^'+3y=0 ?
The solution for y^{''}+16y^'+3y=0 is y=c_{1}e^{(-8+sqrt(61))t}+c_{2}e^{(-8-sqrt(61))t}