What is the solution for 3y^{''}+8y^'+5y=0 ?

The solution for 3y^{''}+8y^'+5y=0 is y=c_{1}e^{-t}+c_{2}e^{-(5t)/3}

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3y^{''}+8y^'+5y=0

3 y^{^{''}} + 8 y^{^{'}} + 5 y = 0

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

Solution steps

3 y^{''} (t) + 8 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

6 x^{^{''}} + 8 x^{^{'}} + 35 x = 0

3 y^{^{''}} + 2 y^{^{'}} + 2 y = 0

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

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 0

\frac{1}{8} x^{^{''}} + x^{^{'}} + 2 x = 0, x (0) = - 1, x^{^{'}} (0) = 14

What is the solution for 3y^{''}+8y^'+5y=0 ?
The solution for 3y^{''}+8y^'+5y=0 is y=c_{1}e^{-t}+c_{2}e^{-(5t)/3}