What is the solution for 3y^{''}+5y^'+2y=0 ?

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

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3y^{''}+5y^'+2y=0

3 y^{^{''}} + 5 y^{^{'}} + 2 y = 0

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

Solution steps

3 y^{''} (t) + 5 y^{'} (t) + 2 y = 0

Solve linear ODE:

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

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

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