What is the solution for 3y^{''}+13y^'+4y=0 ?

The solution for 3y^{''}+13y^'+4y=0 is y=c_{1}e^{-t/3}+c_{2}e^{-4t}

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3y^{''}+13y^'+4y=0

3 y^{^{''}} + 13 y^{^{'}} + 4 y = 0

y = c_{1} e^{- \frac{t}{3}} + c_{2} e^{- 4 t}

Solution steps

3 y^{''} (t) + 13 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{t}{3}} + c_{2} e^{- 4 t}

y = c_{1} e^{- \frac{t}{3}} + c_{2} e^{- 4 t}

\int \frac{e ^{2 x}}{49 + e ^{4 x}} d x

(2 y - 9) d x + x d y = 0

\int \frac{1}{x ^{\frac{1}{3}} + x ^{\frac{1}{4}}} d x

75.6 x + \frac{36590.4}{x ^{2}}

What is the solution for 3y^{''}+13y^'+4y=0 ?
The solution for 3y^{''}+13y^'+4y=0 is y=c_{1}e^{-t/3}+c_{2}e^{-4t}