What is the solution for 3y^{''}+17y^'+10y=0 ?

The solution for 3y^{''}+17y^'+10y=0 is y=c_{1}e^{-(2t)/3}+c_{2}e^{-5t}

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3y^{''}+17y^'+10y=0

3 y^{^{''}} + 17 y^{^{'}} + 10 y = 0

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

Solution steps

3 y^{''} (t) + 17 y^{'} (t) + 10 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (m x^{m - 2} (m - 1))

y = e^{x}, y = e^{- 3 x}, x = ln (6)

\frac{d y}{d t} = t^{2} + 13 t^{2} y

\int (e^{x} + x cos (y)) d y

\int_{3}^{6} \frac{x}{x ^{2} + 4 x + 13} d x

What is the solution for 3y^{''}+17y^'+10y=0 ?
The solution for 3y^{''}+17y^'+10y=0 is y=c_{1}e^{-(2t)/3}+c_{2}e^{-5t}