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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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k = 1 \sum \infty \frac{1}{k ^{4}}

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