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

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

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3y^{''}-8y^'-3y=0

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 0

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

Solution steps

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

Solve linear ODE:

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

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

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