What is the solution for 3y^{''}-2y^'-8y=0 ?

The solution for 3y^{''}-2y^'-8y=0 is y=c_{1}e^{2t}+c_{2}e^{-(4t)/3}

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3y^{''}-2y^'-8y=0

3 y^{^{''}} - 2 y^{^{'}} - 8 y = 0

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

Solution steps

3 y^{''} (t) - 2 y^{'} (t) - 8 y = 0

Solve linear ODE:

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

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

3 y^{^{''}} + 7 y^{^{'}} + 2 y = 3 x^{2} + 3 x + 1, y (0) = 0, y^{^{'}} (0) = 0

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What is the solution for 3y^{''}-2y^'-8y=0 ?
The solution for 3y^{''}-2y^'-8y=0 is y=c_{1}e^{2t}+c_{2}e^{-(4t)/3}