What is the solution for 3y^{''}-8y^'-3y=5 ?

The solution for 3y^{''}-8y^'-3y=5 is y=c_{1}e^{3t}+c_{2}e^{-t/3}-5/3

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3y^{''}-8y^'-3y=5

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 5

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

Solution steps

3 y^{''} (t) - 8 y^{'} (t) - 3 y = 5

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} - 5 = 2 t^{2} - 3 t + 6

y^{^{''}} + 25 y = 6 x, y (0) = 1, y^{^{'}} (0) = 5

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = e^{t} (t^{2} + sin (t))

y^{^{''}} + 16 y = 3, y (0) = 0, y^{^{'}} (0) = 0

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

What is the solution for 3y^{''}-8y^'-3y=5 ?
The solution for 3y^{''}-8y^'-3y=5 is y=c_{1}e^{3t}+c_{2}e^{-t/3}-5/3