What is the solution for -5y^{''}-2y^'+24=0 ?

The solution for -5y^{''}-2y^'+24=0 is y=c_{1}e^{-(2t)/5}+c_{2}+12t

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-5y^{''}-2y^'+24=0

- 5 y^{^{''}} - 2 y^{^{'}} + 24 = 0

y = c_{1} e^{- \frac{2 t}{5}} + c_{2} + 12 t

Solution steps

- 5 y^{''} (t) - 2 y^{'} (t) + 24 = 0

Solve linear ODE:

y = c_{1} e^{- \frac{2 t}{5}} + c_{2} + 12 t

y = c_{1} e^{- \frac{2 t}{5}} + c_{2} + 12 t

y^{^{''}} - 3 y^{^{'}} = cos (5 x), y (0) = 0, y^{^{'}} (0) = 0

\frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{3} e^{3 t}

2 y^{^{'''}} - 2 y^{^{'}} = - 3 t^{2} - 4 t

y^{^{''}} + 2 y^{^{'}} + 392 y = cos (2 t)

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 3 t^{2} - 18

What is the solution for -5y^{''}-2y^'+24=0 ?
The solution for -5y^{''}-2y^'+24=0 is y=c_{1}e^{-(2t)/5}+c_{2}+12t