What is the solution for y^{''}-10y^'+24=0 ?

The solution for y^{''}-10y^'+24=0 is y=c_{1}e^{10t}+c_{2}+(12t)/5

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y^{''}-10y^'+24=0

y^{^{''}} - 10 y^{^{'}} + 24 = 0

y = c_{1} e^{10 t} + c_{2} + \frac{12 t}{5}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 24 = 0

Solve linear ODE:

y = c_{1} e^{10 t} + c_{2} + \frac{12 t}{5}

y = c_{1} e^{10 t} + c_{2} + \frac{12 t}{5}

y^{^{'''}} + 3 y^{^{''}} - 4 = 0

2 y^{^{''}} + 8 y^{^{'}} - y = 8

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 e^{- 4 t}, y (0) = 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = π x^{2} + 3

y^{^{''}} + y^{^{'}} - 2 y = 9 e^{x} - 2

What is the solution for y^{''}-10y^'+24=0 ?
The solution for y^{''}-10y^'+24=0 is y=c_{1}e^{10t}+c_{2}+(12t)/5