What is the solution for y^{''}-9y^'+20=0 ?

The solution for y^{''}-9y^'+20=0 is y=c_{1}e^{9t}+c_{2}+(20t)/9

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y^{''}-9y^'+20=0

y^{^{''}} - 9 y^{^{'}} + 20 = 0

y = c_{1} e^{9 t} + c_{2} + \frac{20 t}{9}

Solution steps

y^{''} (t) - 9 y^{'} (t) + 20 = 0

Solve linear ODE:

y = c_{1} e^{9 t} + c_{2} + \frac{20 t}{9}

y = c_{1} e^{9 t} + c_{2} + \frac{20 t}{9}

y^{^{''}} - y^{^{'}} = e^{- 3 t}

y^{^{''}} + 10 y^{^{'}} + 24 y = 4 e^{- 4 t} (2 t - 1)

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

2 y^{^{''}} - 6 y^{^{'}} + 3 y = 4 sin (3 x) + 6 x

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 7 + 7

What is the solution for y^{''}-9y^'+20=0 ?
The solution for y^{''}-9y^'+20=0 is y=c_{1}e^{9t}+c_{2}+(20t)/9