What is the solution for y^{''}-10y^'+9y=5t ?

The solution for y^{''}-10y^'+9y=5t is y=c_{1}e^{9t}+c_{2}e^t+(5t)/9+50/81

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y^{''}-10y^'+9y=5t

y^{^{''}} - 10 y^{^{'}} + 9 y = 5 t

y = c_{1} e^{9 t} + c_{2} e^{t} + \frac{5 t}{9} + \frac{50}{81}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 9 y = 5 t

Solve linear ODE:

y = c_{1} e^{9 t} + c_{2} e^{t} + \frac{5 t}{9} + \frac{50}{81}

y = c_{1} e^{9 t} + c_{2} e^{t} + \frac{5 t}{9} + \frac{50}{81}

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y^{^{''}} - 8 y^{^{'}} + 20 y = 100 x^{2} - 65 x e^{x}

y^{^{''}} - 6 y^{^{'}} + 9 y = cos (x)

What is the solution for y^{''}-10y^'+9y=5t ?
The solution for y^{''}-10y^'+9y=5t is y=c_{1}e^{9t}+c_{2}e^t+(5t)/9+50/81