What is the solution for y^{''}-10y^'+9y=e^{2x} ?

The solution for y^{''}-10y^'+9y=e^{2x} is y=c_{1}e^{9x}+c_{2}e^x-1/7 e^{2x}

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y^{''}-10y^'+9y=e^{2x}

y^{^{''}} - 10 y^{^{'}} + 9 y = e^{2 x}

y = c_{1} e^{9 x} + c_{2} e^{x} - \frac{1}{7} e^{2 x}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 9 y = e^{2 x}

Solve linear ODE:

y = c_{1} e^{9 x} + c_{2} e^{x} - \frac{1}{7} e^{2 x}

y = c_{1} e^{9 x} + c_{2} e^{x} - \frac{1}{7} e^{2 x}

y^{^{''}} + y^{^{'}} - 6 y = e^{2 t} + t

y^{^{'''}} - 3 y^{^{''}} - y^{^{'}} + 3 y = 1 + e^{x}

y^{^{''}} - y = 2 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 25 y = 6 sin (5 x)

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 10 x^{2} - 4 x - 15

What is the solution for y^{''}-10y^'+9y=e^{2x} ?
The solution for y^{''}-10y^'+9y=e^{2x} is y=c_{1}e^{9x}+c_{2}e^x-1/7 e^{2x}