What is the solution for y^{''}-9y^'+20y=x+1 ?

The solution for y^{''}-9y^'+20y=x+1 is y=c_{1}e^{5x}+c_{2}e^{4x}+x/(20)+29/400

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y^{''}-9y^'+20y=x+1

y^{^{''}} - 9 y^{^{'}} + 20 y = x + 1

y = c_{1} e^{5 x} + c_{2} e^{4 x} + \frac{x}{20} + \frac{29}{400}

Solution steps

y^{''} (x) - 9 y^{'} (x) + 20 y = x + 1

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{4 x} + \frac{x}{20} + \frac{29}{400}

y = c_{1} e^{5 x} + c_{2} e^{4 x} + \frac{x}{20} + \frac{29}{400}

y^{^{''}} - 5 y^{^{'}} + 6 y = cos (2 x)

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

y^{^{''}} + 16 y = 3 sec (4 x)

y^{^{''}} + 4 y = 6 cos (x)

y^{^{''}} + 9 y^{^{'}} = e^{t}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-9y^'+20y=x+1 ?
The solution for y^{''}-9y^'+20y=x+1 is y=c_{1}e^{5x}+c_{2}e^{4x}+x/(20)+29/400