What is the solution for y^{''}-9y^'+8y=x ?

The solution for y^{''}-9y^'+8y=x is y=c_{1}e^{8x}+c_{2}e^x+x/8+9/64

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y^{''}-9y^'+8y=x

y^{^{''}} - 9 y^{^{'}} + 8 y = x

y = c_{1} e^{8 x} + c_{2} e^{x} + \frac{x}{8} + \frac{9}{64}

Solution steps

y^{''} (x) - 9 y^{'} (x) + 8 y = x

Solve linear ODE:

y = c_{1} e^{8 x} + c_{2} e^{x} + \frac{x}{8} + \frac{9}{64}

y = c_{1} e^{8 x} + c_{2} e^{x} + \frac{x}{8} + \frac{9}{64}

y^{^{''}} + 14 y^{^{'}} + 50 y = 371 cos (t)

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

\frac{d ^{3} x}{d t ^{3}} + x = \frac{1}{2} t^{2} e^{t}

y^{^{''}} - 9 y^{^{'}} + 14 y = 3 x^{2} - 5 sin (2 x) + 7 x e^{6 x}

y^{^{''}} + y^{^{'}} + 2 y = x + 3

What is the solution for y^{''}-9y^'+8y=x ?
The solution for y^{''}-9y^'+8y=x is y=c_{1}e^{8x}+c_{2}e^x+x/8+9/64