What is the solution for y^{''}-9y^'+8y=x,y(0)=e^x ?

The solution for y^{''}-9y^'+8y=x,y(0)=e^x is y=c_{1}e^{8x}+c_{2}e^x+x/8+9/64

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y^{''}-9y^'+8y=x,y(0)=e^x

y^{^{''}} - 9 y^{^{'}} + 8 y = x, y (0) = e^{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^{^{''}} + 2 y^{^{'}} + 4 y = t + cos (2 t)

y^{^{''}} - 4 y^{^{'}} + 4 y = 12 t e^{2 t} (t + 1)

y^{^{''}} - 5 y^{^{'}} + 6 y = 4 e^{4 x}, y (0) = 0, y^{^{'}} (0) = 1

(d^{2} + 4) y = cos (2 x)

y^{^{''}} - 14 y^{^{'}} + 49 y = 21 x + 2

What is the solution for y^{''}-9y^'+8y=x,y(0)=e^x ?
The solution for y^{''}-9y^'+8y=x,y(0)=e^x is y=c_{1}e^{8x}+c_{2}e^x+x/8+9/64