What is the solution for y^{''}-7y^'+6y=x ?

The solution for y^{''}-7y^'+6y=x is y=c_{1}e^{6x}+c_{2}e^x+x/6+7/36

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y^{''}-7y^'+6y=x

y^{^{''}} - 7 y^{^{'}} + 6 y = x

y = c_{1} e^{6 x} + c_{2} e^{x} + \frac{x}{6} + \frac{7}{36}

Solution steps

y^{''} (x) - 7 y^{'} (x) + 6 y = x

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{x} + \frac{x}{6} + \frac{7}{36}

y = c_{1} e^{6 x} + c_{2} e^{x} + \frac{x}{6} + \frac{7}{36}

y^{^{''}} - 2 y^{^{'}} + 3 = 0

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

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

3 y^{^{''}} - 5 y^{^{'}} + 2 y = 6 x e^{- 2 x}

y^{^{''}} + 2 y^{^{'}} + y = sin (x) + 7 cos (2 x)

What is the solution for y^{''}-7y^'+6y=x ?
The solution for y^{''}-7y^'+6y=x is y=c_{1}e^{6x}+c_{2}e^x+x/6+7/36