What is the solution for y^{''}-7y^'+6y=x,y^1(0)=ex ?

The solution for y^{''}-7y^'+6y=x,y^1(0)=ex is y=c_{1}e^{6x}+c_{2}e^x+x/6+7/36

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y^{''}-7y^'+6y=x,y^1(0)=ex

y^{^{''}} - 7 y^{^{'}} + 6 y = x, y^{1} (0) = e 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^{^{''}} - 8 y^{^{'}} - 9 y = 10, y (0) = 0, y^{^{'}} (0) = 4

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

y^{^{''}} + 2 y = e^{- 2 x} (x^{2} + 1)

y^{^{''}} + 25 y = sin (7 x)

y^{^{''}} + 2 y^{^{'}} + y = g (t - 1), y (0) = 2, y^{^{'}} (0) = 3

What is the solution for y^{''}-7y^'+6y=x,y^1(0)=ex ?
The solution for y^{''}-7y^'+6y=x,y^1(0)=ex is y=c_{1}e^{6x}+c_{2}e^x+x/6+7/36