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

The solution for y^{''}-7y^'+6y=e^x is y=c_{1}e^{6x}+c_{2}e^x-(e^xx)/5

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

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

y = c_{1} e^{6 x} + c_{2} e^{x} - \frac{e ^{x} x}{5}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{x} - \frac{e ^{x} x}{5}

y = c_{1} e^{6 x} + c_{2} e^{x} - \frac{e ^{x} x}{5}

6 y^{^{''}} - 3 y = x^{2} - x + 7

y^{^{''}} + 4 y = x^{2} + 5 cos (2 x)

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 4 x + 6

y^{^{''}} + y^{^{'}} = 3 x^{2}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - y = sin^{2} (x)

What is the solution for y^{''}-7y^'+6y=e^x ?
The solution for y^{''}-7y^'+6y=e^x is y=c_{1}e^{6x}+c_{2}e^x-(e^xx)/5