What is the solution for y^{''}-2y^'+1=e^x ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 9 y = 5 csc^{2} (3 x)

y^{^{''}} - y - 12 y = e^{4 x}

\frac{d ^{2} y}{d x ^{2}} - y = \frac{1}{x}

6 y^{^{''}} - y^{^{'}} - 2 y = 3 sin (10) x

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