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

The solution for y^{''}-2y^'+y=(e^x)/(1+x^2) is y=c_{1}e^x+c_{2}xe^x-1/2 e^xln(1+x^2)+e^xxarctan(x)

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

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

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{1}{2} e^{x} ln (1 + x^{2}) + e^{x} x arctan (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{1}{2} e^{x} ln (1 + x^{2}) + e^{x} x arctan (x)

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{1}{2} e^{x} ln (1 + x^{2}) + e^{x} x arctan (x)

\int_{0}^{2} x + 2 d x

\int (2 x + 3) sin (x^{2} + 3 x + 1) d x

\frac{\partial}{\partial z} (x e^{6 y} sin (5 z))

\int e^{x^{\frac{1}{2}}} d x

y = (4 x + 3)^{5}

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