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

The solution for y^{''}-2y^'+y=(6e^x)/(1+x^2) is y=c_{1}e^x+c_{2}xe^x-3e^xln(1+x^2)+6e^xxarctan(x)

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y^{''}-2y^'+y=(6e^x)/(1+x^2)

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

y = c_{1} e^{x} + c_{2} x e^{x} - 3 e^{x} ln (1 + x^{2}) + 6 e^{x} x arctan (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = \frac{6 e ^{x}}{1 + x ^{2}}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - 3 e^{x} ln (1 + x^{2}) + 6 e^{x} x arctan (x)

y = c_{1} e^{x} + c_{2} x e^{x} - 3 e^{x} ln (1 + x^{2}) + 6 e^{x} x arctan (x)

What is the solution for y^{''}-2y^'+y=(6e^x)/(1+x^2) ?
The solution for y^{''}-2y^'+y=(6e^x)/(1+x^2) is y=c_{1}e^x+c_{2}xe^x-3e^xln(1+x^2)+6e^xxarctan(x)