What is the solution for y^{''}-2y^'+y=(4e^x)/(x^2+1) ?

The solution for y^{''}-2y^'+y=(4e^x)/(x^2+1) is y=c_{1}e^x+c_{2}xe^x-2e^xln(x^2+1)+4e^xxarctan(x)

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y^{''}-2y^'+y=(4e^x)/(x^2+1)

y^{^{''}} - 2 y^{^{'}} + y = \frac{4 e ^{x}}{x ^{2} + 1}

y = c_{1} e^{x} + c_{2} x e^{x} - 2 e^{x} ln (x^{2} + 1) + 4 e^{x} x arctan (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = \frac{4 e ^{x}}{x ^{2} + 1}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - 2 e^{x} ln (x^{2} + 1) + 4 e^{x} x arctan (x)

y = c_{1} e^{x} + c_{2} x e^{x} - 2 e^{x} ln (x^{2} + 1) + 4 e^{x} x arctan (x)

y^{^{''}} + y^{^{'}} = x + e^{x} + x^{4}

\frac{d ^{2} y}{d x ^{2}} + y = csc (x) cot (x)

y^{^{''}} - 4 y^{^{'}} + 3 = e^{- 5 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 13 y = 5 cos (2 x)

\frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{3} e^{2 t}

What is the solution for y^{''}-2y^'+y=(4e^x)/(x^2+1) ?
The solution for y^{''}-2y^'+y=(4e^x)/(x^2+1) is y=c_{1}e^x+c_{2}xe^x-2e^xln(x^2+1)+4e^xxarctan(x)