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

解

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)

解答ステップ

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

線形 ODE を解く:

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)