y^{''}-2y^'+y=(5e^t)/(1+t^2)

解

y^{^{''}} - 2 y^{^{'}} + y = \frac{5 e ^{t}}{1 + t ^{2}}

y = c_{1} e^{t} + c_{2} t e^{t} - \frac{5}{2} e^{t} ln (1 + t^{2}) + 5 e^{t} t arctan (t)

解答ステップ

y^{''} (t) - 2 y^{'} (t) + y = \frac{5 e ^{t}}{1 + t ^{2}}

線形 ODE を解く:

y = c_{1} e^{t} + c_{2} t e^{t} - \frac{5}{2} e^{t} ln (1 + t^{2}) + 5 e^{t} t arctan (t)

y = c_{1} e^{t} + c_{2} t e^{t} - \frac{5}{2} e^{t} ln (1 + t^{2}) + 5 e^{t} t arctan (t)