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

解

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

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

解答ステップ

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

線形 ODE を解く:

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

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