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

解

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

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

解答ステップ

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

線形 ODE を解く:

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

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