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

解

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

y = c_{1} e^{x} + c_{2} + e^{x} \cdot \int \frac{e ^{2 - x} x}{2 + e ^{x}} d x + \int - \frac{e ^{2} x}{2 + e ^{x}} d x

解答ステップ

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

線形 ODE を解く:

y = c_{1} e^{x} + c_{2} + e^{x} \cdot \int \frac{e ^{2 - x} x}{2 + e ^{x}} d x + \int - \frac{e ^{2} x}{2 + e ^{x}} d x

y = c_{1} e^{x} + c_{2} + e^{x} \cdot \int \frac{e ^{2 - x} x}{2 + e ^{x}} d x + \int - \frac{e ^{2} x}{2 + e ^{x}} d x