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

解答

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

求解步骤

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

求解线性方程:

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

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

so l v e f or d, d (t) = 4000 - 3000 e^{- 0.06 t}

(d^{2} - d - 2) \cdot y = 2 \cdot lo g_{10} (x) + \frac{1}{x} + \frac{1}{( x ^{2} )}

(d - 5)^{2} y = e^{5 x}

(d^{2} - 5 d + 6) \cdot y = x