2dx-e^{y-x}dy=0

解答

2 d x - e^{y - x} d y = 0

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

求解步骤

2 d x - e^{y - x} d y = 0

求解可分离方程:

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

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

y^{2} = x (x + 2) \frac{d y}{d x}

\frac{d S}{d t} = 0.06 - \frac{3 S}{1420}, t = 10

\frac{d x}{d t} = \frac{5}{x e ^{t + 4 x}}

y^{^{'}} = - 5 x^{4} y^{2}

\frac{d y}{d t} = k y + an