x+yxdx-dy=0

解答

x + y x d x - d y = 0

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

求解步骤

x + y x d x - d y = 0

求解可分离方程:

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

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

\frac{d y}{d x} = \frac{y ^{\frac{1}{2}} ln ( x )}{x}

\frac{d y}{d t} = k (y - 2)

y^{^{'}} = \frac{( x y + 3 y )}{x - 3}

\frac{d}{d x} (ln (y)) = \frac{1}{3} ln (x^{2} - 9)

y^{^{'}} = sin (x) sin (y)