xy^3dx-(y+1)e^{-x}dy=0

解答

x y^{3} d x - (y + 1) e^{- x} d y = 0

y = \frac{- 1 + - 2 ( e ^{x} x - e ^{x} + c _{1} ) + 1}{2 ( e ^{x} x - e ^{x} + c _{1} )}, y = - \frac{- 2 ( e ^{x} x - e ^{x} + c _{1} ) + 1 + 1}{2 ( e ^{x} x - e ^{x} + c _{1} )}

求解步骤

x y^{3} d x - (y + 1) e^{- x} d y = 0

求解可分离方程:

y = \frac{- 1 + - 2 ( e ^{x} x - e ^{x} + c _{1} ) + 1}{2 ( e ^{x} x - e ^{x} + c _{1} )}, y = - \frac{- 2 ( e ^{x} x - e ^{x} + c _{1} ) + 1 + 1}{2 ( e ^{x} x - e ^{x} + c _{1} )}

y = \frac{- 1 + - 2 ( e ^{x} x - e ^{x} + c _{1} ) + 1}{2 ( e ^{x} x - e ^{x} + c _{1} )}, y = - \frac{- 2 ( e ^{x} x - e ^{x} + c _{1} ) + 1 + 1}{2 ( e ^{x} x - e ^{x} + c _{1} )}

(1 + x^{2}) y^{^{'}} = - 4 x y

1 = \frac{cos ( y )}{x ^{2} sin ( y )} \frac{d y}{d x}, sin (y) > 0

x^{2} y^{^{'}} = y

\frac{d x}{d t} = 18 - \frac{72 x}{100}

y^{^{'}} + \frac{( 1 0 ^{3} )}{4} y - 2 \cdot 1 0^{3} = 0