(dy)/(dx)=-((2xy+y^2+1))/(x^2+2xy)

Lösung

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

y = \frac{- x ^{2} + x ^{4} - 4 x ( x - c _{1} )}{2 x}, y = \frac{- x ^{2} - x ^{4} - 4 x ( x - c _{1} )}{2 x}

Schritte zur Lösung

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

Solve exact ODE:

y = \frac{- x ^{2} + x ^{4} - 4 x ( x - c _{1} )}{2 x}, y = \frac{- x ^{2} - x ^{4} - 4 x ( x - c _{1} )}{2 x}

y = \frac{- x ^{2} + x ^{4} - 4 x ( x - c _{1} )}{2 x}, y = \frac{- x ^{2} - x ^{4} - 4 x ( x - c _{1} )}{2 x}

d er i v a t i v e \frac{1 x ^{3}}{2 x ^{4}}

\frac{d}{d x} (- 2 e^{- 3 x})

\frac{\partial}{\partial x} (3 x y^{2} - 7)

\int \frac{2}{3} sec^{2} (π x) d x

\int \frac{cos ( x ) - 1}{x} d x