(dy)/(dx)=(((1+y)^2))/((1+x)^3)

Lösung

\frac{d y}{d x} = \frac{( ( 1 + y ) ^{2} )}{( 1 + x ) ^{3}}

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

Schritte zur Lösung

\frac{d y}{d x} = \frac{( ( 1 + y ) ^{2} )}{( 1 + x ) ^{3}}

Solve separable ODE:

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

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

\int e^{- 2 x} cos (e^{- x}) d x

\frac{\partial}{\partial x} (ln (x + \frac{y}{2 x}))

\frac{d}{d x} (x^{2} + \frac{2 a ^{3}}{x})

x \to + 3 - lim (\frac{3}{x ^{2} - 9})

x^{^{'}} = x