y^'(x^2+2)+4x(y^2+2y+1)=0

Lösung

y^{^{'}} (x^{2} + 2) + 4 x (y^{2} + 2 y + 1) = 0

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

Schritte zur Lösung

y^{'} (x) + 4 x (y^{2} + 2 y + 1) = 0

Solve separable ODE:

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

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

d er i v a t i v e (1 + cos (5 t))^{- 4}

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

\int \frac{- 5 e ^{- 5 x}}{( e ^{- 5 x} + 3 ) ^{3}} d x

s \to 2 - lim (s + ∣ 3 s - 6 ∣)

\frac{d}{d x} (\frac{sin ( x ) + cos ( x )}{sin ( x )})