(x+y)^2dx+(2xy+x^2-4)dy=0

Lösung

(x + y)^{2} d x + (2 x y + x^{2} - 4) d y = 0

y = \frac{12 - 3 x ^{2} + 144 - 72 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}, y = \frac{12 - 3 x ^{2} - 144 - 72 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}

Schritte zur Lösung

(x + y)^{2} d x + (2 x y + x^{2} - 4) d y = 0

Solve exact ODE:

y = \frac{12 - 3 x ^{2} + 144 - 72 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}, y = \frac{12 - 3 x ^{2} - 144 - 72 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}

y = \frac{12 - 3 x ^{2} + 144 - 72 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}, y = \frac{12 - 3 x ^{2} - 144 - 72 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}

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

\frac{d}{d x} ((x^{2} + 4) e^{- 2 x})

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

a re a x - y^{2} = 0, x + 2 y^{2} = 3

\frac{\partial}{\partial x} (x^{2} y^{2} z^{2})