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

Lösung

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

y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}, y (2) = 5

\frac{\partial}{\partial y} (e^{x y z^{4}} y z^{4})

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

t an g e n t f (x) = x^{3} - 4 x, (2, 0)

d er i v a t i v e cos (x) - \frac{2}{2} x