y^'=-(xy)/(x^2+y^2)

Lösung

y^{^{'}} = - \frac{x y}{x ^{2} + y ^{2}}

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

Schritte zur Lösung

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

Solve using substitution:

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

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

\int \frac{x ^{2}}{1 + x ^{2}} d x

x y^{^{'}} + 4 y = 8 x^{4}, y (1) = 2

t an g e n t x^{2} + 3 x + 9, at x = 4

\frac{d}{d x} ((x + \frac{1}{x})^{2} - \frac{5}{3 x})

\frac{\partial}{\partial x} (x e^{3 y} sin (5 z))