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

Lösung

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

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

Schritte zur Lösung

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

Solve using substitution:

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

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

s l o p e 2 - \frac{3}{x ^{2}}, at x = 1

t an g e n t f (x) = \frac{7}{x}, (1, 7)

x \to 0 + lim (x - ln (x))

\frac{\partial}{\partial x} (\frac{x ^{3}}{1 - x ^{2}})

\int_{- π}^{π} sin (x) cos (n x) d x