(2y^2+3xy)+(2xy+x^2)(dy)/(dx)=0

Lösung

(2 y^{2} + 3 x y) + (2 x y + x^{2}) \frac{d y}{d x} = 0

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

Schritte zur Lösung

(2 y^{2} + 3 x y) + (2 x y + x^{2}) \frac{d y}{d x} = 0

Solve using substitution:

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

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

\int_{1}^{3} x^{2} e^{x^{3} - 1} d x

\int 2 x sin (x^{2}) + \frac{1}{x ^{2}} sin (\frac{1}{x}) d x

t an g e n t y = \frac{7}{sin ( x ) + cos ( x )}, (0, 7)

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

t an g e n t y = \frac{x}{1 + x ^{2}}, (3, 0.3)