(2xy^2-3)dx+(2x^2y+4)dy=0

Lösung

(2 x y^{2} - 3) d x + (2 x^{2} y + 4) d y = 0

y = \frac{- 2 + - x ^{2} ( - 3 x - c _{1} ) + 4}{x ^{2}}, y = - \frac{- x ^{2} ( - 3 x - c _{1} ) + 4 + 2}{x ^{2}}

Schritte zur Lösung

(2 x y^{2} - 3) d x + (2 x^{2} y + 4) d y = 0

Solve exact ODE:

y = \frac{- 2 + - x ^{2} ( - 3 x - c _{1} ) + 4}{x ^{2}}, y = - \frac{- x ^{2} ( - 3 x - c _{1} ) + 4 + 2}{x ^{2}}

y = \frac{- 2 + - x ^{2} ( - 3 x - c _{1} ) + 4}{x ^{2}}, y = - \frac{- x ^{2} ( - 3 x - c _{1} ) + 4 + 2}{x ^{2}}

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

\frac{d}{d x} (\frac{( x ^{2} )}{2})

t an g e n t (3)^{x}

\frac{\partial}{\partial x} (5 y e^{x} - ln (z))

\int \frac{1}{x ^{4}} - \frac{1}{4 x} - 4 d x