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

Lösung

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

y = \frac{- 8 + 64 - 12 x ^{2} ( - 6 x - c _{1} )}{6 x ^{2}}, y = \frac{- 8 - 64 - 12 x ^{2} ( - 6 x - c _{1} )}{6 x ^{2}}

Schritte zur Lösung

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

Solve exact ODE:

y = \frac{- 8 + 64 - 12 x ^{2} ( - 6 x - c _{1} )}{6 x ^{2}}, y = \frac{- 8 - 64 - 12 x ^{2} ( - 6 x - c _{1} )}{6 x ^{2}}

y = \frac{- 8 + 64 - 12 x ^{2} ( - 6 x - c _{1} )}{6 x ^{2}}, y = \frac{- 8 - 64 - 12 x ^{2} ( - 6 x - c _{1} )}{6 x ^{2}}

\int \frac{( x + 7 ) ^{3}}{2 x} d x

\frac{d P}{d t} = 2 Pt, P (1) = 7

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

t an g e n t y = x^{2} + 4

d er i v a t i v e f (x^{3} - x)