(7xy^2-3)dx+(7x^2y+4)dy=0

Lösung

(7 x y^{2} - 3) d x + (7 x^{2} y + 4) d y = 0

y = \frac{- 8 + 64 - 28 x ^{2} ( - 6 x - c _{1} )}{14 x ^{2}}, y = \frac{- 8 - 64 - 28 x ^{2} ( - 6 x - c _{1} )}{14 x ^{2}}

Schritte zur Lösung

(7 x y^{2} - 3) d x + (7 x^{2} y + 4) d y = 0

Solve exact ODE:

y = \frac{- 8 + 64 - 28 x ^{2} ( - 6 x - c _{1} )}{14 x ^{2}}, y = \frac{- 8 - 64 - 28 x ^{2} ( - 6 x - c _{1} )}{14 x ^{2}}

y = \frac{- 8 + 64 - 28 x ^{2} ( - 6 x - c _{1} )}{14 x ^{2}}, y = \frac{- 8 - 64 - 28 x ^{2} ( - 6 x - c _{1} )}{14 x ^{2}}

\int_{0}^{3} x^{4} + 1 x^{3} d x

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\int \frac{- 4 y - 1}{y + 1} d y

d er i v a t i v e f (x) = x (3 x - 1)

\frac{d}{d x} ((3 x + 4)^{6})