2x+3y+(6x+9y-3)y^'=0

Lösung

2 x + 3 y + (6 x + 9 y - 3) y^{^{'}} = 0

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

Schritte zur Lösung

2 x + 3 y + (6 x + 9 y - 3) y^{'} (x) = 0

Solve using substitution:

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

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

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\frac{\partial}{\partial y} ((x + y) (x - y))

\frac{d}{d x} (x^{e} + \frac{1}{x ^{10}})

\frac{d ^{2}}{d x ^{2}} ((4 x^{2} + 8 x) cos (x))

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