y^'=((2x^2+3y^2))/(2xy)

Lösung

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

y = x - 2 + c_{1} x, y = - x - 2 + c_{1} x

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = x - 2 + c_{1} x, y = - x - 2 + c_{1} x

y = x - 2 + c_{1} x, y = - x - 2 + c_{1} x

\frac{\partial}{\partial v} (tan (\frac{u}{v}))

n = 3 \sum \infty \frac{2}{n ( n + 2 )}

\int - 5 csc^{2} (x) d x

x \to - 3 + lim (\frac{x + 4}{x + 3})

x \to + (- 4) lim (∣ x + 4 ∣)