y^'=((3y^2+9))/(2xy)

Lösung

y^{^{'}} = \frac{( 3 y ^{2} + 9 )}{2 x y}

y = c_{1} x^{3} - 3, y = - c_{1} x^{3} - 3

Schritte zur Lösung

y^{'} (x) = \frac{( 3 y ^{2} + 9 )}{2 x y}

Solve separable ODE:

y = c_{1} x^{3} - 3, y = - c_{1} x^{3} - 3

y = c_{1} x^{3} - 3, y = - c_{1} x^{3} - 3

n = 1 \sum \infty \frac{1}{2 + n}

x \to 0 lim (\frac{1 - e ^{2 x}}{x})

\int_{- \infty}^{8} r e^{\frac{r}{2}} d r

\frac{d}{d t} (sin (π t))

\frac{\partial}{\partial y} (sin^{3} (x^{2} y))