-3y^'+(3y)/x =(y^4)/(x^2)

Lösung

- 3 y^{^{'}} + \frac{3 y}{x} = \frac{y ^{4}}{x ^{2}}

y = \frac{3 2 x}{3 x ^{2} + c _{1}}

Schritte zur Lösung

- 3 y^{'} (x) + \frac{3 y}{x} = \frac{y ^{4}}{x ^{2}}

Solve Bernoulli ODE:

y = \frac{3 2 x}{3 x ^{2} + c _{1}}

y = \frac{3 2 x}{3 x ^{2} + c _{1}}

\frac{\partial}{\partial y} (3 x^{2} y^{2} + 4)

\frac{\partial}{\partial x} (- 8 (x + y + z)^{2})

\int \frac{arcsin ( x )}{( 1 - x ^{2} ) ^{\frac{1}{2}}} d x

x \to 0 lim (x^{15 x})

\int \frac{x ^{3}}{6 - x ^{4}} d x