y^'x^2+2xy=5y^3

Lösung

y^{^{'}} x^{2} + 2 x y = 5 y^{3}

y = \frac{x}{2 + c _{1} x ^{5}}, y = - \frac{x}{2 + c _{1} x ^{5}}

Schritte zur Lösung

y^{'} (x) x^{2} + 2 x y = 5 y^{3}

Solve Bernoulli ODE:

y = \frac{x}{2 + c _{1} x ^{5}}, y = - \frac{x}{2 + c _{1} x ^{5}}

y = \frac{x}{2 + c _{1} x ^{5}}, y = - \frac{x}{2 + c _{1} x ^{5}}

\int \frac{4 x ^{2} + x + 4}{x ^{3} - 1} d x

t an g e n t y = (1 + 4 x)^{10}, (0, 1)

3 \frac{d y}{d x} = e^{y - 6 x}

\frac{d}{d x} (2 (2 x e^{2 x} + e^{2 x} \cdot 2 x^{2}))

x \to - 2 lim (\frac{x}{( x + 2 ) ^{3}})