xy^'+y=-4xy^2

Lösung

x y^{^{'}} + y = - 4 x y^{2}

y = \frac{1}{x ( 4 ln ( x ) + c _{1} )}

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = \frac{1}{x ( 4 ln ( x ) + c _{1} )}

y = \frac{1}{x ( 4 ln ( x ) + c _{1} )}

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

\int_{0}^{10} 1 + x d x

\int_{- \infty}^{0} \frac{1}{6 - 5 x} d x

x^{2} \cdot \frac{d y}{d x} = y - x y

\int (x^{2} y) d y