y^'+4x^3y^2=0

Lösung

y^{^{'}} + 4 x^{3} y^{2} = 0

y = - \frac{1}{- x ^{4} + c _{1}}

Schritte zur Lösung

y^{'} (x) + 4 x^{3} y^{2} = 0

Solve separable ODE:

y = - \frac{1}{- x ^{4} + c _{1}}

y = - \frac{1}{- x ^{4} + c _{1}}

\frac{\partial}{\partial x} (x^{2} - 5 x^{2} y - 6 x y^{2})

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

d er i v a t i v e y = (x^{2} + 1)

\int \frac{36 + ( ln ( 8 x ) ) ^{2}}{x} d x

x \to 0 lim (\frac{x ^{2} - 9}{x + 3})