xy^'+y=x^4y^3

Lösung

x \cdot y^{^{'}} + y = x^{4} \cdot y^{3}

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

Schritte zur Lösung

x \cdot y^{'} (x) + y = x^{4} \cdot y^{3}

Solve Bernoulli ODE:

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

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

\int t^{19} e^{- t^{20}} d t

in v erse l a pl a ce \frac{5}{s ^{3} - 2 s ^{2} - 3 s}

s l o p e in t erce pt (2005, 56), (2010, 59)

\int_{0}^{1} \frac{3}{x ^{2} - x - 2} d x

d er i v a t i v e \frac{3}{x ^{2}}