y^'+2x^{-1}y=6x^4y^2

Lösung

y^{^{'}} + 2 x^{- 1} y = 6 x^{4} y^{2}

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

Schritte zur Lösung

y^{'} (x) + 2 x^{- 1} y = 6 x^{4} y^{2}

Solve Bernoulli ODE:

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

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

\int 3 x^{- \frac{2}{3}} - 2 x^{\frac{5}{3}} d x

\int e^{\frac{- 1}{2} l n (x)} d x

\int \frac{1}{u ^{2} + 9} d u

\frac{d}{d x} (lo g_{e} (sin (x)))

l a pl a ce t r an s f or m t (t cos (t) - sin (t))