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

Lösung

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

y = \int \frac{- x ^{2} + x ^{4} + c _{1}}{2} d x, y = \int \frac{- x ^{2} - x ^{4} + c _{1}}{2} d x

Schritte zur Lösung

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

Solve using substitution:

y = \int \frac{- x ^{2} + x ^{4} + c _{1}}{2} d x, y = \int \frac{- x ^{2} - x ^{4} + c _{1}}{2} d x

y = \int \frac{- x ^{2} + x ^{4} + c _{1}}{2} d x, y = \int \frac{- x ^{2} - x ^{4} + c _{1}}{2} d x

\frac{d}{d x} (\frac{3 + 8 x}{5 - 6 x})

\int (\frac{4}{x}) d x

x \to 1 + lim ((ln (x))^{x - 1})

\frac{d}{d x} (8 x^{3} - 108 x^{2} + 480 x)

x \to 0 lim ((0.5)^{x})