(y+x)y^'=y

Lösung

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

y = \frac{x}{W ( \frac{x}{c _{1}} )}

Schritte zur Lösung

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

Solve using substitution:

y = \frac{x}{W ( \frac{x}{c _{1}} )}

y = \frac{x}{W ( \frac{x}{c _{1}} )}

\int_{0}^{π} x cos (2 x) d x

in v erse l a pl a ce \frac{32}{( s ^{2} + 4 ^{2} ) ( s + 4 ) ^{2}}

x \to 3 lim (8 x + 2)

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d er i v a t i v e 3 e^{x} + \frac{4}{3 x}