x^2-3xy+xy^'=0

Lösung

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

y = \frac{3 x + 1}{9} + c_{1} e^{3 x}

Schritte zur Lösung

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

Solve linear ODE:

y = \frac{3 x + 1}{9} + c_{1} e^{3 x}

y = \frac{3 x + 1}{9} + c_{1} e^{3 x}

\frac{d}{d x} (ln (\frac{5 x + 1}{5 x - 1}))

n = 1 \sum \infty n^{2} + 1

im pl i c i t 2 x y - y^{2} = - 1

\int (3 x^{2} + 4 x - 6) d x

\frac{d}{d x} (- \frac{1}{3} (x^{- 3} - x^{6}))