(4+2y)dx+(2x-5)dy=0

Lösung

(4 + 2 y) d x + (2 x - 5) d y = 0

y = \frac{c _{1}}{2 x - 5} - 2

Schritte zur Lösung

(4 + 2 y) d x + (2 x - 5) d y = 0

Solve separable ODE:

y = \frac{c _{1}}{2 x - 5} - 2

y = \frac{c _{1}}{2 x - 5} - 2

y^{^{'}} + (x + 6) y = 0

\frac{d y}{d x} = \frac{2 x}{y + x ^{2} y}, y (0) = 2

y^{^{'}} = - \frac{y}{a} + b

y^{^{'}} = y - 12

\frac{d x}{d y} = k (150 - 2 x)