Differentialgleichung mit Bruch. y' = (y-3)(x-1)/(x^2-5x+6)

Question

y' = (y-3)(x-1)/(x^2-5x+6)

y'*1/y-3 = x-1/x^2-5x+6

1/y-3 dy = x-1/(x-2)(x-3) dx

ln | y-3 | = - ln | x- 2 | + 2 *ln | x- 3| + c

ln | y -3| = -2 ln | x^2-5x+6 | + c

y =3 + e^c * e^-2ln | x^2-5x+6| mit e^c = K

y = K *(-2x+10x-9)

Answer 1

das ist leider falsch.

y' = (y-3)(x-1)/(x²-5x+6)