Question

solve y'= -(y-x)^2 using the substitution u=y-x

Answer 1

\[y'=-(y-x)^2\quad;\quad u=y-x\quad,\quad u'=y'-1\]\[u'+1=-u^2\]\[u'+1+u^2=0\]\[\frac{u'}{1+u^2}+1=0\]\[(\arctan{u})'=-1\]\[\arctan{u}=-x+C\]\[u=\tan{(-x+C)}=-\tan(x-C)\]\[y=x-\tan{(x-C)}\]