Substitutions — Question 5

Question 5 – Solution

Strategy. Translate the intersection of the two affine lines to the origin, then apply the homogeneous ratio substitution locally.

Step 1: Choose the translation. The equations $h+k-3=0$ and $h-k+1=0$ give $\boxed{h=1,\ k=2}$. Hence $$X=x-1,\quad Y=y-2,\quad \frac{dY}{dX}=\frac{X+Y}{X-Y},\qquad (X,Y)=(1,0)\text{ initially}.$$ A translation does not change the derivative because $dX/dx=1$.

Step 2: Substitute and integrate. With $Y=Xv$, $$v+Xv'=\frac{1+v}{1-v},\qquad \frac{1-v}{1+v^2}\,dv=\frac{dX}{X}.$$ On the local region $X>0$, integration gives $$\arctan v-\frac 12\ln(1+v^2)=\ln X+C.$$ At $(1,0)$, $C=0$. Substituting $v=Y/X$ and simplifying yields $$\boxed{\arctan(Y/X)-\frac 12\ln(X^2+Y^2)=0}$$ for the local branch through $(1,0)$ with $X>0$.

Step 3: Verify and find the slope. For the displayed left side $F$, differentiation gives $$F_X=-\frac{X+Y}{X^2+Y^2},\qquad F_Y=\frac{X-Y}{X^2+Y^2}.$$ Thus $F_X+F_Y Y'=0$ recovers the shifted equation wherever $X-Y\ne 0$. At $(1,0)$, $F_Y=1$, so a local graph exists and $\boxed{y'(2)=1}$.

Step 4: Separate coordinate and equation restrictions. The ratio $Y/X$ excludes $X=0$, namely $x=1$, but the original equation is defined there whenever $y\ne 2$. Its actual excluded line is $\boxed{x-y+1=0}$, or $y=x+1$. An alternative local coordinate or angle branch can be needed when the ratio chart fails; that failure alone does not prove that an original solution ends. No global branch claim is made by this local arctangent formula.