If the solution curve of dydx=x+y-2x-y passes through (2,1) and (k+1,2), k0, then:

Question

If the solution curve of $\dfrac{dy}{dx}=\dfrac{x+y-2}{x-y}$ passes through $(2,1)$ and $(k+1,2)$, $k>0$, then:

Accepted Answer

With $X=x-1,Y=y-1$: $Y=tX \implies \tan^{-1}t-\dfrac12\ln(1+t^2)=\ln X+c$.

Through $(2,1)$: $c=0$. Through $(k+1,2)$: $t=1/k$, $X=k$:
$$\tan^{-1}\frac1k-\frac12\ln\left(1+\frac{1}{k^2}\right) = \ln k \implies 2\tan^{-1}\frac1k = \ln(k^2+1).$$

Solutions · 1