If the solution curve of dydx=x+y-2x-y passing through (2,1) is [^-1!((y-1)/(x-1))-(1)/()_…

Question

If the solution curve of $\dfrac{dy}{dx}=\dfrac{x+y-2}{x-y}$ passing through $(2,1)$ is
$$\tan^{-1}\!\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta}\log_e\!\left[\alpha+\left(\frac{y-1}{x-1}\right)^2\right] = \log_e|x-1|,$$
then $5\beta+\alpha$ is equal to \_\_\_\_\_\_.

Answer: 11

Accepted Answer

With $X=x-1,Y=y-1$: $\dfrac{dY}{dX}=\dfrac{X+Y}{X-Y}$; substituting $Y=vX$ gives
$$\tan^{-1}v-\frac12\ln(1+v^2) = \ln|X|+C,$$

matching form gives $\beta=2,\alpha=1$, so $5\beta+\alpha=11$.

Solutions · 1