Number of solutions to a differential equation

Question

The number of solutions of $$\frac{dy}{dx} = \frac{y+1}{x-1}$$ when $y(1) = 2$ is

Options are a) none b) one c) two d) infinite

Now this can be solved via variable separable method which gives me $$\frac{dy}{y+1} = \frac{dx}{x-1}$$ And on integrating both sides I got, $\log (y + 1) = \log(x - 1) + \log c$ which gave me $y + 1 = c(x - 1)$ On putting the given values as $x = 1$ and $y = 2$ gives $3 = 0$ or $c = \frac{3}{0}$ which means no solution.

I just want to ask if my solution is correct because the book gave the answer as one solution. I checked few other places and they for some reason are solving by integrating as $\log (y + 1) = \log(x - 1) - \log c$ which of course would give them the desired one solution as $x = 1$.

I am confused.

Answer 1

If you consider $x(y)$ not $y(x)$ you'll get $x = 1 + (1 + y)\cdot c$, and therefore substituting $y=2$ you'll get $c=0$ if you want $x=1$. The solutions is therefore $x(y)=1$.

Edit: also mind that $\int dx/x=\ln|x|$.

Answer 2

Your general solution $y=c(x-1)-1$ is correct by direct check

$$\frac{dy}{dx} =\frac{y+1}{x-1}=\frac{c(x-1)-1+1}{x-1}=cx$$

but since the equation is defined for $x\neq 1$ we have no solution for this particular initial condition.

Answer 3

The problem is ill-posed since $x=1$ is a singular point. But even if you consider the given point to be a limit, you can see from the general solution that

$$ \lim_{x\to 1} y(x) = \lim_{x\to1} \big[c(x-1)-1 \big] = -1 $$

Clearly, there are no such functions where $\lim_{x\to1} y(x) = 2$