Find a set of solutions for $y'=\sqrt{y^2-1},\ y(0)=1$.

Question

For the IVP $y'=\sqrt{y^2-1},\ y(0)=1$ I am supposed to find a set of solutions depending on $2$ parameters.

While I can easily find 2 different solutions $y_1(x)=1$ and $y_2(x)=\frac{1}{2}(e^{-x}+e^x)$ that solve the IVP I don't know how to bring parameters into this problem. How can any solution depend on a parameter when the starting value is given? And if no solution depends on any parameter, what's the trick to create a set of solutions depending on 2 parameters?

Answer 1

If $y=1$, $y'=0$ yields $y=1$, is consistent with the initial condition.

Then if $y\ne1$,$$\frac{y'}{\sqrt{y^2-1}}=1$$ and

$$\text{arcosh }y=x+c,$$ $$y=\cosh(x+c).$$

As said by Robert, we can switch (twice) between the two solutions:

$$y=\begin{cases}x<c_-\to\cosh(x-c_-),\\c_-<x<c_+\to1,\\x>c_+\to\cosh(x-c_+)\end{cases}.$$

where $c_-\le 0\le c_+$. These are your two constants.

Solutions with a single or with no switch are also possible.

Answer 2

Hint: your solution can be $1$ for a while and then switch to a translate of $y_2$.