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Here is the differential equation I am given.

$$\frac{dy}{dx}=y(y^{2}-1)^{\frac{1}{2}}$$

This is a seperable differential equation and I managed to figure out the solution to this equation is

$$y =±sec(x-c)$$

I also found out that the equation had singular solutions at $y = -1,1$. What I don't get is why the differential has infinitely many solutions at $y =1,-1$.

Wouldn't the solution at something like $y=1$ be

$$-arcsec(1)+x=c$$

Since only one value of $c$ works wouldn't that be unique?

user262291
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  • My answer matches with the one in the back of the textbook. You can always try plugging the equation into the differential equation and verify whether it is correct. – user262291 Sep 18 '16 at 02:20

1 Answers1

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$\pm\sec(x-c)$ is not a solution everywhere -- only the part between $c$ and $c+\frac\pi2$ where it moves away from $0$ is a solution. Between $c-\frac\pi2$ and $c$ the derivative is minus what the differential equation asks for.

A fuller solution would be the piecewise defined $$ y(x) = \begin{cases} 1 & \text{when } x \le c \\ \sec(x-c) & \text{when } c < x < c+\frac\pi2 \end{cases} $$

If your boundary condition is $y(0)=1$, then each of the infinitely many combinations of choices of $c\ge 0$ will yield a solution.

hmakholm left over Monica
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