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Homework Statement:

The differential equation of a hanging chain supported at its ends is:
$$y"^2=k^2(1+y'^2)$$.
Solve the equation to find the shape of the chain.
Relevant Equations:
 One can modify this kind of ODE problem by changing variable y' to p. So, by the chain rule, $$y"=p \frac{dp}{dy}$$.
With the new variable, I got:
$$p^2 (p'_y)^{2}=k^2(1+p^2)$$ where ##p'_y## is ##\frac{dp}{dy}##.
I modified the equation so the variable p and dp can be separated from dy. Here what I got:
$$\frac{p}{\sqrt{p^2+1}} dp=k dy$$
I substitute ##p^2+1=u## so I got
$$\sqrt{u}=ky+c_1$$
Back substitution
$$1+p^2=k^2y^2 +(c_2)^2$$
$$p^2=k^2y^2+((c_2)^21)$$
My Question, can I eliminate arbitrary constant so I get:
$$1+p^2=k^2y^2$$
It seems easier to solve in terms of p as dy/dx. Thanks
$$p^2 (p'_y)^{2}=k^2(1+p^2)$$ where ##p'_y## is ##\frac{dp}{dy}##.
I modified the equation so the variable p and dp can be separated from dy. Here what I got:
$$\frac{p}{\sqrt{p^2+1}} dp=k dy$$
I substitute ##p^2+1=u## so I got
$$\sqrt{u}=ky+c_1$$
Back substitution
$$1+p^2=k^2y^2 +(c_2)^2$$
$$p^2=k^2y^2+((c_2)^21)$$
My Question, can I eliminate arbitrary constant so I get:
$$1+p^2=k^2y^2$$
It seems easier to solve in terms of p as dy/dx. Thanks