- #1

agnimusayoti

- 240

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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

$$1+p^2=k^2y^2 +(c_2)^2$$

$$p^2=k^2y^2+((c_2)^2-1)$$

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

$$1+p^2=k^2y^2 +(c_2)^2$$

$$p^2=k^2y^2+((c_2)^2-1)$$

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