Another Second Order ODE Problem (ML Boas, Ch 8, Sec 7, Prob 5)

  • #1
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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
 

Answers and Replies

  • #2
vela
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I substitute ##p^2+1=u## so I got
$$\sqrt{u}=ky+c_1$$
Back substitution
$$1+p^2=k^2y^2 +(c_2)^2$$
Are you really saying that ##(a+b)^2 = a^2+b^2##?
 
  • #3
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It should be ##(a+b)^2=a^2+2ab+b^2##Oh there is a missing term namely ##2kc_1y##. Wait I will make a correction
 
  • #4
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But the form is become more messy I think. Here what I got:
$$p=\frac{dy}{dx}=\sqrt{k^2y^2+2kc_1y+c_2}$$.
This is actually a separable equation. Define ##2kc_1 = c_3## so I got:
$$\frac{dy}{\sqrt{k^2y^2+c_3y+c_2}}=dx$$
How I solve this separable equation, especially on integrating right side of the equation?
 
  • #5
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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}##.
Could you explain how ##(y')^2## becomes ##p^2## on the right-hand side of your differential equation?
 
  • #6
haruspex
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Could you explain how ##(y')^2## becomes ##p^2## on the right-hand side of your differential equation?
Because p=y'.
 
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  • #7
haruspex
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But the form is become more messy I think. Here what I got:
$$p=\frac{dy}{dx}=\sqrt{k^2y^2+2kc_1y+c_2}$$.
This is actually a separable equation. Define ##2kc_1 = c_3## so I got:
$$\frac{dy}{\sqrt{k^2y^2+c_3y+c_2}}=dx$$
How I solve this separable equation, especially on integrating right side of the equation?
A change of variable gets rid of the c3y term, then you can use a trig substitution to get rid of the square root.
 
  • #8
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Because p=y'.
Oh boy, I'm losing it. The expression for ##y''## looked odd to me and not equivalent to ##y'=p##.
 
  • #9
pasmith
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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.
This is already first-order and separable in [itex]u = y'[/itex]: [tex]
u' = \pm|k| \sqrt{1 + u^2}.[/tex]
 
  • #10
There are two tricks to solve this ODE. The first is to notice that
$$(y^{\prime\prime})^2 = k^2\big[1+(y^\prime)^2\big]$$
is nothing more than a 1st-order ODE in ##y^\prime##. Thus, let ##p = y^\prime## (and thus ##p^\prime = y^{\prime\prime}##) and the original ODE becomes
$$(p^\prime)^2 = k^2(1+p^2).$$
The second (and last trick) is to differentiate this ODE w.r.t. the independant variable ##x##, the result of which reads
$$2p^\prime p^{\prime\prime} = 2k^2pp^\prime.$$
This (after at bit of cleaning) is simply the familiar homogeneous 2nd order linear ODE
$$p^{\prime\prime} - k^2p = 0$$
to which e.g. ##\cosh(kx)## and ##\sinh(kx)## constitute a complete set of solutions.
 
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  • #11
haruspex
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There are two tricks to solve this ODE. The first is to notice that
$$(y^{\prime\prime})^2 = k^2\big[1+(y^\prime)^2\big]$$
is nothing more than a 1st-order ODE in ##y^\prime##. Thus, let ##p = y^\prime## (and thus ##p^\prime = y^{\prime\prime}##) and the original ODE becomes
$$(p^\prime)^2 = k^2(1+p^2).$$
The second (and last trick) is to differentiate this ODE w.r.t. the independant variable ##x##, the result of which reads
$$2p^\prime p^{\prime\prime} = 2k^2pp^\prime.$$
This (after at bit of cleaning) is simply the familiar homogeneous 2nd order linear ODE
$$p^{\prime\prime} - k^2p = 0$$
to which e.g. ##\cosh(kx)## and ##\sinh(kx)## constitute a complete set of solutions.
Very neat.
But I feel it is also valuable to address any issues with the actual method attempted; in this case, how to continue.
 
  • #12
pasmith
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But the form is become more messy I think. Here what I got:
$$p=\frac{dy}{dx}=\sqrt{k^2y^2+2kc_1y+c_2}$$.
This is actually a separable equation. Define ##2kc_1 = c_3## so I got:
$$\frac{dy}{\sqrt{k^2y^2+c_3y+c_2}}=dx$$
How I solve this separable equation, especially on integrating right side of the equation?
You have too many constants; they aren't all independent. It's also best not to expand quadratics which you are about to integrate, since the first thing you'll end up doing is reversing that in order to complete the square.

So: From [tex]
\sqrt{1 + p^2} = k(y + c_1)[/tex] you know that [itex]c_1 = k^{-1} - y_0[/itex] where [itex]y_0[/itex] is the minimum height of the chain, which is attained at [itex]x = x_0[/itex]. Hence [tex]
k(y + c_1) = k(y - y_0) + 1[/tex] and [tex]
p^2 = \left(\frac{dy}{dx}\right)^2 = (k(y - y_0)^2 + 1)^2 - 1.[/tex] Now the subsitution [itex]k(y - y_0) + 1 = \cosh(u)[/itex] suggests itself, since [itex]\sinh^2 u = \cosh^2 u - 1[/itex] and [tex]\frac{d}{dx}\cosh u = u'\sinh u.[/tex]
 
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