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Question regarding an integral

  • #1

Homework Statement


I've got an equation which I need to integrate. However, integrating it and checking with the solutions, I get two different results. I get the same result as using wolfram alpha, but a different result from the book.
If I differentiate both results, I get back to the orginal equation, so I don't know why they are different.

Homework Equations


Equation which I need to Integrate by seperation of variables: [tex]\frac{d\dot{x}^2}{dx} = \frac{2l^4 a^2 x}{(12x^2+l^2)^2}[/tex]

Result from Book: [tex]\dot{x}^2 = \frac{ l^2 a^2 x^2}{12x^2+l^2}[/tex]
Result from Wolfram-Alpha: [tex]\dot{x}^2 = -\frac{(a^2 l^4)}{(12 (l^2+12 x^2))}[/tex]

The Attempt at a Solution


If I differentiate both, I'll both get the same equation back which I integrated, I just don't understand why thats the case, since it doesn't look like they differ by a additive constant.
 
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Answers and Replies

  • #2
Ray Vickson
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Homework Statement


I've got an equation which I need to integrate. However, integrating it and checking with the solutions, I get two different results. I get the same result as using wolfram alpha, but a different result from the book.
If I differentiate both results, I get back to the orginal equation, so I don't know why they are different.

Homework Equations


Equation which I need to Integrate by seperation of variables: [tex]\frac{d\dot{x}^2}{dx} = \frac{2l^4 a x}{(12x^2+l^2)^2}[/tex]

Result from Book: [tex]\dot{x}^2 = \frac{ l^2 a^2 x^2}{12x^2+l^2}[/tex]
Result from Wolfram-Alpha: [tex]\dot{x}^2 = -\frac{(a^2 l^4)}{(12 (l^2+12 x^2))}[/tex]

The Attempt at a Solution


If I differentiate both, I'll both get the same equation back which I integrated, I just don't understand why thats the case, since it doesn't look like they differ by a additive constant.
(1) Please clarify: does
[tex] \frac{d\dot{x}^2}{dx} [/tex]
mean
[tex] \frac{d( d(x^2)/dt)}{dx} [/tex]
or
[tex] \frac{d (dx/dt)^2}{dx} [/tex]
or
[tex] \left( \frac{d (dx / dt) }{dx}\right)^2 \:? [/tex]
(2) Show the steps you used in obtaining your answer.
 
Last edited:
  • #3
It means :[tex]\frac{d\dot{x}^2}{dx} = \frac{d}{dx}\left(\frac{dx}{dt}\right)^2[/tex]

For the result from the book there where no steps shown.

For my result: [tex]\frac{d\dot{x}^2}{dx} = \frac{2l^4a^2x}{(12x^2+l^2)^2} \Longrightarrow \int d\dot{x}^2 =2 l^4a^2\int \frac{x}{(12x^2+l^2)^2} dx[/tex]

Apply Substitution: [tex]u = 12x^2 +l^2 \,\,\,\,\,\,\, \frac{du}{dx} = 24x \Rightarrow dx = \frac{1}{24x}du[/tex]

[tex]\int d\dot{x}^2 = \dot{x}^2 = l^4a^2 \int\frac{1}{24u^2}du = -2l^4a^2\frac{1}{24} u^{-1} + c = -l^4a^2\frac{1}{12}\frac{1}{12x^2+l^2}+c[/tex]

There were some starting conditions which would lead to c = 0
 
  • #4
Ray Vickson
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It means :[tex]\frac{d\dot{x}^2}{dx} = \frac{d}{dx}\left(\frac{dx}{dt}\right)^2[/tex]

For the result from the book there where no steps shown.

For my result: [tex]\frac{d\dot{x}^2}{dx} = \frac{2l^4a^2x}{(12x^2+l^2)^2} \Longrightarrow \int d\dot{x}^2 =2 l^4a^2\int \frac{x}{(12x^2+l^2)^2} dx[/tex]

Apply Substitution: [tex]u = 12x^2 +l^2 \,\,\,\,\,\,\, \frac{du}{dx} = 24x \Rightarrow dx = \frac{1}{24x}du[/tex]

[tex]\int d\dot{x}^2 = \dot{x}^2 = l^4a^2 \int\frac{1}{24u^2}du = -2l^4a^2\frac{1}{24} u^{-1} + c = -l^4a^2\frac{1}{12}\frac{1}{12x^2+l^2}+c[/tex]

There were some starting conditions which would lead to c = 0
You are correct, and the book is wrong (as can be checked by differentiating the answer---a step you should always perform).
 
  • #5
I did perform that step as I wrote in my post and I get the same equation, though my derivative might be wrong.

Let's differentiate the solution from the book with respect to x:

[tex]\frac{d}{dx}\left(\frac{l^2a^2x^2}{12x^2+l^2}\right) = l^2a^2 \frac{d}{dx}\left(\frac{x^2}{12x^2+l^2}\right)[/tex]

Using the quotient rule I get: [tex]l^2a^2 \frac{2x(12x^2+l^2)-24x^3}{(12x^2+l^2)^2} = l^2a^2\frac{24x^3-2xl^2-24x^3}{(12x^2+l^2)^2} = \frac{2l^4a^2x}{(12x^2+l^2)^2}[/tex] which is the same as the orginal function which I integrated
 
  • #6
vela
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Even though it doesn't look like it at first glance, the two results do, in fact, differ by just an additive constant:
$$\frac{a^2l^2}{12}-\frac{a^2l^4}{12(12x^2+l^2)} = \frac{l^2a^2x^2}{12x^2+l^2}$$
 
  • #7
Ray Vickson
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Even though it doesn't look like it at first glance, the two results do, in fact, differ by just an additive constant:
$$\frac{a^2l^2}{12}-\frac{a^2l^4}{12(12x^2+l^2)} = \frac{l^2a^2x^2}{12x^2+l^2}$$
Oops, my mistake in #4.
 
  • #8
Ah I see, thanks very much both of you!
 

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