Simple diff.eq

  • Thread starter sony
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  • #1
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d/dx(y*x^2)=x^2/(1+x^2)

I guess I can't write this as:
dyx^2=x^2dx/(1+x^2)
dy=dx/(1+x^2)
beacuse I don't get the right answer...

So what do I do?
 

Answers and Replies

  • #2
Tide
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You can't just divide out the [itex]x^2[/itex] in the differential. I suggest direct integration of the DE with respect to x.
 
  • #3
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well, it's the d/dx(y*x^2) that confused me.
we have only learned to solv the types:
dy/dx=f(x)g(y)

what is direct integration?

Thanks
 
  • #4
Tide
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Direct integration means this:
[tex]\int_{x_0}^{x} \frac {d}{dx'}\left(x'^2 y\right) dx' = \int_{x_0}^{x} \frac {x'^2}{1+x'^2} dx'[/tex]
 
  • #5
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Then I don't think we have learned about direct integration...
 
  • #6
dextercioby
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Well, Tide said that, simply

[tex] \int du = u +\mathcal{C} [/tex]

You must understand what i've written, else why attempt to solve diff. eqns. ?

Daniel.
 
  • #7
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okey, I understand what you just wrote :P

but I still don't know how to solve my problem...
 
  • #8
HallsofIvy
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sony said:
d/dx(y*x^2)=x^2/(1+x^2)
I guess I can't write this as:
dyx^2=x^2dx/(1+x^2)
dy=dx/(1+x^2)
beacuse I don't get the right answer...
So what do I do?

Tide's point was just that you need to divide both sides by x2 to get simply [itex]dy= \frac{dx}{1+x^2}[/itex].

"Direct integration" just meant regular integration!

[tex]y= \int dy= \int \frac{dx}{1+ x^2}[/itex]
 
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  • #9
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Oh, but I didn't think I could do that when y*x^2 was encapsulated by parantheses.

When I solve what you posted I get y=tan^-1(x) but the answer is supposed to be:
(pi/4)(1/x^2)-(1/x)-tan^-1(x)/x^2

(the pi/4 comes from the start value...)
 
  • #10
HallsofIvy
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sony said:
Oh, but I didn't think I could do that when y*x^2 was encapsulated by parantheses.
When I solve what you posted I get y=tan^-1(x) but the answer is supposed to be:
(pi/4)(1/x^2)-(1/x)-tan^-1(x)/x^2
(the pi/4 comes from the start value...)
Oops! Sorry about that. I misread the original equation!

It's not that the yx2 was "encapsulated by parentheses" but that here, specifically, the parentheses mean that the entire function yx2 is being differentiated. I read it as [itex]x^2\frac{dy}{dx}[/itex] rather than [itex]\frac{dx^2y}{dx}[/itex].

Given that, Tide meant that
[tex]x^2y= \int d(x^2y)= \int \frac{x^2dx}{x^2+ 1}[/tex].

That right hand side is a little more complicated. First divide it out:
[tex]\frac{x^2}{x^2+1}= 1- \frac{1}{x^2+1}[/itex]
and it becomes easy to integrate.
 
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  • #11
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Oh, thanks! Now I see it :)
 
  • #12
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Bah, sorry. I don't see what the integral of "dx^2y" evaluates to :(
 
  • #13
HallsofIvy
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sony said:
Bah, sorry. I don't see what the integral of "dx^2y" evaluates to :(

Fundamental Theorem of Calculus!

[tex] \int d(Anything)= Anything+ C[/tex]

[tex]\int d(x^2y)= x^2y + C[/tex]
 
  • #14
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Oh! sorry! I see it! bah

hehe, the yx^2 messed up my head :P
 

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