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Integration and Partial Differentiation Problem

  • Thread starter Mastur
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  • #1
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Homework Statement


(A)
[itex]\int{\frac{(v^2+2v+4)dv}{v^3+v^2+2v+4}}[/itex]

(B)
[itex]\frac{\partial{M}}{\partial{y}}=(1-xy)^{-2}[/itex]

[itex]\frac{\partial{N}}{\partial{x}}=y^2+x^2(1-xy)^{-2}[/itex]

Homework Equations


(A) How can I integrate this?

(B)After getting the partial derivatives, are they equal?

The Attempt at a Solution


(A)This is actually I stopped since I cannot integrate it. I tried factoring the denominator so I can use integration by partial fractions. Unfortunately, I cannot. Any hint in integrating it?

(B)
[itex]\frac{\partial{M}}{\partial{y}}=\frac{2x}{1-xy}[/itex]

[itex]\frac{\partial{N}}{\partial{x}}=\frac{2x+2x^2y-2x^3y^2}{(1-xy)^2}[/itex]

Can I still factor the second equation so I can get the same answer like the [itex]\frac{\partial{M}}{\partial{y}}[/itex]?
 

Answers and Replies

  • #2
LCKurtz
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[itex]\frac{\partial{M}}{\partial{y}}=(1-xy)^{-2}[/itex]

[itex]\frac{\partial{N}}{\partial{x}}=y^2+x^2(1-xy)^{-2}[/itex]


(B)After getting the partial derivatives, are they equal?
Do they look equal if you try, for example, x = 0 and y = 0 in both?
 
  • #3
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Do they look equal if you try, for example, x = 0 and y = 0 in both?
Yes, they will be equal after differentiating.

But that does not work all the time. I've tried checking my other answers in other problem, tried substituting x & y by 0, but not all are equal, but the equation is still exact after differentiating.
 
  • #4
LCKurtz
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We are obviously mis-communicating. Please state the original problem you are working on that involves the M and N in part B of your question. Presumably there is a differential equation involved.

Also you have said

[itex]
\frac{\partial{M}}{\partial{y}}=(1-xy)^{-2}
[/itex]
in your question and

[itex]
\frac{\partial{M}}{\partial{y}}=\frac{2x}{1-xy}
[/itex]

in your solution. They can't both be My
 
Last edited:
  • #5
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Sorry for the confusion.

The equation is:

[itex]M=(1-xy)^{-2}[/itex]

[itex]N=y^2+x^2(1-xy)^{-2}[/itex]

And I got this function after differentiating.

[itex]\frac{\partial{M}}{\partial{y}}=\frac{2x}{1-xy}[/itex]

[itex]\frac{\partial{N}}{\partial{x}}=\frac{2x+2x^2y-2x^3y^2}{(1-xy)^2}[/itex]

My actual question is, is there any way to make the [itex]\frac{\partial{M}}{\partial{y}} = \frac{\partial{N}}{\partial{x}}[/itex]??
 
  • #6
LCKurtz
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Sorry for the confusion.

The equation is:

[itex]M=(1-xy)^{-2}[/itex]

[itex]N=y^2+x^2(1-xy)^{-2}[/itex]

And I got this function after differentiating.

[itex]\frac{\partial{M}}{\partial{y}}=\frac{2x}{1-xy}[/itex]

[itex]\frac{\partial{N}}{\partial{x}}=\frac{2x+2x^2y-2x^3y^2}{(1-xy)^2}[/itex]

My actual question is, is there any way to make the [itex]\frac{\partial{M}}{\partial{y}} = \frac{\partial{N}}{\partial{x}}[/itex]??
Yes, there is, but you need to calculate them correctly. Neither of your partials is correct. Given that both functions have a quantity to the -2 power, you would expect a denominator with exponent of 3 in your simplified answers. Perhaps if you write both M and N as quotients instead of using negative exponents and use the quotient rule, you might have better luck calculating the partials. Just guessing here, since you didn't show your work.
 

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