Integration and Partial Differentiation Problem

In summary, the conversation involves a question about integrating a complicated expression and a question about finding the equality of two partial derivatives. The solution involves factorizing the expressions and using integration by partial fractions. The final solution is obtained by calculating the partial derivatives correctly and using the quotient rule.
  • #1
41
0

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]?
 
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  • #2
Mastur said:
[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
LCKurtz said:
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
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
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
Mastur said:
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.
 

1. What is integration and partial differentiation?

Integration and partial differentiation are mathematical techniques used to solve problems involving functions. Integration involves finding the area under a curve, while partial differentiation involves finding the rate of change of a function with respect to one of its variables.

2. What are the applications of integration and partial differentiation?

Integration and partial differentiation have wide-ranging applications in fields such as physics, engineering, economics, and statistics. They are used to model and analyze real-world situations, such as calculating the velocity of an object, finding the optimal solution to a problem, or predicting future trends.

3. How do you solve integration and partial differentiation problems?

To solve integration and partial differentiation problems, you need to follow certain rules and techniques. For integration, you need to use integration formulas, such as the power rule, substitution rule, and integration by parts. For partial differentiation, you need to use the chain rule, product rule, and quotient rule, among others.

4. What are the differences between integration and partial differentiation?

Integration and partial differentiation have some key differences. Integration involves finding the area under a curve, while partial differentiation involves finding the rate of change of a function. Integration deals with indefinite and definite integrals, while partial differentiation deals with single and multiple variables. Also, integration involves addition, while partial differentiation involves multiplication.

5. Why are integration and partial differentiation important in scientific research?

Integration and partial differentiation play a crucial role in scientific research. They are used to model and analyze complex systems and processes, making it easier to understand and predict their behavior. Additionally, they allow scientists to calculate important quantities, such as volume, speed, and acceleration, which are necessary for making accurate predictions and developing effective solutions.

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