Solving Partial Differential Equations: Understanding a Key Step

  • Thread starter Google_Spider
  • Start date
In summary, the conversation is about the steps involved in solving a partial differential equation and the use of multipliers to get a specific fraction in the solution. The concept of adding multiples on top and bottom to maintain the same ratio is mentioned. The speaker thanks someone for their help and expertise.
  • #1
Google_Spider
85
0
[SOLVED] Whoa! How did they do this ?

Hi,

I am trying to teach myself partial diff equations and I'm stuck with this small step which I do not understand. I hope someone will help me.

Homework Statement
Solve the Partial Differential Equation
[tex](x^2-y^2-z^2)p+2xyq=2xz[/tex] Solution given in the book

Subsidiary equations are

[tex]\frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}[/tex]

The step which I do not understand is this :

Using multipliers x, y, z. we get:
each fraction=[tex]\frac{xdx+ydy+zdz}{x(x^2+y^2+z^2)}[/tex]

Rest of the solution I understand but I don't get that how they did this particular step. How did they get this fraction ?

Thanks for your time and effort:smile:
 
Physics news on Phys.org
  • #2
… ratios …

Google_Spider said:
The step which I do not understand is this :

Using multipliers x, y, z. we get:
each fraction=[tex]\frac{xdx+ydy+zdz}{x(x^2+y^2+z^2)}[/tex]

Rest of the solution I understand but I don't get that how they did this particular step. How did they get this fraction ?

Hi Google_Spider! :smile:

It's easy algebra - but it looks extremely unlikely the first time you see it! :eek:

If the ratio X/A = Y/B = a, then you can add any multiples together on top and bottom, and the ratio is the same:

(pX + qY)/(pA + qB) = (paA + qaB)/(pA + qB) = a(pA + qB)/(pA + qB) = a.​

And the same works for X/A = Y/B = Z/C = a, and so on.

In this case, put p = x, q = y, r = z, and see what happens … :smile:
 
  • #3
Thanks tiny-tim!:smile:
 

1. How did they create this new technology?

The creation of new technology often involves a team of scientists and engineers who conduct research, testing, and development to come up with innovative solutions. They use a variety of tools and techniques, such as computer simulations, prototyping, and experimentation, to bring their ideas to life.

2. Who is responsible for the discovery/invention?

In most cases, a group of scientists or engineers are responsible for a discovery or invention. However, it is also common for multiple people to contribute to the development of new technology, and credit is often shared among the team.

3. What was the inspiration behind this breakthrough?

Inspiration for new technology can come from a variety of sources, such as observations in nature, previous research, or even a problem that needs to be solved. Often, scientists and engineers build upon existing knowledge and use creativity to come up with new and innovative solutions.

4. How long did it take to develop this technology?

The time it takes to develop new technology can vary greatly depending on the complexity of the project, the resources available, and the team's expertise. Some breakthroughs may take years or even decades to fully develop and implement, while others may come together more quickly.

5. What are the potential applications of this technology?

This question is often asked because people are curious about how new technology can be used to improve their lives. The potential applications of a new technology can range from improving healthcare and communication to advancing transportation and energy efficiency. Scientists and engineers carefully consider the potential impact and benefits of their work to determine the best applications for their technology.

Similar threads

  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
772
  • Calculus and Beyond Homework Help
Replies
1
Views
444
  • Calculus and Beyond Homework Help
Replies
2
Views
677
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
760
  • Calculus and Beyond Homework Help
Replies
7
Views
688
Replies
4
Views
648
  • Calculus and Beyond Homework Help
Replies
4
Views
693
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
Back
Top