How can you solve for y in a difficult separable equation?

  • Thread starter Jamin2112
  • Start date
  • Tags
    Separable
In summary, the conversation is about solving a differential equation dy/dx = (y cos x) / (1+y^2). The attempt at a solution involved manipulating the equation to get ln(y) + y^2/2 = sinx + C, but the problem arises when trying to solve for y. Theoretical solutions using the Lambert W function were suggested, but it is unlikely that the problem would require its use. The conversation ends with a mention of checking wolframalpha for alternative solutions, and ultimately solving for y as y= \pm \sqrt{W\left( e^{2 sin(x)+2C}\right)}.
  • #1
Jamin2112
986
12

Homework Statement



dy/dx = (y cos x) / (1+y2)

Homework Equations



Meh

The Attempt at a Solution



I've made it to this point:

ln(y) + y2/2 = sinx + C.

But we can't figure out to solve for y. It seems impossible with the ln(y) hanging around.
 
Physics news on Phys.org
  • #2
As far as the differential equation goes, you've solved it. Does your problem statement requre you do express the solution as y=f(x) ?
 
  • #3
Theoretically, you can solve for y using the Lambert W function, defined as the inverse function to x*e^(x). But I doubt they'd make you do that.
 
  • #4
Char. Limit said:
Theoretically, you can solve for y using the Lambert W function, defined as the inverse function to x*e^(x). But I doubt they'd make you do that.

Is their another way, brah?
 
  • #5
PAllen said:
As far as the differential equation goes, you've solved it. Does your problem statement requre you do express the solution as y=f(x) ?

PAllen = Paul Allen?
 
  • #6
jamin2112 said:
pallen = paul allen?

myob
 
  • #7
Jamin2112 said:
Is their another way, brah?

Not really. I checked wolframalpha, and that's the only way.

[tex]log(y^2)+y^2 = 2 sin(x) + 2C[/tex]

[tex]e^{log(y^2)+y^2} = e^{2 sin(x) + 2C}[/tex]

[tex]y^2 e^{y^2} = e^{2 sin(x)+ 2C} [/tex]

[tex]y^2 = W\left( e^{2 sin(x)+2C}\right)[/tex]

[tex]y= \pm \sqrt{W\left( e^{2 sin(x)+2C}\right)}[/tex]
 

1. What is a "Difficult separable equation"?

A difficult separable equation is a type of mathematical equation that involves both variables and constants, and is difficult to solve using traditional algebraic methods. These equations typically require advanced techniques such as integration or substitution to find a solution.

2. How do you know if an equation is "Difficult separable"?

An equation is considered difficult separable if it cannot be easily solved by separating the variables and integrating. This can be determined by looking at the form of the equation and identifying any complex terms or functions that cannot be easily integrated.

3. What are some common techniques used to solve "Difficult separable equations"?

Some common techniques used to solve difficult separable equations include substitution, integration, and the use of special functions such as trigonometric or logarithmic functions. These techniques require a strong understanding of calculus and advanced algebraic concepts.

4. Are there any tips for solving "Difficult separable equations"?

One tip for solving difficult separable equations is to carefully examine the form of the equation and identify any patterns or relationships that can be used to simplify the equation. It can also be helpful to break the equation down into smaller, more manageable parts and solve them separately before combining the solutions.

5. What are some real-world applications of "Difficult separable equations"?

Difficult separable equations are commonly used in physics, engineering, and other fields of science to model complex systems and phenomena. They can be used to solve problems related to motion, heat transfer, fluid dynamics, and many other areas of science and engineering.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
755
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
811
  • Calculus and Beyond Homework Help
Replies
11
Views
957
Replies
14
Views
9K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
824
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
Back
Top