Hi how to solve? y(y-2x(y'))^3=(y')^2

  • Thread starter math_addict
  • Start date
  • Tags
    Hi
In summary, the author attempted to solve the differential equation y'=-\frac{1}{2y}, but was unsuccessful. He suggests that the equation be rewritten in terms of y and y' and then treated as a first-order DE. After integrating, any factors that were eliminated need to be looked at carefully, as they may be singular solutions.
  • #1
math_addict
5
0
Hello Everybody! Have any suggestions how to solve this differential equation, I haven't.. :(
y(y-2x(y'))^3=(y')^2
 
Physics news on Phys.org
  • #2
The only thing I can suggest is that you first treat it as a cubic equation for y' and try to solve for y'. But that obviously is going to be difficult!
 
  • #3
I tried this way, but it didn't give me any positive result. Moreover even Maple can't solve it exact.. it gives me huge expression..
 
  • #4
math_addict said:
I tried this way, but it didn't give me any positive result. Moreover even Maple can't solve it exact.. it gives me huge expression..

Are numeric or series solutions okay?
 
  • #5
The right solutions are y^2=2*(C^3)*x + C^2 and 27*(x^2)*(y^2)=1. Hmm.. Lagrange method doesn't fit properly here
 
  • #6
math_addict said:
Hello Everybody! Have any suggestions how to solve this differential equation, I haven't.. :(
y(y-2x(y'))^3=(y')^2

Hello math_addict,

This equation can be solved by the following method. If a DE of the first order but not first degree can be written as [itex]x-\phi(y,y')=0[/itex] the way to solve it is by taking the derivative of it and consider y and y' as variables. Then you need to substitute dy/y' for dx and you get a first degree DE in y and y'. After integrating this equation you need to eliminate y' between this solution and the original equation to obtain the general solution. In this process you need to make sure that any factors that are eliminated are to be looked upon with care afterwards because they can be singular solutions. Let's try this on your problem.

OK, first rewriting the original DE gives:

[tex]x=\frac{y}{2y'}-\frac{1}{2y^{1/3}y'^{1/3}}[/tex]

The derivative of this is:

[tex]dx=\frac{1}{2}\left[\frac{1}{y'}+\frac{1}{3y'^{1/3}y^{4/3}}\right]dy+
\frac{1}{2}\left[-\frac{y}{y'^2}+\frac{1}{3y^{1/3}y'^{4/3}}\right]dy'[/tex]

Substituting [itex]\frac{dy}{y'}=dx[/itex] gives after rewriting:

[tex]\frac{1}{2y'}\left[1-\frac{y'^{2/3}}{3y^{4/3}}\right]dy=
-\frac{y}{2y'^2}\left[1-\frac{y'^{2/3}}{3y^{4/3}}\right]dy'[/tex]

The factor between brackets can be eliminated but is important for later on to see if it is a singular solution. We have thus so far:

[tex]\frac{dy'}{y'}=-\frac{dy}{y}[/tex]

Which has as solution:

[tex]y'=\frac{C}{y}[/tex]

Substituting this in the original equation gives you:

[tex]y\left(y-2x\frac{C}{y}\right)^3=\frac{C^2}{y^2}[/tex]

Or, when we set C=K^3:

[tex]y^2=K^2+2xK^3[/tex]

This is the general solution to the DE. Now the factor that was eliminated needs to be looked upon. Setting it equal to zero:

[tex]1-\frac{y'^{2/3}}{3y^{4/3}}=0[/tex]

Or, after rewriting:

[tex]y'=3\sqrt{3}y^2[/tex]

Which has the solution:

[tex]-\frac{1}{y}=3\sqrt{3}x[/tex]

Or:

[tex]27x^2y^2=1[/tex]

And it is not possible to get this by setting K to any number and thus is a singular solution because it is also a solution to the original DE.

Hope this helps, coomast
 
  • #7
Thanks, Coomast, for help.
 

1. What is the equation "Hi how to solve? y(y-2x(y'))^3=(y')^2" used for?

This equation is used in mathematical modeling and is known as a nonlinear differential equation. It is often used to describe the relationship between multiple variables in a system.

2. What do the variables y and x represent in the equation?

The variable y represents the dependent variable, while x represents the independent variable. This means that the value of y depends on the value of x.

3. How can I solve this equation?

This equation can be solved using various techniques, such as separation of variables, substitution, or integration. It is recommended to consult a math textbook or seek assistance from a math tutor for step-by-step instructions on how to solve this specific equation.

4. Can this equation be solved analytically or numerically?

This equation can be solved both analytically and numerically. Analytical solutions involve finding an exact, algebraic expression for the solution, while numerical solutions involve using numerical methods to approximate the solution.

5. What applications does this equation have in real life?

This equation can be used to model various real-life situations, such as the spread of diseases, population growth, and chemical reactions. It can also be used in engineering and physics to describe the behavior of complex systems.

Similar threads

  • Differential Equations
Replies
2
Views
2K
  • Differential Equations
Replies
2
Views
916
  • Differential Equations
Replies
2
Views
802
  • Differential Equations
Replies
4
Views
1K
  • Differential Equations
Replies
10
Views
1K
  • Differential Equations
Replies
3
Views
1K
  • Differential Equations
Replies
3
Views
993
  • Differential Equations
Replies
3
Views
1K
  • Differential Equations
Replies
5
Views
1K
  • Differential Equations
Replies
8
Views
2K
Back
Top