Differential equations substitution method

In summary: In this problem, it is totally useless. :wink:In summary, the problem involves solving the differential equation y'=y+y^3. The attempt at a solution involves setting y=v and substituting back into the original equation, resulting in dv/(v+v^3) = dx. Integrating both sides and using partial fractions yields the final answer y^2 = 1/(C(e^-2x)-1). One should recognize that the first steps of the solution are not necessary and can be skipped.
  • #1
cue928
130
0

Homework Statement



y'=y+y^3

Homework Equations





The Attempt at a Solution



I set y=v, dy/dx = dv/dx. Substituted back into original equation ST dv/dx = v + v^3. Cross multiply, then divide yielded dv/(v+v^3) = dx. After that, I have no clue. The book gives the following answer: y^2 = 1/(C(e^-2x)-1)

Any guidance would be appreciated.
 
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  • #2
You should (saving yourself time) recognise that your first steps would have yielded you exactly nothing, it is like you have just changed a symbol and called "y" "v" instead.

The differential equation part of this is nothing at all really; dy/dx = f(x) or dy/dx = f(y) are nothing different. Just divide by the RHS and mytiply by dx and you will see you have got dx = something involving only y and dy, and so x = the integral of this something. That integration therefore is the hard part, (plus maybe getting the result into nicest form) however you have surely done things like it and know what to do. :smile:

Oops reread your post and apart from the useless renaming y as v you have done what I said. I am sure you previously did harder exercises on this sort of expression than this before. The denominator (v+v3) is a product OK? :wink:
 
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  • #3
epenguin said:
You should (saving yourself time) recognise that your first steps would have yielded you exactly nothing, it is like you have just changed a symbol and called "y" "v" instead.

The differential equation part of this is nothing at all really; dy/dx = f(x) or dy/dx = f(y) are nothing different. Just divide by the RHS and mytiply by dx and you will see you have got dx = something involving only y and dy, and so x = the integral of this something. That integration therefore is the hard part, (plus maybe getting the result into nicest form) however you have surely done things like it and know what to do. :smile:

Oops reread your post and apart from the useless renaming y as v you have done what I said. I am sure you previously did harder exercises on this sort of expression than this before. The denominator (v+v3) is a product OK? :wink:

The renaming was not "useless" - it was, however, required by our professor. It's easier to do what they ask rather than lose points.
 
  • #4
cue928 said:
yielded dv/(v+v^3) = dx. After that, I have no clue..


So you have:

[tex]\frac{dv}{v(1+v^2)}=dx[/tex]

So integrate both sides. How about trying partial fractions on the left.
 
  • #5
cue928 said:
The renaming was not "useless" - it was, however, required by our professor. It's easier to do what they ask rather than lose points.

Any substitution y = v is useless. As epenguin already pointed out, all this does is change any occurrence of y (and dy) with v and dv, to no advantage that I can see.

I can't believe that your professor would require such a useless step.

The original DE can be separated without using this substitution, as
dy/(y^3 + y) = dx

Using jackmell's suggestion, you have
dy/(y(y^2 + 1)) = dx
 
  • #6
cue928 said:
The renaming was not "useless" - it was, however, required by our professor. It's easier to do what they ask rather than lose points.
I suggest you ask your professor about that. I suspect he/she was referring to a problem where such a substitution did make a difference.
 

1. What is the substitution method in differential equations?

The substitution method is a technique used to solve differential equations by substituting a new variable for the dependent variable. This helps to simplify the equation and make it easier to solve.

2. When should the substitution method be used in solving differential equations?

The substitution method is most useful when the differential equation is non-linear and cannot be solved using traditional methods such as separation of variables or integrating factors.

3. How does the substitution method work?

The substitution method works by replacing the dependent variable with a new variable, usually denoted as u or v, and then solving for this new variable. Once the solution for the new variable is found, it can be substituted back into the original equation to find the solution for the dependent variable.

4. What are the advantages of using the substitution method in solving differential equations?

The substitution method allows for the solution of non-linear differential equations, which cannot be solved using traditional methods. It also simplifies the equation, making it easier to solve and understand the behavior of the system described by the equation.

5. Are there any limitations to using the substitution method?

One limitation of the substitution method is that it may not always lead to an exact solution. In some cases, it may only provide an approximation or an implicit solution. Additionally, it may be more time-consuming compared to other methods for solving differential equations.

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