# Differential equations substitution method

1. Sep 9, 2010

### cue928

1. The problem statement, all variables and given/known data

y'=y+y^3

2. Relevant equations

3. 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.

2. Sep 9, 2010

### epenguin

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.

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?

Last edited: Sep 9, 2010
3. Sep 10, 2010

### cue928

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. Sep 10, 2010

### jackmell

So you have:

$$\frac{dv}{v(1+v^2)}=dx$$

So integrate both sides. How about trying partial fractions on the left.

5. Sep 10, 2010

### Staff: Mentor

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. Sep 11, 2010

### HallsofIvy

Staff Emeritus
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.