# Modeling epidemics - solving differential equation

1. Apr 22, 2017

### Schaus

1. The problem statement, all variables and given/known data
Solve for y: $\frac {dy}{dx} = \frac {1+y^6}{xy^5}$ , where y(1) = 1.
Answer $y = \sqrt[6] {2x-1}$
2. Relevant equations

3. The attempt at a solution
$\frac {dy}{dx} = \frac {1+y^6}{xy^5}$
$\frac{dy (y^5)}{1+y^6} = dx \frac {1}{x}$
u= 1+y6
$\frac {du}{y^5}=dx$
$\int \frac{1}{u}du = \int \frac {1}{x}dx$
$\ln|1+y^6| = \ln|x| + C$
The natural logs cancel out.
Substituting in my (1,1)
$1+1^6 = 0+C$
$C=2$
This is where I'm a bit lost. I'm not sure where I messed up but I don't know how to get the 2x-1. Any help would be greatly appreciated!

Last edited: Apr 22, 2017
2. Apr 22, 2017

### Staff: Mentor

You lost me at the substitution. Shouldn't there be a $\frac{1}{6}$ somewhere? I also got a different result than the one you expect as an answer. Did you differentiate it? Plus $\ln |1| = 0$, not $1$.

3. Apr 22, 2017

### Schaus

Yes I definitely messed up on the $\ln |1| = 0$ but I don't understand where this $\frac {1}{6}$ is coming from.

4. Apr 22, 2017

### Staff: Mentor

If $u=1+y^6$ then $\frac{du}{dy}=6y^5$. Where did the factor go to?

5. Apr 22, 2017

### Schaus

Oh I see. Would the integral look like this then
$\ln |6(1+y^6)|$? Or do I put the $\frac{1}{6}$ on the outside of the natural log?

6. Apr 22, 2017

### Staff: Mentor

As I couldn't follow your calculation beyond the substitution I did it step by step:
$\frac{dy}{dx}=\frac{u}{xy^5}$ and $y^5=\frac{1}{6}\frac{du}{dy}$. Now the $dy$ canceled out and leaves $\frac{dx}{x}=\frac{1}{6}\frac{du}{u}$. The factor in front of the logarithm becomes an exponent in the logarithm and thus the $6-$th root. Next I replaced $u$ and calculated the constant $C$. The rest was some algebra. What's always a good idea in such cases, is to differentiate the result again and check whether the original equation pops up.

7. Apr 22, 2017

### Schaus

Sorry, I'm still a little confused. After integrating should it look like this then?
$\frac {1}{6} \int \frac {1}{u}$
$= \frac {1}{6} \ln |1+y^6| = \ln|x| + C$

8. Apr 22, 2017

### Staff: Mentor

Yes. Or likewise $\ln|x| + C = \ln \sqrt[6]{1+y^6}$ and $C = \frac{1}{6} \ln (1+1^6) - \ln 1$.

9. Apr 22, 2017

### Schaus

I'm still not understanding how to get the answer. If $\frac {1}{6} \ln(2)-0 = C$ then C ≈ 0.1155... or at least that is what my calculator says.

10. Apr 22, 2017

### Staff: Mentor

You don't need to calculate it. Just drop the $\ln$ as you did in your solution. It's probably more convenient to first multiply by $6$ and get $\ln |1+y^6| = \ln 2|x|^6$ and then drop the $\ln$ as you did before.

11. Apr 22, 2017

### Schaus

Now I got my answer I'm just curious where the highlighted 2 came from.

12. Apr 22, 2017

### Staff: Mentor

The $2$ is from $\ln 2 + \ln |x|^6 = \ln (2x^6)$ or what did you mean? And it isn't your answer, it is $y = \sqrt[6]{2x^6-1}$, not $y = \sqrt[6]{2x-1}$. At least if I made no mistake, but I also differentiated it again.

13. Apr 22, 2017

### Schaus

The answer in my course is 2x-1 but it could be wrong, it is easily the worst course I've done. Thank you for all your help though.

14. Apr 23, 2017

### Ray Vickson

The given solution $y(x) = \sqrt[6]{2x-1}$ is incorrect: if you substitute it into the DE you will see that it does not work.

15. Apr 23, 2017

### epenguin

Unless the problem was miscopied at some stage and the x should have been in the numerator.

16. Apr 24, 2017

### Ray Vickson

In that case the DE would be $y' = x(1+y^6)/y^5, \; y(1) = 1$, whose solution is
$$y(x) = \left( 2 e^{3x^2-3}-1\right)^{1/6}.$$

17. Apr 24, 2017

### epenguin

OK, OK.
Just Unless the problem was miscopied at some stage....