# Modeling epidemics - solving differential equation

## Homework Statement

Solve for y: ##\frac {dy}{dx} = \frac {1+y^6}{xy^5}## , where y(1) = 1.
Answer ## y = \sqrt {2x-1}##

## 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:

fresh_42
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##.

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

fresh_42
Mentor
Yes I definitely messed up on the ##\ln |1| = 0## but I don't understand where this ##\frac {1}{6}## is coming from.
If ##u=1+y^6## then ##\frac{du}{dy}=6y^5##. Where did the factor go to?

• 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?

fresh_42
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.

• 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##

fresh_42
Mentor
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##
Yes. Or likewise ##\ln|x| + C = \ln \sqrt{1+y^6}## and ##C = \frac{1}{6} \ln (1+1^6) - \ln 1##.

• 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.

fresh_42
Mentor
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.
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.

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

fresh_42
Mentor
Now I got my answer I'm just curious where the highlighted 2 came from.
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{2x^6-1}##, not ##y = \sqrt{2x-1}##. At least if I made no mistake, but I also differentiated it again.

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.

Ray Vickson
Homework Helper
Dearly Missed
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.
The given solution ##y(x) = \sqrt{2x-1}## is incorrect: if you substitute it into the DE you will see that it does not work.

• Schaus
epenguin
Homework Helper
Gold Member
The given solution ##y(x) = \sqrt{2x-1}## is incorrect: if you substitute it into the DE you will see that it does not work.

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

Ray Vickson
Homework Helper
Dearly Missed
Unless the problem was miscopied at some stage and the x should have been in the numerator.

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

epenguin
Homework Helper
Gold Member
Unless the problem was miscopied at some stage and the x should have been in the numerator.

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

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