# Homework Help: Going from dy/dx + P(x)y = Q(x)y^n to du/dx + (1-n)P(x) u = (1-n)Q(x)

1. Jul 13, 2013

### s3a

1. The problem statement, all variables and given/known data
There's this problem I'm presented with that has a part before asking the question which states a fact and, I want to be able to derive that fact.

I want to go from

dy/dx + P(x) y = Q(x) y^n

to

du/dx + (1 - n) P(x) u = (1 - n) Q(x).

2. Relevant equations
Bernoulli differential equation process, u = y^(1 - n)

3. The attempt at a solution
u = y^(1 – n), du/dx = (1 – n) y^(-n) dy/dx, u^(n – 1) = y, y^n = u^(n^2 – n)

dy/dx + P(x) y = Q(x) y^n

du/dx 1/(n – 1) y^n + P(x) u^(n – 1) = Q(x) u^(n^2 – n)

du/dx y^n + (1 – n) P(x) u^(n – 1) = (1 – n) Q(x) u^(n^2 – n)

du/dx u^(n^2 – n) + (1 – n) P(x) u^(n – 1) = Q(x) u^(n^2 – n)

du/dx + (1 – n) P(x) u^(-n^2 + 2n – 1) = (1 – n) Q(x)

but, instead, I should be getting

du/dx + (1 – n) P(x) u = (1 – n) Q(x).

What am I doing wrong?

Any input would be greatly appreciated!

Last edited: Jul 13, 2013
2. Jul 13, 2013

### CAF123

Hi s3a,

The error is there.

3. Jul 13, 2013

### s3a

Thanks for pointing out which part was wrong but, via "reverse engineering", I think it should be u^(n^2 - n + 1) but, looking at u = y^(1 - n), I get u^(1/(1 - n)) = y.

What do I do now?

4. Jul 13, 2013

### CAF123

$y = u^{\frac{1}{1-n}}$ is correct. Now you have $\frac{dy}{dx},\,\,y\,\,\text{and}\,\,y^n$ all in terms of $u$ and $\frac{du}{dx}$, so sub them into your original equation and simplify.

Last edited: Jul 13, 2013
5. Jul 13, 2013

### s3a

Sorry for still not getting it.

Here's my latest work.:

dy/dx + P(x) y = Q(x) y^n

du/dx y^n / (1 – n) + P(x) u^(1/(1 – n)) = Q(x) u^(n^2 – n)

du/dx u^(n^2 – n) + (1 – n) P(x) ^(1/(1 – n)) = (1 – n) Q(x) u^(n^2 – n)

du/dx + (1 – n) P(x) u^[(n^2 – n)/(1 – n)] = (1 – n) Q(x)

du/dx + (1 – n) P(x) u^[-n(1 – n)/(1 – n)] = (1 – n) Q(x)

du/dx + (1 – n) P(x) u^(-n) = (1 – n) Q(x)

6. Jul 13, 2013

### CAF123

You appear to still have the expression for y in terms of u incorrect.
If $u = y^{1-n}$ then $u^{1/(1-n)} = \left(y^{1-n}\right)^{1/(1-n)} = y$ as you mentioned in your last post.

How did you get $y = u^{n^2 -n}$?

7. Jul 14, 2013

### s3a

Actually, I solved it, now!

Thanks! :)