# Separating the variables or some other way?

• Apr 24th 2010, 01:03 PM
DangerousDave
Separating the variables or some other way?
Hello all,

I'm being confused by a differential equation problem. Solve:

(2x + 4x^2) dy/dx = e^y

My approach has been to transfer the function of x to the right i.e. 1/it and then the e^y to the right yielding e^-y, and then the usual separating the variables method with partial fractions to integrate the right hand side. However, to find y, I ultimately end up with the log of a log, and I was wondering if this is bad news, and whether I should perhaps be using an entirely different method.

Any help appreciated.
• Apr 24th 2010, 02:11 PM
AllanCuz
Quote:

Originally Posted by DangerousDave
Hello all,

I'm being confused by a differential equation problem. Solve:

(2x + 4x^2) dy/dx = e^y

My approach has been to transfer the function of x to the right i.e. 1/it and then the e^y to the right yielding e^-y, and then the usual separating the variables method with partial fractions to integrate the right hand side. However, to find y, I ultimately end up with the log of a log, and I was wondering if this is bad news, and whether I should perhaps be using an entirely different method.

Any help appreciated.

$\displaystyle (2x + 4x^2) \frac{dy}{dx} = e^y$

$\displaystyle e^{-y} dy = \frac{1}{x(2+4x)} dx$

Your original conclusion that this is a seperable equation and that you must use partial fractions on the R.H.S is correct.

Let's just work through it.

Partial Fractions,

$\displaystyle \frac{A}{x} + \frac{B}{2+4x} = \frac{A(2+4x) + B(x)}{x(2+4x)}$

$\displaystyle 2A = 1$

$\displaystyle 4A + B = 0$

$\displaystyle A= \frac{1}{2}$ and $\displaystyle B = -2$

So the R.H.S becomes

$\displaystyle \frac{1}{2} \int \frac{1}{x} dx - 2 \int \frac{1}{2+4x}dx$

$\displaystyle \frac{1}{2} lnx - \frac{1}{2} ln(2+4x) + C$

So far so good.

Now the L.H.S

$\displaystyle \int e^{-y} dy = - e^{-y}$

L.H.S = R.H.S

$\displaystyle -e^{-y}=\frac{1}{2} lnx - \frac{1}{2} ln(2+4x) + C$

$\displaystyle e^{-y}=\frac{1}{2} ln(2+4x) - \frac{1}{2} lnx - C$

Clearly to bring down Y we would use ln as you suggested. Is there any problem with having (as an example)

$\displaystyle ln ( \frac{1}{2} ln(2+4x) )$

$\displaystyle \frac{1}{2} ln(2+4x)$

Well, this will evaluate to a number! So essentially, we'll have ln(some number) which is clearly no problem. So to answer your question, this is not a problem.

But you should note for

$\displaystyle ln ( \frac{1}{2} ln(2+4x) )$

We have the restriction that,

$\displaystyle \frac{1}{2} ln(2+4x) \ge 1$

Other then that, we are fine!
• Apr 24th 2010, 02:16 PM
dwsmith
Quote:

Originally Posted by DangerousDave
Hello all,

I'm being confused by a differential equation problem. Solve:

(2x + 4x^2) dy/dx = e^y

My approach has been to transfer the function of x to the right i.e. 1/it and then the e^y to the right yielding e^-y, and then the usual separating the variables method with partial fractions to integrate the right hand side. However, to find y, I ultimately end up with the log of a log, and I was wondering if this is bad news, and whether I should perhaps be using an entirely different method.

Any help appreciated.

$\displaystyle \frac{dx}{2x+4x^2}=\frac{dy}{e^Y}\Rightarrow$

$\displaystyle \int\frac{dx}{2x+4x^2}=\int\frac{dy}{e^y}$

By partial fractions, we obtain:

$\displaystyle \frac{1}{2}\int\frac{1}{x}dx-\int\frac{1}{1+2x}dx=-e^{-y}$

Try finishing from here.
• Apr 24th 2010, 02:20 PM
AllanCuz
Quote:

Originally Posted by dwsmith
$\displaystyle \frac{dx}{2x+4x^2}=\frac{dy}{e^Y}\Rightarrow$

$\displaystyle \int\frac{dx}{2x+4x^2}=\int\frac{dy}{e^y}$

By partial fractions, we obtain:

$\displaystyle \frac{1}{2}\int\frac{1}{x}dx-\int\frac{1}{1+2x}dx=-e^{-y}$

Try finishing from here.

He can solve the problem. He wanted to know if there's any implication of having

$\displaystyle ln(ln(something))$

Which is what you'll get when you finish this problem.
• Apr 25th 2010, 07:35 PM
DangerousDave
Many thanks for your help. Is it definitely going to be a log of a log in the answer then? There is no alternative method that will produce an alternative neater formulation? (I only ask because these problems usually seem to pan out a bit neater, and when it's messy(ish) then I start to think I did it wrong)