# Separation of Variables

• Aug 7th 2012, 12:31 AM
Separation of Variables
Hello, I'm a little stuck on this problem:

Find a solution to the differential equation subject to the initial conditions.

dz

dt = tez , through the origin.

Here is how I tried to solve the problem:

dz

dt = tez

multiply both sides by dt, and then divide both sides by ez to get like variables with like variables:

dz

ez = tdt

Take the integral of both sides:

(integral sign) 1/ez dz = (integral sign) t dt

substitution
w = ez
dw = ez dz

(integral sign)1/w dz = (integral sign) t dt

ln|w| = t2/2 + C

ln|ez| =
t2/2 + C

exponentiate both sides to get rid of the natural log:
eln|ez| = et2/2 + C

the e and the natural log cancel out
ez = e(t^2)ec

ez = Ce(t^2) ----(where C is just a constant)

Now this is where I am stuck. Have I done it correctly so far? What do I need to do next? Please help. Thank you.
• Aug 7th 2012, 01:12 AM
jgv115
Re: Separation of Variables
Hmm...
$\displaystyle \int e^{-z} dz = -e^{-z}$

You can't do that w substitution as you will get $\displaystyle dw =e^{z} dz$ but there is no $\displaystyle e^{z}$ in the integral to cancel it out
• Aug 7th 2012, 01:23 AM
a tutor
Re: Separation of Variables
I would agree that you should just use the result given by jgv115 but you can use the substitution.

You should end up with $\displaystyle \displaystyle \int \frac{1}{w^2}dw$
• Aug 7th 2012, 01:33 AM
Prove It
Re: Separation of Variables
Quote:

Originally Posted by a tutor
I would agree that you should just use the result given by jgv115 but you can use the substitution.

You should end up with $\displaystyle \displaystyle \int \frac{1}{w^2}dw$

You can use the result IF you mutliply the top and bottom by \displaystyle \displaystyle \begin{align*} e^z \end{align*} first.
• Aug 7th 2012, 01:36 AM
a tutor
Re: Separation of Variables
Quote:

Originally Posted by Prove It
You can use the result IF you mutliply the top and bottom by \displaystyle \displaystyle \begin{align*} e^z \end{align*} first.

I see, you need to multiply by 1 before it works..:p
• Aug 7th 2012, 01:54 AM
tom@ballooncalculus
Re: Separation of Variables
Just in case a picture also helps...

http://www.ballooncalculus.org/draw/sep/three.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case t), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
• Aug 7th 2012, 09:44 AM
Re: Separation of Variables
I'm sorry. I am still not sure what substitution to use. How would I end up with my substitution as 1/w2 ​??
• Aug 7th 2012, 10:10 AM
a tutor
Re: Separation of Variables
Quote:

I'm sorry. I am still not sure what substitution to use. How would I end up with my substitution as 1/w2 ​??

$\displaystyle \displaystyle \int \frac{1}{w^2}dw$ is not your substitution it is the result of doing the substitution you started.

You had $\displaystyle w=e^z$

so

$\displaystyle dw=e^z \ dz$

and then

$\displaystyle \displaystyle \int \frac{1}{e^z}dz=\int \frac{1}{w}\frac{dw}{w}=\int\frac{1}{w^2}dw$
• Aug 7th 2012, 03:26 PM
Prove It
Re: Separation of Variables
Quote:

Originally Posted by a tutor
$\displaystyle \displaystyle \int \frac{1}{w^2}dw$ is not your substitution it is the result of doing the substitution you started.

You had $\displaystyle w=e^z$

so

$\displaystyle dw=e^z \ dz$

and then

$\displaystyle \displaystyle \int \frac{1}{e^z}dz=\int \frac{1}{w}\frac{dw}{w}=\int\frac{1}{w^2}dw$

$\displaystyle \displaystyle = \int{w^{-2}\,dw}$, which should be easy to integrate...