exponential problem

[TheLivingMartyr]

I was solving some differentials for practice, and the final question I was given was

[math]

(x^2-1) \frac{dy}{dx} = xy

[/math]

 

I went about solving this by doing (some steps missed out):

[math]

\int\frac{1}{y}dy = \int x^3-x dx

[/math]

 

[math]

ln(y) = \frac{2x^4-4x^2}{8}+k

[/math]

 

And then, I was left with this, extremely irritating answer, which didn't look right at all, and incidently, wasn't right:

[math]

y = Ae^\frac{2x^4-4x^2}{8}

[/math]

 

Note: A = e^k

 

I don't think I've done anything wrong with the integration, I just have a feeling there's a way to simplify this that I don't know. Any suggestions please: I need to learn!

 

ADDITION: It would be helpful if you could explain how to get to the answer, rather than just telling me, so I can do it with other similar problems

 

ADDITION II: I don't think dividing the whole fraction by 2 will really make much difference, as you'll still have a big fraction as the exponent

Edited August 23, 2011 by TheLivingMartyr

[Schrödinger's hat]

I was solving some differentials for practice, and the final question I was given was

[math] (x^2-1) \frac{dy}{dx} = xy [/math]

 

I went about solving this by doing (some steps missed out):

[math] \int\frac{1}{y}dy = \int x^3-x dx [/math]

I'm not entirely sure this is the right time of night for me to be dividing things, but isn't that:

[math] \int\frac{1}{y}dy = \int \frac{x}{x^2 - 1} dx [/math]

Which will have a somewhat different solution (if you're stuck, I'd probably try u substitution first).

[math] ln(y) = \frac{2x^4-4x^2}{8}+k [/math]

 

And then, I was left with this, extremely irritating answer, which didn't look right at all, and incidently, wasn't right:

[math] y = Ae^\frac{2x^4-4x^2}{8} [/math]

 

Note: A = e^k

 

I don't think I've done anything wrong with the integration, I just have a feeling there's a way to simplify this that I don't know. Any suggestions please: I need to learn!

 

ADDITION: It would be helpful if you could explain how to get to the answer, rather than just telling me, so I can do it with other similar problems

 

ADDITION II: I don't think dividing the whole fraction by 2 will really make much difference, as you'll still have a big fraction as the exponent

 

As to simplifying the exponential, I'd write that something along the lines of [math]Ae^{\frac{x^2}{4}(x^2-2)}[/math]

Which is still a bit more messy than is usual for contrived exercises.

Looking at the integral formed above the logs will probably cancel and you'll get a nice polynomial.

[TheLivingMartyr]

Oh god, I've done it again. yes, you're right, it does make

[math]

\int\frac{1}{y}dy = \int\frac{x}{x^2-1}dx

[/math]

once again, I have made a simple tiny error that has messed up everything. If people listened to me, they'd agree that:

When

[math]

xy = z

[/math]

,

[math]

y = xz

[/math]

God i'm a fool

 

oh dear, that looks even worse though!

 

the final solution of that is:

[math]

Ae^\frac{log(x+1)+log(x-1)}{2}

[/math]

 

well, according to an integral calc. I don't really know how to integrate something like

[math]

\int\frac{x}{x^2-1}dx

[/math]

 

Wait, is this below correct?

[math]

\frac{log(x+1)+log(x-1)}{2} = \frac{log(x^2-1)}{2}

[/math]

 

because if it is, the the solution looks alot more tidy

[Fuzzwood]

Jup, adding logs means multiplying of the terms inbetween the bracket, which in your case comes down to a² - b² = (a+b)(a-b)

Edited August 23, 2011 by Fuzzwood

[TheLivingMartyr]

sorted, the solution is:

[math]

y = A(x^2-1)

[/math]

where A = c²

 

thanks for helping

[Schrödinger's hat]

God i'm a fool

 

oh dear, that looks even worse though!

Don't stress about it, we all make silly errors.

It doesn't seem correlated with intelligence at all (you should see my mathematician friends trying to do arithmetic sometimes -- esp after an all nighter).

If it's bad enough to be a problem then the only ways I know of to improve it are to do lots and lots of practise or marking assignments. Even then the effects are only temporary.

If you want some practise exercises one place to get auto-generated (basic arithmetic and algebra) ones is the practise section of khan academy. Alternatively just go through your textbook and do all the exercises and get algebra practise that way (doesn't really matter if they're calculus exercises, there's still algebra practise).

http://www.khanacademy.org

sorted, the solution is:

[math] y = A(x^2-1) [/math]

where A = c²

 

thanks for helping

Double check that (what happened to that two?). There should be a square or square root somewhere.

Tip: Always put the function you get back through your ODE to see if it works.

 

Also, if you want to know how to integrate something like that, use u substitution.

The general theme is look for something that looks like:

[math]\int\frac{du(x)}{dx}f(u(x)) dx[/math]

Then we can use the chain rule to turn it into:

[math]\int f(u)du[/math]

 

So we see that the derivative of [math]x^2-1[/math] is [math]2x[/math]

To get a 2 we can multiply by 2/2, then take the 1/2 out of the integral to get:

[math]\frac{1}{2}\int\frac{2x}{x^2 - 1} dx[/math]

Set [math]u=x^2-1[/math] which leaves us with:

[math]\frac{1}{2}\int\frac{1}{u} du[/math]

or [math]\frac{1}{2}\ln{u}+C=\frac{1}{2}\ln({x^2-1})+C[/math]

You may want to use a few more steps (or if you don't follow I can explain further -- this is a bit terse).

Edited August 24, 2011 by Schrödinger's hat

[Fuzzwood]

Putting 1/2 in front of the integral is wrong. You would only need that if the numerator would be x instead of 2x. Now you simply have to replace the 2x by du

 

u = x² - 1, du = 2x dx

Edited August 24, 2011 by Fuzzwood

[Schrödinger's hat]

Putting 1/2 in front of the integral is wrong. You would only need that if the numerator would be x instead of 2x. Now you simply have to replace the 2x by du

 

u = x² - 1, du = 2x dx

 

Pardon? I don't quite understand (was this even directed at me?).

[Fuzzwood]

Well you are talking about putting 1/2 in front of the integral. Try to differentiate your integral again. I'm quite sure you will be missing a factor 2.

[Schrödinger's hat]

The original integral was [math]\int \frac{x}{x^2 - 1} dx[/math]

I added the 2 and the half by multiplying by 1=2/2, take the 1/2 out the front then you have 2x.

[Fuzzwood]

Ah didn't notice that. I will shut up now

[Schrödinger's hat]

Good good. I was getting a little worried there.

If I was that clueless on a separable ODE, something somewhere had gone terribly terribly wrong.