Initial value problem

• Dec 7th 2009, 08:23 AM
bigroo
Initial value problem
Hello all,

Can you see if this makes sense please.

$\frac{dy}{dx} = \frac{e^{-4x} - x}{e^{-4x} + 2x^{2} + 1} when y(0)=4$

$\int \frac{f'(x)}{f(x)}=ln(f(x))+c$

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

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

$y(0)=4$

$4=(1)C=4$

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

I'm not sure but I think it all goes wrong when I put all equal to y.

• Dec 7th 2009, 09:44 AM
Jester
Quote:

Originally Posted by bigroo
Hello all,

Can you see if this makes sense please.

$\frac{dy}{dx} = \frac{e^{-4x} - x}{e^{-4x} + 2x^{2} + 1} when y(0)=4$

$\int \frac{f'(x)}{f(x)}=ln(f(x))+c$

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

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

$y(0)=4$

$4=(1)C=4$

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

I'm not sure but I think it all goes wrong when I put all equal to y.

It looks like some sign problems here. Differentiate y and see if you get back your DE.
• Dec 7th 2009, 10:10 AM
bigroo
Thanks Danny,

I'm sure it's a sign issue too but I'm pulling my hair out. Where do you think the error occurs?

I can't see anything online to help with initial value problems so any help or solutions & their stages will be gratefully appreciated.
• Dec 7th 2009, 10:29 AM
Jester
Quote:

Originally Posted by bigroo
Thanks Danny,

I'm sure it's a sign issue too but I'm pulling my hair out. Where do you think the error occurs?

I can't see anything online to help with initial value problems so any help or solutions & their stages will be gratefully appreciated.

If you put in an extra step it might help

Quote:

Originally Posted by bigroo
Hello all,

Can you see if this makes sense please.

$\frac{dy}{dx} = \frac{e^{-4x} - x}{e^{-4x} + 2x^{2} + 1} when y(0)=4$

$\int \frac{f'(x)}{f(x)}=ln(f(x))+c$

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

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

$y(0)=4$

$4=(1)C=4$

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

I'm not sure but I think it all goes wrong when I put all equal to y.

Your separated DE is $y = \int \frac{e^{-4x} - x}{e^{-4x} + 2x^{2} + 1} \,dx$

If you let $u = e^{-4x} + 2x^2 + 1$ then

$du = -4\left(e^{-4x} - x\right)\,dx$

and so

$
y = -\frac{1}{4} \int \frac{du}{u} + c
$

integrate, then back substitute. Then you can use your IC to find c.
• Dec 7th 2009, 11:38 AM
bigroo
Why would you let $u = e^{-4x}$ minus $2x^2 + 1$ when it is a $+$

Is this an error or correct? Thanks
• Dec 7th 2009, 12:09 PM
Jester
Quote:

Originally Posted by bigroo
Why would you let $u = e^{-4x}$ minus $2x^2 + 1$ when it is a $+$

Is this an error or correct? Thanks

Sorry - it's a typo. It should be plus. (Giggle)
• Dec 7th 2009, 01:12 PM
bigroo
Quote:

Originally Posted by Danny
Sorry - it's a typo. It should be plus. (Giggle)

haha no worries, trying to confuse me even more!

$y=-\frac{1}{4}ln(u)+c$

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

$y(0)=4$

$c=\frac{1}{4}ln(2)+4$

So this makes:

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

Is this correct now?
• Dec 7th 2009, 01:29 PM
Jester
Quote:

Originally Posted by bigroo
haha no worries, trying to confuse me even more!

$y=-\frac{1}{4}ln(u)+c$

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

$y(0)=4$

$c=\frac{1}{4}ln(2)+4$

So this makes:

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

Is this correct now?

Yes - it looks real good!
• Dec 7th 2009, 01:36 PM
bigroo
Great stuff, I've added a thanks for all of your inspiration. (Bow)