Initial value problem

**bigroo** · December 7th 2009, 07:23 AM

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.
Would appreciate any help please.

Thanks in advance.

**Danny** · December 7th 2009, 08:44 AM

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.
Would appreciate any help please.

Thanks in advance.

It looks like some sign problems here. Differentiate y and see if you get back your DE.

**bigroo** · December 7th 2009, 09:10 AM

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.

**Danny** · December 7th 2009, 09:29 AM

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

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.
Would appreciate any help please.

Thanks in advance.

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

$<br /> y = -\frac{1}{4} \int \frac{du}{u} + c<br />$

integrate, then back substitute. Then you can use your IC to find c.

**bigroo** · December 7th 2009, 10:38 AM

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

Is this an error or correct? Thanks

**Danny** · December 7th 2009, 11:09 AM

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.

**bigroo** · December 7th 2009, 12:12 PM

Originally Posted by Danny

Sorry - it's a typo. It should be plus.

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?

**Danny** · December 7th 2009, 12:29 PM

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!

**bigroo** · December 7th 2009, 12:36 PM

Great stuff, I've added a thanks for all of your inspiration.

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