1. ## Initial value problem

Hello all,

Can you see if this makes sense please.

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

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

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

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

$\displaystyle y(0)=4$

$\displaystyle 4=(1)C=4$

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

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

2. Originally Posted by bigroo
Hello all,

Can you see if this makes sense please.

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

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

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

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

$\displaystyle y(0)=4$

$\displaystyle 4=(1)C=4$

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

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.

3. 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.

4. 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.

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

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

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

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

$\displaystyle y(0)=4$

$\displaystyle 4=(1)C=4$

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

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

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

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

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

and so

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

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

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

Is this an error or correct? Thanks

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

Is this an error or correct? Thanks
Sorry - it's a typo. It should be plus.

7. Originally Posted by Danny
Sorry - it's a typo. It should be plus.
haha no worries, trying to confuse me even more!

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

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

$\displaystyle y(0)=4$

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

So this makes:

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

Is this correct now?

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

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

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

$\displaystyle y(0)=4$

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

So this makes:

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

Is this correct now?
Yes - it looks real good!

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