# Homework Help: First order linear ODE's.

1. Sep 8, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

Solve this first order linear ODE

x (dy/dx)-2y=6x5

2. Relevant equations

3. The attempt at a solution

x (dy/dx)-2y=6x5

divide through by x

dy/dx -2y/x = 6x4

I= e∫-2/x dx

=e-ln(2) =-2

∴ -2(dy/dx) + 4y/x = -12x4

d/dx(-2y) = -12x5/5 + C

y= 6/5(x5) - C/2

this is wrong but where did I go wrong?

Thanks

2. Sep 8, 2012

### vela

Staff Emeritus
You didn't calculate the integrating factor correctly. Recheck the integral.

3. Sep 8, 2012

### charmedbeauty

ohh it should be a half; thanks

so it should be y = y=-6/5x5 - 2C

the answer in my book has 2x5 + Ax2

4. Sep 8, 2012

### vela

Staff Emeritus
No, it should have an x in it.

EDIT: It refers to the integrating factor.

Last edited: Sep 8, 2012
5. Sep 8, 2012

### LCKurtz

The book is correct. Show your steps and we can help you find where you went astray.

6. Sep 9, 2012

### Bohrok

This isn't true; take a look at log/exponential rules.

7. Sep 9, 2012

### charmedbeauty

okbthanks I figured it out now../.

I=1/x2

if you work out the integral of -2/x

-2ln(x)

so I thought e^-2ln(x)

which is e^1/-2ln(x)

should be -1/2x

and not -1/x^2

??

thanks.

8. Sep 9, 2012

### HallsofIvy

You seem to be making one algebra mistake after another. Yes, the integral of -2/x is -2ln(x) and the exponential of that is $e^{-2\ln(x)}$ but that is NOT $e/(-2\ln(x))$. I can't imagine where you got that. It is, rather $1/e^{2\ln(x)}$.

But what you should have done first was simplify -2 ln(x). Do you know how to take that -2 'into' the logarithm?

Last edited by a moderator: Sep 9, 2012
9. Sep 9, 2012

### charmedbeauty

yeah I made a stupid mistake of course e-2ln(x) = 1/e2ln(x)

its been a while so first I should of done

-2ln(x) = -ln(x2)

Thanks

Last edited by a moderator: Sep 9, 2012