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First order linear ODE's.

  • #1

Homework Statement



Solve this first order linear ODE

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


Homework Equations





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
 

Answers and Replies

  • #2
vela
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You didn't calculate the integrating factor correctly. Recheck the integral.
 
  • #3
You didn't calculate the integrating factor correctly. Recheck the integral.
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
vela
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No, it should have an x in it.

EDIT: It refers to the integrating factor.
 
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  • #5
LCKurtz
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ohh it should be a half; thanks

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

the answer in my book has 2x5 + Ax2
The book is correct. Show your steps and we can help you find where you went astray.
 
  • #6
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  • #7
This isn't true; take a look at log/exponential rules.
okbthanks I figured it out now../.

I=1/x2

although I am alittle confused about this integral

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
HallsofIvy
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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 [itex]e^{-2\ln(x)}[/itex] but that is NOT [itex]e/(-2\ln(x))[/itex]. I can't imagine where you got that. It is, rather [itex]1/e^{2\ln(x)}[/itex].

But what you should have done first was simplify -2 ln(x). Do you know how to take that -2 'into' the logarithm?
 
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  • #9
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 [itex]e^{-2\ln(x)}[/itex] but that is NOT [itex]e/(-2\ln(x))[/itex]. I can't imagine where you got that. It is, rather [itex]1/e^{2\ln(x)}[/itex].

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

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
 
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