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March 11th, 2011, 10:00 AM   #1
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Linear differential equations

Problem:
Use the integrating factor method to solve the following linear differential equation

x dy/dx - y = x

I used the same method as I did last night with the other equations and found the integrating factor to be ?(x)=e^?-1/x dx = e^-ln(x) = -x (or in this case would it be just (x) ? )

Then Multiplying the standard form equation by (-x) ,

-x(dy/dx)-y= -x

Integrating to get y=1/2 x + A / x Im a bit unsure on it
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March 11th, 2011, 10:12 AM   #2
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Re: Linear differential equations



Now try it.
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March 11th, 2011, 10:26 AM   #3
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Re: Linear differential equations

y= - (ln(x) +A / x)
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March 11th, 2011, 10:36 AM   #4
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Re: Linear differential equations

We have in standard form:



Multiplying through by gives:





Integrate w.r.t. x:



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March 11th, 2011, 11:38 AM   #5
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Re: Linear differential equations

oh yes, i made the mistake on multiplying by the integrating factor.

my next equation

cos(t) dx/dt + x sin(t) = cos^2 (t)

Re write in Std.form, divide by cos(t)
dx/dt + x (sin(t)/cos(t)) = cos(t)

Obtain Integrating factor,

?(x)=e^?sin(t)/cos(t) = e^-ln(cos(t)) = -cos(t) or is it 1/cos(t)
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March 11th, 2011, 12:37 PM   #6
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Re: Linear differential equations

It is , since by the properties of logarithms, we have .

This transforms your equation in standard form to:



Then you have:

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March 11th, 2011, 01:09 PM   #7
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Re: Linear differential equations

oh ok, yes i thought the laws of logs would be involved.

so then i integrate d/dx (x sect)=1 ?
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March 11th, 2011, 01:19 PM   #8
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Re: Linear differential equations

You integrate:



w.r.t. t (Notice the slight difference in what I wrote and what you wrote. )
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March 11th, 2011, 01:32 PM   #9
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Re: Linear differential equations

oh yes it was suposed to say t.

x sec(t) = x + C , do i divide by sec(t)
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March 11th, 2011, 02:21 PM   #10
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Re: Linear differential equations

Or multiply through by cos(t), same thing.

Although your right side should be t + C (recall we integrated with respect to t).
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