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February 9th, 2012, 02:34 AM   #1
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Integrating Factor Method

I am trying to use the Integrating Factor Method to find the general solution to the differential equation

x^3 dy/dx = y+x^2e^4x

After dividing through by x and rearranging I think I get

dy/dy - 1/x y = xe^4x

Not sure this is correct

Can anyone help?
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February 9th, 2012, 05:30 AM   #2
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Re: Intergating Factor Method

To solve any first order linear ordinary differential equations by using integrating factor, we need to rewrite the given ODE into the standard form of
.
Notice that we need to have 1 as the coefficient of the term .

Hence, from the given ODE , I find we need to divide the left and right sides of the given ODE by :







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February 9th, 2012, 10:09 AM   #3
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Re: Integrating Factor Method

Method for solving linear first order ODEs:

a) Write the equation in the standard form



b) Calculate the integrating factor by the formula:



c) Multiply the equation in standard form by and, recalling that the left-hand side is just , obtain





d) Integrate the last equation and solve for y by dividing by .

The given ODE in standard form is:



The integrating factor is then:

and multiplying through, we have:







The integral on the right cannot be expressed in elementary terms.
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February 9th, 2012, 06:50 PM   #4
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Re: Intergating Factor Method

Argh...By re-reading this post, to leave off at this step is what's killing me, but never mind, I'll blame my weary head in the middle of last night, .

I really should write it in the form , i.e.







Sorry and Mark's complete workout is great!
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February 10th, 2012, 06:09 AM   #5
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Re: Integrating Factor Method

Thanks for your help guys but I made a mistake, I should have written

x dy/dx = y +x^2e^4x

After dividing through by x we get

dy/dx = y/x + x^2e^4x /x

rearranging give

dy/dx - y/x = x^2e^4x /x

not sure about the rest.

Can anyone help?
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February 10th, 2012, 06:20 AM   #6
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Math Focus: Calculus/ODEs
Re: Integrating Factor Method

Writing the ODE in standard linear form, we have:



Now, we calculate the integrating factor:

and multiplying through:









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February 11th, 2012, 01:20 AM   #7
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Re: Integrating Factor Method

Thank you so much for your help.

I now need to find particular solution of the differential equation that satisfies the initial condition y(1)=0.

So far I have

For y(1)=0

0=1(1/4e^4*1 + C)

Not sure about the algebra here to find out what C is.

Can anyone help?
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February 11th, 2012, 01:26 AM   #8
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Re: Integrating Factor Method

You would simply have, upon solving for C, by subtracting from both sides:

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February 11th, 2012, 08:19 AM   #9
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Re: Integrating Factor Method

Would this then leave

y = x (1/4^4x - 1/4e^4)

as the particular solution.
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February 11th, 2012, 08:45 AM   #10
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Re: Integrating Factor Method

No, it would be:

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