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March 10th, 2011, 11:06 AM   #11
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Re: Exact form, differential equations

When you divide through by x, the (1/4)x^4 becomes (1/4)x^3
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March 10th, 2011, 11:51 AM   #12
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Re: Exact form, differential equations

Yes ofcourse.

Show that x^2 dy/dx + 3xy = x^3 can be made exact and hence obtain its general solution, please can you check my answer

Multiply by integrating factor x,

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

d/dx (x^3 y) = x^4 Exact form

Now integrate wrt x,

x^3 y = (1/5)x^5 + A

y = (1/5)x^2 + A/x^3 General solution
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March 10th, 2011, 11:53 AM   #13
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Re: Exact form, differential equations

Flawless!
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March 10th, 2011, 11:56 AM   #14
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Re: Exact form, differential equations

There's hope yet my friend
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March 10th, 2011, 01:06 PM   #15
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Re: Exact form, differential equations

Similar question,

Find the integrating factor for the following equation and find the general solution

x^2 dy/dx + xy =1
Rewrite in std form.... divide by x^2

dy/dx + 1/x^2 y =1/x^2

Obtain integrating factor, I got e^-1/x this may be incorrect as I have a feeling the std form may be wrong..!


Before I go any further can you advise me on the integrating factor.
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March 10th, 2011, 01:17 PM   #16
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Re: Exact form, differential equations

When you divide through by x. you should have:



Now we find the integrating factor is...?
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March 10th, 2011, 01:34 PM   #17
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Re: Exact form, differential equations

integrating factor,

?(x) = e^?1/x = e^ln(x) = x

integrating factor = x yes?
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March 10th, 2011, 01:37 PM   #18
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Re: Exact form, differential equations

Yes. The same result could be achieved by dividing through the original equation by x. However, it's not always so easy to spot what you need, and the formula you are using is a great help in those cases.
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March 10th, 2011, 01:40 PM   #19
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Re: Exact form, differential equations

Yes, not so easy to see from the start, best to work through it to make sure.

so, multiply by x

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

d/dx (x^2 y) = x exact form

integrate wrt to x

x^2 y =(1/2)x^2 + A

y= (1/2)x + A/x^2
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March 10th, 2011, 01:52 PM   #20
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Re: Exact form, differential equations

No, you must multiply the equation in standard form by the integrating factor x:



Now integrate and solve for y.

Notice you have:

which is not true.
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