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 June 4th, 2013, 04:37 PM #1 Newbie   Joined: May 2013 Posts: 3 Thanks: 0 Need help differential equations? How do you solve a differential equation with variable coefficients, like the equation y''-4xy'+2y=0 for example? None of the ways I know to solve it work.
June 4th, 2013, 10:36 PM   #2
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Re: Need help differential equations?

Quote:
 Originally Posted by Antuanne How do you solve a differential equation with variable coefficients, like the equation y''-4xy'+2y=0 for example? None of the ways I know to solve it work.
Try infinite series.
The solutions of many differential équations cannot be expressed with a finite number of elementary functions. Some solutions can be expressed on the form of infinite series. Some infinite series are known on closed form as special functions.
For example, the closed forms of the solutions of the equation y''-4xy'+2y=0 requires to know the confluent hypergeometric functions of first and second kind (Kummer and Tricomi functions)

 June 6th, 2013, 09:10 AM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Need help differential equations? Let $y= \sum_{n=0}^\infty a_n x^n$ as JJacqueline suggested. Then $y'= \sum_{n=1}^\infty na_nx^{n-1}$ (Do you see that the n= 0 terms is 0?) $y''= \sum_{n=2}^\infty n(n- 1) a_nx^{n-2}$ So the equation becomes $\sum_{n=2}^\infty n(n- 1)a_n x^{n-2}- \sum_{n= 1}^\infty 4na_n x^n+ \sum_{n=0}^\infty 2a_nx^n= 0$ The "n" in each of these sums is a "dummy index" that has meaning only in the sum. So we can change each index at will In the first sum, let i= n- 2, Then n= 2 gives i= 0, and n= i+ 2 so $a_n= a_{i+2}$ and $x^{n-2}= x^i$. In the last two sums, let i= n so we have $\sum_{i=0}^\infty a_{i+2} x^i- \sum_{i=1}^\infty 4i a_i x^i+ \sum_{i=0}^\infty 2a_i x^i= 0$ For those sums to be 0, the sum for each separate i must be 0. For i= 0 that means we must have $a_2+ 2a_0= 0$ For i> 0 we must have $a_{i+2}- (4i+ 2) a_i= 0$. Those are recursive equations for the coefficients.

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