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### Solving a Differential Equation by Series

```
Date: 06/13/99 at 17:55:06
From: Gene Ko
Subject: Solving differential equations by series

Solve y' = 2xy by series.

I know that this is a separable equation whose solution is y =
C*e^(x^2). So if this is solved by series, I should be able to get the
same y = C*e^(x^2). But I'm not getting that answer. I'm not sure as
to why it doesn't work.

This is how I tried to do the problem:

"inf" represents "infinity."

inf                            inf
y =  Sum ((c_n)*(x^n))         y' = Sum (n*(c_n)*(x^(n-1)))
n=0                            n=1

inf                              inf
Sum (n*(c_n)*(x^(n-1)))  =  2x * Sum ((c_n)*(x^n))
n=1                              n=0

inf                                  inf
Sum ((n+1)*(c_(n+1))*(x^n))  =  2x * Sum ((c_n)*(x^n))
n=0                                  n=0

(n+1)*(c_(n+1))                       2x*(c_n)
c_n = ---------------     ---->   c_(n+1) = --------
2x                                n+1

n=0: c_1 = (2x*(c_0)) / 1
n=1: c_2 = (2x*(c_1)) / 2  =  ((2x)^2 * (c_0)) / 2
n=2: c_3 = (2x*(c_2)) / 3  =  ((2x)^3 * (c_0)) / 6
n=3: c_4 = (2x*(c_3)) / 4  =  ((2x)^4 * (c_0)) / 24
etc.

Therefore:
(2x)^k * c_0
c_k = ------------ , k >= 1
k!

Then:
2x   4x^2   8x^3   16x^4
y = C * (1 + -- + ---- + ---- + ----- + ... )
1!    2!     3!      4!

= C * e^(2x).      [FINAL INCORRECT SOLUTION]

Any help on where I messed up, and how I can fix it so that I get the
correct answer of C * e^(x^2), would be greatly appreciated.
```

```
Date: 06/13/99 at 18:48:37
From: Doctor Anthony
Subject: Re: Solving differential equations by series

If we assume a solution of the form

y = a0 + a1.x + a2.x^2 + a3.x^3 + ...

y' = a1 + 2.a2.x + 3.a3.x^2 + ...

Substituting into the differential equation

a1 + 2a2.x + 3.a3.x^2 + ... = 2x[a0 + a1.x + a2.x^2 + a3.x^3 + ...]

a1 + 2a2.x + 3.a3.x^2 + ... = 2a0.x + 2a1.x^2 + 2a2.x^3 + 2a3.x^4
+ ...

Equating coefficients on the two sides of the equation:

a1 = 0

2.a2 = 2a0, so a2 = a0

3.a3 = 2a1 = 0

4.a4 = 2.a2, so a4 = a2/2

If we continue in this way we find that a1 = a3 = a5 = a7 = ... = 0,
and a2 = a0, a4 = a0/2!, a6 = a0/3!, and so on.

Therefore

y = a0[1 + x^2 + x^4/2! + x^6/3! + ...]

=  a0.e^(x^2)   which agrees with the usual analytical solution.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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