# Series solution of D.E.

1. Nov 26, 2005

### asdf1

how do you solve
xy`-3y=k(constant)
using the power series method?

2. Nov 26, 2005

### Fermat

Write y as a polynomial in x : y = a_0 + a_1.x + a_2.x² + a_3.x^3 + ...

differentiate y to get y'

Substitute for y and y' in the original de, analyse and solve !

3. Nov 27, 2005

### asdf1

@@ but there's an extra constant! usually don't you use that method only if the right side=0?

4. Nov 28, 2005

### Fermat

It comes out OK.

Write y as a poynomial
$$y = a_0 + a_1.x + a_2.x^2 + a_3.x^3 + a_4.x^4 + ...$$

Differentiate
$$y' = a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3 + ...$$

Substitute
$$xy' - 3y = a_1x + 2a_2x^2 + 3a_3x^3 +4a_4x^4 + ... - 3a_0 - 3a_1x - 3a_2x^2 - 3a_3x^3 - 3a_4x^4 + ...$$
$$xy' - 3y = (-3a_0) + (a_1 -3a_1)x + (2a_2 - 3a_2)x^2 + (3a_3 - 3a_3)x^3 + (4a_4 - 3a_4)x^4 + ...$$

$$\mbox{substituting for } xy' - 3y = k,$$

$$k = (-3a_0) + (a_1 -3a_1)x + (2a_2 - 3a_2)x^2 + (0)x^3 + (4a_4 - 3a_4)x^4 + ... ----------------------(1)$$

Analysis
$$\mbox{For the lhs to equal the rhs, } k = -3a_0 \mbox{ (a constant) and all the other coefficents must be zero: }a_n = 0, n \in N, n \neq 0,3.$$

ergo,

$$y = a_0 + a_3x^3$$

or

$$y = -k/3 + Cx^3$$
============

As long as the original DE is made up of powers of x only (I think it would be difficult for trig functions and exponentials) then you're just manipulating the coefficents in the rhs of (1) to give your answer.

Last edited: Nov 28, 2005
5. Nov 28, 2005

### HallsofIvy

Staff Emeritus
You don't have to. Just write the right hand side as a power series also and set coefficients of corresponding powers of x equal. In the simple case that k is constant, you are just setting the constant terms equal as Fermat did.

If the right hand side were sin(x) or ex, you would write those as Taylor's series and set corresponding coefficients equal.

6. Nov 29, 2005

### asdf1

thank you very much for writing the steps out!!! it makes things very clear~ :)