Power Series Method for Solving xy`-3y=k Differential Equation

[asdf1]

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

 

[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 !

 

[asdf1]

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

 

[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.

 

[HallsofIvy]

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

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 e^x, you would write those as Taylor's series and set corresponding coefficients equal.

 

[asdf1]

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