Power series method for DE

[siddharth]

Question: Find a power series solution in powers of x for the following differential equation

$$xy' - 3y = k$$

My attempt:
Assume
$$y = \sum_{m=0}^{\infty} a_m x^m$$

So,
$$xy' = \sum_{m=0}^{\infty}m a_m x^m$$

$$xy'-3y-k=0$$

implies

$$\sum_{m=0}^{\infty}m a_m x^m - 3\sum_{m=0}^{\infty} a_m x^m - k = 0$$

and

$$\left(a_1x+2a_2x^2 +3a_3x^3 +... \right) - \left(k + 3a_0 + 3a_1x+3a_2x^2+3a_3x^3+... \right) = 0$$

Which means

$$a_0=-k/3$$
$$a_1-3a_1=0, a_1=0$$
$$2a_2-3a_2=0, a_2=0$$
$$3a_3 - 3a_3=0, a_3=?$$
$$... a_n=0, n>3$$

The Question: Now, how do I find $$a_3$$?

 

[0rthodontist]

It could be anything-substitute back.

 

[siddharth]

It could be anything-substitute back.

Of course.What was I thinking? I think my brain degenerated over the summer hols
Thanks for the help.

 

[nocturnal]

Just to add to this, for a n'th order linear DE you should expect to find n arbitrary constants. So it shouldn't be too surprising that one of the coeffiecients is arbitrary given that this is a first order linear DE.

 

[WigneRacah]

Just to add to this, for a n'th order linear DE you should expect to find n arbitrary constants. So it shouldn't be too surprising that one of the coeffiecients is arbitrary given that this is a first order linear DE.

Perfect!

In fact the general solution for the given equation has the form

$$y = C x^3 - \frac{k}{3}$$.

 

[basurcekirge]

i got the same question; but i am not sure that the general solution is a power series representation