# find the general power series solution to the following differentialfind the general power series solution to the following differential equation up to the x^5 term of the power series solution:...

find the general power series solution to the following differential

find the general power series solution to the following differential equation up to the x^5 term of the power series solution: 2y'' - xy' = 0
lfryerda | Certified Educator

We start by assuming a power series solution to the differential equation.  That is, let `y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+cdots`

Now differentiating, we get

`y'=a_1+2a_2x+3a_3x^2+4a_4x^3+5a_5x^4+cdots`

and again to get

`y''=2a_2+6a_3x+12a_4x^2+20a_5x^3+cdots`

where the dots represent higher order terms that we can ignore.

Combining both of these into the left side of the differential equation, we get

`4a_2+12a_3x+24a_4x^2+40a_5x^3+cdots`

`- a_1x-2a_2x^2-3a_3x^3-4a_4x^4-5a_5x^5+cdots=0`

And now compare powers of x to zero to solve for the coefficients.

Constant: `a_2=0`

linear: `12a_3-a_1=0`

cubic: `40a_5-3a_3=0`

We don't need to go any further since the question is only asking up to `a_5` .

This means that `a_2=a_4=0` , `a_0` and `a_1` are arbitrary, `a_3=1/12 a_1` and `a_5=3/40 a_3=1/160 a_1` .

Therefore, up to the `x^5` term, the power series solution is

`y=a_0+a_1(x+1/12x^3+1/160 x^5)`