Solve $y'' - xy' + x^2 = 0$ with taylor series

Question

I am forced to solve $y'' - xy' + x^2 = 0$ with a taylor series approach.

I was able to convert this ODE with a power series approach into an equation of the form $y' = f(x)$ (such that I would be able to integrate it), but this is not the requested way to solve the exercise.

Simply substituting the taylor series into the equation did not work, since there were infinitely many terms not proven to be zero. Additionally, I was not able to find a closed formula for the $n$-th derivative.

Differentiating the ODE three times did yield a formula for $y^5$ not being homogenous linear.

What do I have to do to use a taylor series approach successfully?

(It is an initial value problem, hence the solution might be dependent on a parameter.)

And another question out of pure interest: Does this ODE play an important role in physics?

Many thanks in advance!

Answer 1

By differentiation of

$$y''-xy'+x^2=0$$

we have

$$y'''-xy''-y'+2x=0$$

then

$$y''''-xy'''-2y''+2=0$$

then

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

then

$$y''''''-xy'''''-4y''''=0$$

and more generally

$$y^{(n+2)}-xy^{(n+1)}-ny^{(n)}=0.$$

Now, when you evaluate the last identity at $x=0$, you get the simple recurrence

$$y^{(n+2)}(0)=ny^{(n)}(0).$$

Now you have all you need to find the first and general terms.

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The homogeneous solution is

$$y'=ce^{x^2/2}$$ so that $y$ has no analytical expression. This fast growing function has no particular signifiance and the ODE is just an exercise.

Answer 2

Considering the equation $$y'' - xy' + x^2 = 0$$ let $$y=\sum_{n=0}^\infty a_n x^n \implies y'=\sum_{n=0}^\infty na_n x^{n-1}\implies y''=\sum_{n=0}^\infty n(n-1)a_n x^{n-2}$$ This makes $$\sum_{n=0}^\infty n(n-1)a_n x^{n-2}-\sum_{n=0}^\infty na_n x^{n}+x^2=0$$ Except for degree $2$, for a given degree $m$ you need $$(m+2)(m+1) a_{m+2}-m a_m=0$$ which gives the recurrence relation.

For the case of power $2$, you need $$12a_4-2a_2+1=0$$ and you have the two arbitrary constant $a_0$ and $a_1$.