MHB Solve Differential Equation w/ Power Series Method

[evinda]

Hello! (Wave)

The differential equation $y''+xy=0$ is given.
Find the general solution of the differential equation (with the power series method).

That's what I have tried:

We are looking for a solution of the form $y(x)=\sum_{n=0}^{\infty} a_n x^n$, where the radius of convergence is $R>0$.
$xy(x)=\sum_{n=1}^{\infty} a_{n-1}x^n$.
We have $y'(x)= \sum_{n=0}^{\infty} (n+1)a_{n+1} x^n$.
$y''(x)=\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n$.

So it has to hold:

$$2a_2 + \sum_{n=1}^{\infty} \left[ (n+2)(n+1) a_{n+2}+a_{n-1}\right]x^n=0, \forall x \in (-R,R)$$

Thus:

$$a_2=0$$

$$(n+2)(n+1)a_{n+2}+a_{n-1}=0, \forall n=1,2,3, \dots$$
$$3 \cdot 2 \cdot a_3+a_2=0 \Rightarrow a_3=0$$
$$4 \cdot 3 \cdot a_4 +a_1=0 \Rightarrow a_4=-\frac{a_1}{12}$$

Is it right so far? Or have I done something wrong?

 

[chisigma]

Hello! (Wave)

The differential equation $y''+xy=0$ is given.
Find the general solution of the differential equation (with the power series method).

That's what I have tried:

We are looking for a solution of the form $y(x)=\sum_{n=0}^{\infty} a_n x^n$, where the radius of convergence is $R>0$.
$xy(x)=\sum_{n=1}^{\infty} a_{n-1}x^n$.
We have $y'(x)= \sum_{n=0}^{\infty} (n+1)a_{n+1} x^n$.
$y''(x)=\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n$.

So it has to hold:

$$2a_2 + \sum_{n=1}^{\infty} \left[ (n+2)(n+1) a_{n+2}+a_{n-1}\right]x^n=0, \forall x \in (-R,R)$$

Thus:

$$a_2=0$$

$$(n+2)(n+1)a_{n+2}+a_{n-1}=0, \forall n=1,2,3, \dots$$
$$3 \cdot 2 \cdot a_3+a_2=0 \Rightarrow a_3=0$$
$$4 \cdot 3 \cdot a_4 +a_1=0 \Rightarrow a_4=-\frac{a_1}{12}$$

Is it right so far? Or have I done something wrong?

The solution of the ODE...

$\displaystyle y^{\ ''} + x\ y = 0\ (1)$

... requires the knowledege of $y(0)$ and $y^{\ '} (0)$...

... suppose to have a solution of the form $\displaystyle y(x)= \sum_{n=0}^{\infty} a_{n}\ x^{n}$ You have...

$\displaystyle y^{\ '} (x) = \sum_{n=0}^{\infty} n\ a_{n}\ x^{n-1}\ (2)$

$\displaystyle y^{\ ''} (x) = \sum_{n=0}^{\infty} n\ (n-1)\ a_{n}\ x^{n-2}\ (3)$

... so that is...

$\displaystyle y^{\ ''} (x) = \sum_{n=0}^{\infty} n\ (n-1)\ a_{n}\ x^{n-2} = - \sum_{n=0}^{\infty} a_{n}\ x^{n+1}\ (4)$

Here $a_{0}$ and $a_{1}$ are given from the initial conditions, and the successive coefficients can be found with (4) ...

$\displaystyle y^{\ ''} = 2\ a_{2} + 6\ a_{3}\ x + 12\ a_{4}\ x^{2} + ... = - a_{0}\ x - a_{1}\ x^{2} - ... \implies a_{2} = 0$

$\displaystyle y^{\ '''} = 6\ a_{3} + 24\ a_{4}\ x + ... = - a_{0} - 2\ a_{1}\ x - ... \implies a_{3} = - \frac{a_{0}}{6}$...

... and so one...

Kind regards

$\chi$ $\sigma$

 

[evinda]

$\displaystyle y^{\ ''} (x) = \sum_{n=0}^{\infty} n\ (n-1)\ a_{n}\ x^{n-2}\ (3)$

I found that $y''(x)= \sum_{n=1}^{\infty} (n+1) a_{n+1} n x^{n-1}= \sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n$.

Am I wrong? (Thinking)

 

[chisigma]

I found that $y''(x)= \sum_{n=1}^{\infty} (n+1) a_{n+1} n x^{n-1}= \sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n$.

Am I wrong? (Thinking)

Is...

$\displaystyle y(x) = \sum_{n = 0}^{\infty} a_{n}\ x^{n}\ (1)$

... and if the series (1) uniformely converges in an interval You can derive it 'term by term' obtaining...

$\displaystyle y^{\ '} (x) = \sum_{n=0}^{\infty} n\ a_{n}\ x^{n-1}\ (2)$

The same is for the series (2) so that is...

$\displaystyle y^{ ''} (x) = \sum_{n=0}^{\infty} n\ (n-1)\ a_{n}\ x^{n - 2}\ (3)$

Kind regards

$\chi$ $\sigma$