MHB Help on a 2nd Order Linear Differential Equation

[frank1234]

Hi, I need help solving this ODE. I know the answer is a Bessel function but I need help on the process of getting there.

Initial conditions y(0)=1 and y'(0)=0

xy''+y'+xy=0

I have made it this far...

${x}^{2}*\sum_{n=0}^{inf} [n(n+1)*{C}_{n+2}+(n+1)*{C}_{n+1}+{C}_{n}]*{x}^{n}=0$

 

[Siron]

I did not check if the solution you already have is entirely correct. To proceed, you know the sum in the LHS must be equal to zero hence the coefficient of $x^n$ is zero:
$$\forall n \geq 0: n(n+1)C_{n+2}+(n+1)C_{n+1}+C_n = 0$$
the above can be written as
$$\forall n \geq 0: C_{n+2} = -\frac{C_n+(n+1)C_{n+1}}{n(n+1)}$$

To solve this recursion you can put $n=0, 1,2,..$ successively and try to notice a pattern in the coefficients. I mean, it's clear that $c_0$ and $c_1$ can't be determined and thus they have to be chosen arbitrary. Now you have to write the other coefficients in function of $c_0$ and/or $c_1$.

 

[chisigma]

Hi,I need help solving this ODE. I know the answer is a Bessel functionbut I need help on the process of getting there.Initial conditionsy(0)=1 and y'(0)=0 xy''+y'+xy=0

Writing the equationin a little different way ...$\displaystyle y^{\ ''} = - \frac{y^{\ '}}{x} - y\ (1)$... we can say thatits the general solution is ...$\displaystyle y =c_{1}\ u(x) + c_{2}\ v(x)\ (2)$ ... where u(x) isanalytic in x=0 and v(x) isn't, so that is ...$\displaystyle u(x)= \sum_{n=0}^{\infty} a_{n}\ x^{x}\ (3)$Since the initial conditions are $y(0)=1$ and $y^{\ '} (0)=0$ it will be in (2)$c_{2}=0$ and in (3) $a_{0}=1$ and $a_{1}=0$. The value of $a_{2}$ isfound by observing that for (1) is... $\displaystyle y^{\ ''} (0) = \lim_{x \rightarrow 0} (- \frac{y^{\ '}}{x} - y) = - 2\ a_{2} - 1 = 2\ a_{2} \implies a_{2}= - \frac{1}{4}\ (4)$

The coefficient $a_{3}$ is obtained deriving (1)...$\displaystyle y^{\ '''} (0) = \lim_{x \rightarrow 0} (- \frac{y^{\ ''}}{x} + \frac{y^{\ '}}{x^{2}} - y^{\ '}) = \lim_{x \rightarrow 0} (- \frac{2\ a_{3}}{x}+ \frac{2\ a_{3}}{x} + 3\ a_{3}) = 6\ a_{3} \implies a_{3}=0\ (5) $... and $a_{4}$deriving (5)...

$\displaystyle y^{\ ''''} (0) = \lim_{x \rightarrow 0} (- \frac{y^{\ '''}}{x} + 2\ \frac{y^{\ ''}}{x^{2}} - y^{\ ''} - 2\ \frac{y^{\ '}}{x^{3}}) = \lim_{x \rightarrow 0} (- 6\ \frac{a_{3}}{x} - 24\ a_{4} + 4\ \frac{a_{2}}{x^{2}} + 12\ \frac{a_{3}}{x} + 24\ a_{4} - 2\ \frac{a_{3}}{x} - 8\ a_{4}) = 24\ a_{4} \implies a_{4} = - \frac{a_{2}}{16} = \frac{1}{64}\ (6)$

We have obtained till now...

$\displaystyle y = 1 - \frac{x^{2}}{4} + \frac{x^{4}}{64} + ...\ (7)$

Proceeding in the same way we arrive at the solution...

$\displaystyle y = \sum_{n = 0}^{\infty} (-1)^{n} \ \frac{x^{2\ n}}{2^{2\ n}\ (n!)^{2}}\ (8)$

... which is known as Bessel Function of the first kind of order 0...

Kind regards

$\chi$ $\sigma$