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The initial value problem $xy'' + y' + xy = 0; y(0)=1, y'(0) = 0$ has_____

Options:

a) a unique solution.

b) no solution

c) infinitely many solutions

d) two linearly independent solutions

(Is there any other way around to do it other than power series method?)

Ans) I started with y(x) = $x^r$ but fails to continue. Then I choose $$y(x) =\sum_{n=0}^{\infty} a_n x^n$$

$$y'(x) =\sum_{n=0}^{\infty} na_nx^{n-1}$$ $$y''(x) = \sum_{n-0}^{\infty} n(n-1)a_nx^{n-2}$$ Substitution leads to

$$ \sum_{n=0}^{\infty} [n(n-1) + n]a_nx^{n-1} + \sum_{n=0}^{\infty} a_n x^{n+1}=0$$

Change of index leads to

$$ \sum_{n=0}^{\infty} [n(n-1) + n]a_nx^{n-1} + \sum_{n=2}^{\infty} a_{n-2} x^{n-1}=0$$

Rewrite as

$$ a_1 + \sum_{n=2}^{\infty} (n^2a_n + a_{n-2})x^{n-1}=0$$ which implies $a_1=0$.

Recursive relation is

$$n^2a_n = -a_{n-2}$$ Take $a_0 = a_0$ and $a_1 = a_1$

$a_2 = \displaystyle\frac{-a_0}{2^2}$ $a_3 = \displaystyle\frac{-a_1}{3^2}$

$a_4 = \displaystyle\frac{a_0}{2^2.4^2}$ $a_5 = \displaystyle\frac{a_1}{3^2.5^2}$

$a_6 = \displaystyle\frac{-a_0}{2^2.4^2.6^2}$ $a_7 = \displaystyle\frac{-a_1}{3^2.5^2.7^2}$

$$y(x) = a_0 + a_0\sum_{k=1}^{\infty} \frac{(-1)^kx^{2k}}{\Pi_{n=1}^{k} (2n)^2} + a_1\sum_{k=0}^{\infty} \frac{(-1)^kx^{2k+1}}{\Pi_{n=0}^{k} (2n+1)^2}$$

already we got $a_1 = 0$ hence solution will be $$y(x) = a_0 + a_0\sum_{k=1}^{\infty} \frac{(-1)^kx^{2k}}{\Pi_{n=1}^{k} (2n)^2}$$ Given condition $y(0) = 1$ gives $a_0 = 1$

Final solution will be $$y(x) = 1 + \sum_{k=1}^{\infty} \frac{(-1)^kx^{2k}}{\Pi_{n=1}^{k} (2n)^2}$$ and clearly $y'(0) = 0$ satisfies. So concludes a unique solution.

Is there any other way to do it smartly?

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  • $\begingroup$ $x=0$ is a singular point. You have to use the Frobenius method. $\endgroup$ – user170231 Nov 10 '15 at 13:04
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    $\begingroup$ Multiply with $x$ and you obtain Bessel's equation, en.wikipedia.org/wiki/Bessel_function. $\endgroup$ – Urgje Nov 10 '15 at 14:28
  • $\begingroup$ This might be solvable using the Laplace transform $\endgroup$ – Dylan Nov 11 '15 at 1:54
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Let $Y(s)$ be the Laplace transform of $y(x)$, then $y' = sY - y(0)$ and $y'' = s^2Y - sy(0) - y'(0) $

Taking the transform of the equation $$ -(s^2Y - s)' + (sY-1) - Y' = 0 $$ $$ -(2sY + s^2Y' - 1) + sY - 1 - Y' = 0 $$ $$ (1+s^2)Y' + sY = 0 $$

Solving the above first-order ODE gives $$ Y = \frac{C}{\sqrt{1+s^2}}$$

Using a table lookup, we find that $$ y(x) = J_0(x) $$

As @Urgje pointed out, you can obtain the same solution by multiplying both sides by $x$ $$ x^2y'' + xy' + x^2y = 0 $$ which is the Bessel equation

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It should be noted that taking the Laplace Transform of this differential equation as @Dylan did, the solution J0 (The Bessel Function) is a solution, generalized for any initial conditions, as they will cancel algebraically after the Laplace Transform is taken. For example: $$-(s^2Y-sy(0)-y'(0))'+(sY-y(0))+Y'=0$$ $$-(2sY+s^2Y'-y(0)-0)+(sY-y(0))+Y'=0$$ $$-s^2Y'-sY-Y'=0$$Which then becomes a separable differential equation that can be solved using the same method, and will result in a solution of the form J0 using a lookup table.

Source: Dylan (https://math.stackexchange.com/users/135643/dylan), Solve $xy'' + y' + xy = 0, y(0)=1, y'(0)=0$, URL (version: 2015-11-11): https://math.stackexchange.com/q/1523454

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