How can I solve a Bessel equation with Reduction of order?

Question

If $y_{1}(x) = \frac{\sin(x)}{\sqrt(x)}$ is one solution of the differential equation $$x^2y'' +xy' + (x^2-\frac{1}{4})y = 0$$ find the second solution $y_{2}(x)$.

My effort using Wronskian

The general form of the homogeneous linear 2nd order differential equation is :

$$ y'' + p(x)y' +q(x)y =0$$

The given differential eqaution is a Bessel equation with $n =\frac{1}{4}$ :

$$x^2 y'' +xy' + (x^2 - \frac{1}{4})y = 0$$

Rearranging :

\begin{align*} x^2 y'' +xy' + (x^2 =\frac{1}{4})y = 0 \\ y'' +\frac{x}{x^2}y' + \frac{(x^2 -\frac{1}{4})}{x^2}y = 0 \\ y'' +\frac{x}{x^2}y' + \frac{(x^2 -\frac{1}{4})}{x^2}y = 0 \\ y'' +\frac{1}{x}y' + (1 - \frac{\frac{1}{4}}{x^2}) y = 0 \\ y'' +\frac{1}{x}y' + (1 - \frac{1}{4x^2}) y = 0 \\ \end{align*}

Therefore $$p(x) = \frac{1}{x}$$

Using Abel's Theorem we have :

$$y_{2}(x) = y_{1}(x) \int \left( \frac{ e^{-\int p(x)dx}}{ y_{1}^2(x)} \right) dx$$

Substituting : $$\begin{align*} y_{2}(x) &= \frac{sin(x)}{\sqrt(x)} \int \left( \frac{ e^{-\int \frac{1}{x}dx}}{ \left(\frac{sin(x)}{\sqrt(x)}\right)^2} \right) dx \\ &= \frac{sin(x)}{\sqrt(x)} \int \left( \frac{ e^{-log(|x|)}}{ \left(\frac{sin^2(x)}{\sqrt(x)^2}\right)} \right) dx \\ &= \frac{sin(x)}{\sqrt(x)} \int \left( \frac{ |x|}{ \left(\frac{sin^2(x)}{x}\right)} \right) dx \\ &= \frac{sin(x)}{\sqrt(x)} \int \frac{ x}{ sin^2(x)|x|}dx \\ &= \frac{sin(x)}{\sqrt(x)} - \frac{ x cot(x)}{|x|} \\ &= \frac{sin(x)}{\sqrt(x)} - \frac{ x cos(x)}{|x|sin(x)} \\ &= \frac{cos(x)}{\sqrt(x)} \end{align*}$$

The Wronskian matrix :

$$W = \begin{bmatrix} y_{1} & y_{2} \\ y_{1}' & y_{2}' \end{bmatrix}$$

becomes :

$$W = \begin{bmatrix} \frac{\sin(x)}{\sqrt{x}}& \frac{\cos(x)}{\sqrt{x}} \\ \frac{\sin'(x)}{\sqrt{x}} & \frac{\cos'(x)}{\sqrt{x}} \end{bmatrix} \neq 0$$

Thus the determinant is :

\begin{align*} W(x) &= \frac{\sin(x)cos'(x)}{\sqrt{x}} - \frac{\sin'(x) cos(x)}{\sqrt{x}} \\ &= - \frac{sin^2(x) + cos^2(x)}{\sqrt(x)} \\ &= - \frac{1}{\sqrt(x)} \end{align*}

With Reduction of order

Given that $$y_1(x) = \frac{\sin(x)}{\sqrt{x}}$$ is a solution to the Bessel equation $$x^2y'' + xy' + (x^2 - \frac{1}{4})y = 0$$, i need to find the second linearly independent solution, denoted as $y_2(x)$.

I'll use the method of reduction of order. Assuming $y_2(x) = v(x)y_1(x)$, where $v(x)$ is a function to be determined.

I have: $$ y_2(x) = v(x)y_1(x) = v(x)\frac{\sin(x)}{\sqrt{x}} $$

Now, I find $y_2'(x)$ and $y_2''(x)$: $$ y_2'(x) = v'(x)y_1(x) + v(x)y_1'(x) $$ $$ y_2''(x) = v''(x)y_1(x) + 2v'(x)y_1'(x) + v(x)y_1''(x) $$

Substituting the above into the given differential equation: $$ x^2y_2'' + xy_2' + (x^2 - \frac{1}{4})y_2 = 0 $$

i have :

$$ x^2(v''(x)y_1(x) + 2v'(x)y_1'(x) + v(x)y_1''(x)) + x(v'(x)y_1(x) + v(x)y_1'(x)) + (x^2 - \frac{1}{4})v(x)y_1(x) = 0 $$

I substitute $y_1(x) = \frac{\sin(x)}{\sqrt{x}}$ and its derivatives: $$ x^2(v''(x)\frac{\sin(x)}{\sqrt{x}} + 2v'(x)\frac{\cos(x)}{\sqrt{x}} - v(x)\frac{\sin(x)}{2x\sqrt{x}}) + x(v'(x)\frac{\sin(x)}{\sqrt{x}} + v(x)\frac{\cos(x)}{\sqrt{x}}) + (x^2 - \frac{1}{4})v(x)\frac{\sin(x)}{\sqrt{x}} = 0 $$

Now, i want to choose $v(x)$ such that the terms involving $v(x)$ cancel each other out. I can do this by setting the coefficient of $\frac{\sin(x)}{\sqrt{x}}$ to zero. This gives me the following differential equation for $v(x)$: $$ x^2(v''(x) + 2v'(x)\frac{1}{x} - v(x)\frac{1}{2x}) + x(v'(x)\frac{1}{\sqrt{x}} + v(x)\frac{1}{\sqrt{x}}) + (x^2 - \frac{1}{4})v(x) = 0 $$

How I proceed further ? Any help ?

Answer 1

Let $y_2=\frac{\sin(x)}{\sqrt x}u$ be a solution. Then $$ y_2'=\frac{2 x \sin (x) u'(x)+u(x) (2 x \cos (x)-\sin (x))}{2 x^{3/2}}$$ and $$ y_2''=\frac{u(x) \left(\left(3-4 x^2\right) \sin (x)-4 x \cos (x)\right)+4 x \left(x \sin (x) u''(x)+u'(x) (2 x \cos (x)-\sin (x))\right)}{4 x^{5/2}}. $$ Now putting them in the equation gives $$ x^{3/2}(2\cos(x)u'+\sin(x)u'')=0 $$ which gives $$ \frac{u''}{u'}=-\frac{2\cos(x)}{\sin(x)}. $$ Integrating both sides gives $$ u'=\frac{c_1}{\sin^2(x)} $$ from which one has $$ u=-c_1\cot(x)+c_2. $$ Choose $c_1=-1,c_2=0$ and then $u=\cot(x)$. So the second solution is $$ y_2=\frac{\sin(x)}{\sqrt x}\cot(x)=\frac{\cos(x)}{\sqrt{x}}. $$

Answer 2

$\textbf{Hint:}$ Rewrite the differential equation using

$$v = \sqrt{x}y \implies \begin{cases} xy' = \sqrt{x}v' - \frac{1}{2\sqrt{x}} v \\ x^2y'' = \sqrt{x^3}v'' -\sqrt{x}v' + \frac{3}{4\sqrt{x}}v\end{cases}$$

which simplifies the differential equation to

$$x^2y'' + xy' + \left(x^2 - \frac{1}{4}\right)y = \sqrt{x^3}v'' + \sqrt{x^3}v = 0$$

or $$v'' + v =0$$

This is a more easily solvable equation now.