0
$\begingroup$

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 ?

$\endgroup$
3
  • $\begingroup$ You didn't differentiate $\dfrac{\sin(x)}{\sqrt{x}}$ correctly. $\endgroup$ Commented Mar 13, 2024 at 21:13
  • $\begingroup$ @CheeHan where? can you be more specific?In which method ? $\endgroup$ Commented Mar 13, 2024 at 21:14
  • $\begingroup$ You claimed that $y_1' = \frac{\cos x}{\sqrt{x}}$ but you need to use the Product Rule to compute $y_1'$. $\endgroup$ Commented Mar 13, 2024 at 21:17

2 Answers 2

Reset to default
2
$\begingroup$

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 $$ y_2=\frac{\sin(x)}{\sqrt x}\cot(x)=\frac{\cos(x)}{\sqrt{x}}. $$

$\endgroup$
1
  • 1
    $\begingroup$ Sorry, it is a small mistake. Thanks $\endgroup$ Commented Mar 15, 2024 at 14:00
1
$\begingroup$

$\textbf{Hint:}$ 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.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.