O.D.E. Problem

**JonathanEyoon** · April 1st 2009, 10:48 AM

I need some help on knowing how to tackle this problem. I missed the class where we discussed this so i'm a bit clueless on how to do it

.

One solution of the differential equation

(4x^2)y'' +(4x)y' + ((4x^2) -1)y = 0

is y1(x) = x^(-1/2)sinx. Find a second independent solution y2(x)

Any help is appreciated

**bobak** · April 1st 2009, 10:49 AM

you tired using the wronskian?

Bobak

**JonathanEyoon** · April 1st 2009, 11:13 AM

have no idea what that is but i'm gonna look into it

**bobak** · April 1st 2009, 11:30 AM

here is a quick summary of things you need to know about it.

for a second order equation $W(x) = \left|\begin{array}{cc}y_{1} & y_{2} \\y'_{1} & y'_{2} \end{array}\right| = y_{1} y'_{2} - y_{2}y'_{1}$

if you write your differential equation in the form $y'' + p(x) y' + q(x) y = 0$ then you can show that $W'(x) + p(x)W = 0$ so $W(x) = e^{ - \int p(x) dx}$

so given $y_{1}$ you can contruct a first order differential equation in $y_{2}$

Bobak

**JonathanEyoon** · April 1st 2009, 11:35 AM

Oh wow that is so helpful

. I'll give it a try and report back~

**chisigma** · April 1st 2009, 08:01 PM

Let's write the equation as...

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

The general solution of (1) is...

$y(x)= c_{1}\cdot u(x) + c_{2}\cdot v(x)$ (2)

... where $u(*)$ and $v(*)$ are two linearly independent sulution of (1), $c_{1}$ and $c_{2}$ two arbitrary constants.

If $u(*)$ and $v(*)$ are both solution of (1) then...

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

$v^{''}+ \frac{v^{'}}{x} + (1-\frac{1}{4\cdot x^{2}})\cdot v=0$ (3)

If we multiply the first equation by v and the second by u and take the difference we obtain...

$x\cdot (u^{''}\cdot v - v^{''}\cdot u) + u^{'}\cdot v - v^{'}\cdot u = \frac {d}{dx} [x\cdot (u^{'}\cdot v - v^{'}\cdot u)]=0$ (4)

The solution of (4) is 'very easy'...

$x\cdot (u^{'}\cdot v - v^{'}\cdot u)= c_{2}$ (5)

... where $c_{2}$ is an arbitrary constant. Deviding both terms of (5) by $x\cdot v^{2}$ we obtain...

$\frac{u^{'}\cdot v - v^{'}\cdot u}{v^{2}}= \frac{d}{dx} \frac{u}{v}= \frac{c_{2}}{x\cdot v^{2}}$ (6)

Also the (6) is relatively easy to solve and we obtain...

$\frac{u}{v}= c_{1} + c_{2}\cdot \int \frac{dx}{x\cdot v^{2}}$ (7)

... and from it...

$u=c_{1}\cdot v + c_{2}\cdot v\cdot \int \frac{dx}{x\cdot v^{2}}$ (8)

The (8) allows us to obtain, if you know a solution $v(*)$ of the (1), an other solution $u(*)$ independent from it as follows...

$u= v \cdot \int \frac{dx}{x\cdot v^{2}}$ (9)

In our case is...

$v(x)= \frac{\sin x}{\sqrt x}$ (10)

... so that what we have to do now is to compute the integral in (9) and then find $u(*)$

Kind regards

$\chi$ $\sigma$

**chisigma** · April 2nd 2009, 12:16 AM

Some useful comments… remembering the so called ‘Bessel equation’ …

$x^{2}\cdot y^{''} + x \cdot y^{'} + (x^{2}- n^{2})\cdot y = 0$ (1)

… it is easy to verify that the proposed ODE is the Bessel equation with $n=\frac{1}{2}$ , the solution of which is …

$y(x)= c_{1}\cdot J_{\frac{1}{2}} (x) + c_{2}\cdot J_{-\frac{1}{2}} (x)$ (2)

With a little of patience you can find that is…

$J_{\frac{1}{2}} (x)= \frac {\sin x}{\sqrt x}$

$J_{-\frac{1}{2}}(x) = \frac{\cos x}{\sqrt x}$ (3)

… functions that are represented here…

Setting $J_{- \frac{1}{2}} (x)= u(x)$ and $J_{\frac{1}{2}} (x)= v(x)$ you can verify that is [unless the sign] …

$u(x) = v(x)\cdot \int \frac{dx}{x\cdot v^{2}(x)}$

$v(x) = u(x)\cdot \int \frac{dx}{x\cdot u^{2}(x)}$ (4)

In is interesting to note that, if $u(*)$ and $v(*)$ are linearly independent solution of the equation…

$y^{''} + \frac{y^{'}}{x} + g(x)\cdot y=0$ (5)

… with $g(*)$ an arbitrary function of x, the relations (4) hold in any case…

Kind regards

$\chi$ $\sigma$

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