Thread: O.D.E. Problem

1. O.D.E. Problem

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

2. you tired using the wronskian?

Bobak

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

4. 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

5. Oh wow that is so helpful . I'll give it a try and report back~

6. 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

$\chi$ $\sigma$

7. 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

$\chi$ $\sigma$