# Homogeneous diff. eqn help!

1. Aug 26, 2009

### jahnybrew

Hi guys,

I've been out of the loop with differentials for some time, and was hoping to get some direction with these..
Given that:

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

A) Explain why the general theory doesn't guarantee a unique soln to the equation satisfying the initial conditions y(0) = 0, y'(0) = 1. And show that y1(x)=x is a solution of the equation.

B) Using the method of reduction of order, find a second linearly independent solution y2(x) of the equation.

C) Is there a unique solution to the eqn satisfying the intial conditions y(0)=0, y'(0)=1?

Any help or clues with this would really be helpful because I'm not very familiar with it, and if someone could talk through any part of the question would be really appreciated.

Thanks:)

2. Aug 26, 2009

### mbisCool

If you simply plug in $$y_1{(x)}$$=x you should notice that this solves the differential equation. This is your first solution. Are you familiar with the method of reduction of order? Your second solution will be of the form $$y_2{(x)}$$=u(x)$$y_1{(x)}$$. Plug this into your differential equation and see if you can solve for u(x) which then allows you to solve for your second solution. After solving the equation simply plug in your initial conditions.

3. Aug 26, 2009

### HallsofIvy

Staff Emeritus
In order to answer A, you will need to tell why the "existence and uniqueness" theorem for differential equations does not apply here. The first "existence and uniqueness" theorem you learn is usually for first order differential equations. Here, if you let u= y and v= y', your second order equation becomes $x^2v'+ 2x(x-1)v- (2x-1)u= 0$ or
[tex]v'= \frac{2x(x-1)v- (2x-1)u}{x^2}[itex]
which, together with u'= v, gives two first order equations which you can think of as a single first order equation for the vector <u, v>. And the proof of the basic "existence and uniqueness" theorem goes over easily to vector valued functions.

Now, what "existence and uniqueness" theorems does your text book have?

4. Aug 27, 2009

### jahnybrew

hi,

I'm not entirely sure how to apply them. all we have learnt so far is the Wronskian that checks for linear independence..?

The theorems in the book are,
1. if p(x), q(x) are continuous functions on I and x is in I, then the IVP hasa unique solution y(x) on I.

2. Suppose the IVP has continuous coefficients p(x), q(x) on I. Then two solutions y1 and y2 of the IVP on I are linearly dependent on I, if and only if their Wronskian is zero at some x in I. If there is an x in I at which W is not zero, then y1, y2 are linearly independent on I.

3. If coefficients p(x), q(x) of IVP are continuous on some open I, then the IVP has a general solution on I.

4. Suppose the IVP has continuous coefficients p(x), q(x) on open interval I. Then every solution y=Y(x) or the IVP of I is of the form Y(x)=c1y1(x)+c2y2(x), where y1, y2 are the basis of solution, and c1, c2 are suitable constants.

I thought that if i were able to show the DEQ isnt continuous at 0, then there would be no unique solution to the equation with the initial conditions given?