Second order DE from my midtearm

  • Thread starter Thread starter Danny B
  • Start date Start date
  • Tags Tags
    Second order
Danny B
Messages
5
Reaction score
0

Homework Statement



Consider the ODE:

y''+p(x)y'+q(x)y=g(x)

It is given that the functions y=x^{2}, y=x and y\equiv1
are solutions of the equation.
Find the general solution of the equation.

Homework Equations



The Attempt at a Solution



Well, given the three solutions, and taking into account that they are mutually linearly independent, I would say that any couple of these functions will give me a set that forms the general solution.

If i take y=x and y\equiv1, I get the general solution:

y(x)=c1+c2*x

Now, I know that x^{2} is also a solution, but it can't be derived from the general solution, so obviously I'm doing something wrong.

I also thought of the possibility that y=1 is a singular solution and the general solution is:

y(x)=c1*x+c2*x^2

does this makes sense?

This is a question from my midterm and i still can't figure it out. So... help?
 
Physics news on Phys.org
Welcome to PF!

Hi Danny B! Welcome to PF! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)

Each of those three are particular solutions …

Hint: what relation is there between any two particular solutions (ie any two solutions)? :wink:
 
Thank you.

I think I got it now.

My mistake was to use the principle of superposition with a non-homogeneous equation.

So the general solution should be:

(Considering that any difference of the particular solutions is a homogeneous solution)

y_{(x)}=c_{1}(x-1)+c_{2}(x^{2}-1)+Y_{p}

But how do I choose Y_{p}?

Can I just use any of the three particular solutions?
 
Hi Danny B! :smile:

Let's write it out in full, so you can see how it works …

the general solution can be written in three ways

A(x2 - 1) + B(x - 1) + 1

A(x2 - x) + x + B(1 - x)

x2 + A(x - x2) + B(1 - x2)​

the homogeneous general solution can be written in three ways

A(x2 - 1) + B(x - 1)

A(x2 - x) + B(1 - x)

A(x - x2) + B(1 - x2)​

now you need to spend ten minutes convincing yourself that they're the same! :rolleyes:

he he :biggrin:
 
Thanks a lot!

You've been a great help!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top