# Solving polynomial using matrix methods

1. May 27, 2013

### unscientific

1. The problem statement, all variables and given/known data

Let y(x) = a0 + a1x + a2x2 + a3x3 + a4x4

Represent the differential equation

(d/dx)[ (1-x2)(dy/dx) ] + λy = 0

in matrix form. Find the values of λ for which there is a solution to the matrix equation, and find the solutions for the smallest and largest values of λ. Normalize these solutions with the condition y(1) = 1.

2. Relevant equations

3. The attempt at a solution

I got as far as finding the values of λ = 0, 2, 6, 12, 20. Am i supposed to plug in the values of λ into the differential equation and solve it by finding the characteristic equation by assuming y = Aeβx? But the question gives y as a polynomial: y(x) = a0 + a1x + a2x2 + a3x3 + a4x4

2. May 27, 2013

### Dick

Check your matrix. You are missing a λ in the first row. That will give you another value for λ. And you are supposed to substitute the values of λ into your matrix and find corresponding a's.

3. May 28, 2013

### unscientific

Oops i wrote my matrix wrongly from my working on paper. The smallest value of λ = 0.

I don't understand what you mean by "finding the corresponding a's".

Say i sub in λ = 0. All i get is: BA = 0, where B is the matrix with the subbed value of λ and A is the column with all the a0, a1, ....

For example, let's consider [5,5].

(-20)(a4) = 0.

This doesn't mean anything other than a4 = 0. Then it would mean a0 = a1 = a2 = a3 = a4 = 0

4. May 28, 2013

### Dick

You are jumping to conclusions. Just because a4=0 doesn't mean ALL of they a's have to be zero. They don't. Which one doesn't?

5. May 28, 2013

### unscientific

Taking λ = 0,

-20a4 = 0,

=> a4 = 0

-12a3 = 0

=> a3 = 0

-6a2 + 12a4 = 0

=> a2 = 0

-2a1 + 6a3 = 0

=> a1 = 0

and finally 2a2 = 0

6. May 28, 2013

### Dick

You forgot a0.

7. May 28, 2013

### unscientific

I put in λ = 0 into the matrix at [1,1]. This gives a0 = 0 as well..

8. May 28, 2013

### vela

Staff Emeritus
No, it doesn't.

9. May 28, 2013

### unscientific

0 * a0 + 2a1 = 0

does this mean a0 can be any real number?

10. May 28, 2013

### vela

Staff Emeritus
To satisfy that equation, yes. But it needs to be non-zero because you're looking for a non-trivial solution to the system of equations.

11. May 28, 2013

### unscientific

Then when i sub in λ = 20,

I have a4 = any real number,

a3 = 0,

a2 = -(6/7)a4

a1 = 0

a0 = 0

is this right?

12. May 28, 2013

### vela

Staff Emeritus
You shouldn't get $a_0=0$. The rest are right.

13. May 29, 2013

### unscientific

For λ = 0

-20a4 = 0,

=> a4 = 0

-12a3 = 0

=> a3 = 0

-6a2 + 12a4 = 0

=> a2 = 0

-2a1 + 6a3 = 0

a1*0 = 0

a1 $\subset$ ℝ

is it because from the condition:

y(1) = 1

a0 + a1 + a2 + a3 + a4 = 1

So, a0 = 1

For λ = 20

=> a4 $\subset$ ℝ,

=> a3 = 0

14a2 + 12a4 = 0
=> a2 = (-6/7) a4

18a1 + 6a3 = 0
=> a1 = 0

20a0 + 2a2 = 0
=> a0 = (-1/10)a2 = (3/35)a4

Using a0 + a1 + a2 + a3 + a4 = 1

a0 = 3/8
a1 = 0
a2 = -48/245
a3 = 0
a4 = 35/8

Last edited: May 29, 2013
14. May 29, 2013

### Dick

All correct, except your final a's for λ=20 don't sum to 1.