Solving polynomial using matrix methods

[unscientific]

Homework Statement

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.

Homework Equations

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 = Ae^βx? But the question gives y as a polynomial: y_(x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴

 

[Dick]

Homework Statement

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.

Homework Equations

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 = Ae^βx? But the question gives y as a polynomial: y_(x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴

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.

 

[unscientific]

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.

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

 

[Dick]

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

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?

 

[unscientific]

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?

Taking λ = 0,

Let's start with the bottom row.

-20a₄ = 0,

=> a₄ = 0

-12a₃ = 0

=> a₃ = 0

-6a₂ + 12a₄ = 0

=> a₂ = 0

-2a₁ + 6a₃ = 0

=> a₁ = 0

and finally 2a₂ = 0

 

[Dick]

Taking λ = 0,

Let's start with the bottom row.

-20a₄ = 0,

=> a₄ = 0

-12a₃ = 0

=> a₃ = 0

-6a₂ + 12a₄ = 0

=> a₂ = 0

-2a₁ + 6a₃ = 0

=> a₁ = 0

and finally 2a₂ = 0

You forgot a0.

 

[unscientific]

You forgot a0.

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

 

[vela]

No, it doesn't.

 

[unscientific]

No, it doesn't.

0 * a₀ + 2a₁ = 0

does this mean a₀ can be any real number?

 

[vela]

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.

 

[unscientific]

a

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?

 

[vela]

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

 

[unscientific]

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

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

 

[Dick]

For λ = 0

-20a₄ = 0,

=> a₄ = 0

-12a₃ = 0

=> a₃ = 0

-6a₂ + 12a₄ = 0

=> a2 = 0

-2a₁ + 6a₃ = 0

a₁*0 = 0

a₁ \subset ℝis it because from the condition:

y(1) = 1

a₀ + a₁ + a₂ + a₃ + a₄ = 1

So, a₀ = 1For 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

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