- #1

- 1

- 0

I think it would help if an example were solved:

2(d^2y/dx^2)-5(dy/dx)+2y = 0

I don't even know how to start

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter gruffins
- Start date

- #1

- 1

- 0

I think it would help if an example were solved:

2(d^2y/dx^2)-5(dy/dx)+2y = 0

I don't even know how to start

- #2

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

Hi gruffins! Welcome to PF!

(try using the X

I think it would help if an example were solved:

2(d^2y/dx^2)-5(dy/dx)+2y = 0

I don't even know how to start

Matrices I dunno …

but if you mean using

you write it (2D

and then you factor that into (D - a)(D - b)y = 0,

so the solutions are linear combinations of the solutions to Dy = ay

- #3

- 118

- 0

It's kind of a neat technique I think. Since this is a pretty simple ODE I will use the operator form for solving it as well.

You have 2y'' - 5y' + 2y = 0. We seek forms of e^{x} in the solutions.

Using operator form we can rewrite this as: (2D^{2} - 5D +2)y = 0

with auxiliary polynomial 2r^{2} - 5r +2 = 0

containing roots r = 1/2 and 2

and linearly independent solutions e^{2x} and e^{x/2}.

Therefore by the superposition principle we can say: y(x) = c_{1}e^{2x} + c_{2}e^{x/2} where c_{1} and c_{2} are our constants.

Now if we want to rewrite this as a vector formulation, in our original ODE we can let z_{1} = y and z_{2} = y'

so our ODE becomes 2z_{2}' - 5z_{2} + 2z_{1} = 0.

Now we can write a system:

z_{1}' = z_{2}

z_{2}' = -z_{1} + (5/2)z_{2}

And letting a vector z = (z_{1}, z_{2}) and a matrix A = [0, 1; -1, 5/2]

we have the vector differential equation: z' = Az. (Note: in A, the 0 and 1 are the first row and -1 and 5/2 are the second row. I am new to this forum and haven't had time to see if the forum has tex or anything.)

We must now examine A and find its eigenvalues, which happen to be: 1/2 and 2. (When you find the characteristic polynomial of A to find the eigenvalues, notice how similar it is to the auxiliary equation in solving the ODE in operator form.)

Using the eigenvalue of r_{1} = 2, we have an eigenvector v_{1} = (1, 2) and since we require solutions in the form of e^{x} our first solution is e^{r1x}v_{1} = e^{2x}(1, 2).

With eigenvalue r_{2} = 1/2, an eigenvector is v_{2} = (1, 1/2) and our second linearly independent solution is e^{x/2}(1, 1/2).

And our solution to our vector DE is z(x) = c_{1}e^{2x}(1, 2) + c_{2}e^{x/2}(1, 1/2).

And since z = (z_{1}, z_{2}) and z_{1} = y, we only need the first component of this vector solution. The second component is a solution to y' since z_{2} = y'.

Therefore y(x) = z_{1} = c_{1}e^{2x} + c_{2}e^{x/2}.

Few comments:

The solution to the vector differential equation can vary depending on how you pick your eigenvectors.

Also, extra steps need to be taken if the separate solutions are not linearly independent (i.e. same eigenvectors, etc..), which goes into cycles of generalized eigenvectors.

You have 2y'' - 5y' + 2y = 0. We seek forms of e

Using operator form we can rewrite this as: (2D

with auxiliary polynomial 2r

containing roots r = 1/2 and 2

and linearly independent solutions e

Therefore by the superposition principle we can say: y(x) = c

Now if we want to rewrite this as a vector formulation, in our original ODE we can let z

so our ODE becomes 2z

Now we can write a system:

z

z

And letting a vector z = (z

we have the vector differential equation: z' = Az. (Note: in A, the 0 and 1 are the first row and -1 and 5/2 are the second row. I am new to this forum and haven't had time to see if the forum has tex or anything.)

We must now examine A and find its eigenvalues, which happen to be: 1/2 and 2. (When you find the characteristic polynomial of A to find the eigenvalues, notice how similar it is to the auxiliary equation in solving the ODE in operator form.)

Using the eigenvalue of r

With eigenvalue r

And our solution to our vector DE is z(x) = c

And since z = (z

Therefore y(x) = z

Few comments:

The solution to the vector differential equation can vary depending on how you pick your eigenvectors.

Also, extra steps need to be taken if the separate solutions are not linearly independent (i.e. same eigenvectors, etc..), which goes into cycles of generalized eigenvectors.

Last edited:

- #4

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

Hi pbandjay! Welcome to PF!

Just two comments …

i] for LaTeX, type [noparse][tex] before and [/tex] after[/noparse] (or use the ∑ tag just above the Reply box)

- #5

- 118

- 0

ii] "axillary" means "relating to the armpits"![/INDENT]

Oops heh.. thanks for noticing that. I think I just accidentally spelled auxiliary wrong and automatically clicked axillary in the spell check.

Share:

- Replies
- 5

- Views
- 6K