# Homogeneous Differential Equations.

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• Oct 4th 2008, 11:00 AM
Showcase_22
Homogeneous Differential Equations.
$\displaystyle \frac{d^2 x}{dt^2}+4\frac{dx}{dt}+9x=2cos3t$

Is this ordinary, linear differential equation homogeneous? I was thinking about this because 2cos3t can =0 at $\displaystyle \frac {\pi}{2}+n\pi$ so at these points could it be considered homogeneous?

$\displaystyle \frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)}$

This is an ordinary, non-linear differential equation. However, is it homogeneous?

My book defines a homogenous equation as "a differential equation that can be written as $\displaystyle \frac{dy}{dx}=F(\frac{y}{x})$". $\displaystyle \frac{y}{x}$ is present on the RHS, however $\displaystyle \frac{2-3x}{1-3y}$ cannot be written in this way. Therefore the equation is non-homogeneous.

Alternatively this can be rearranged to $\displaystyle \frac{dy}{dx}+\frac{y(2-3x)}{x(1-3y)}=0$. Since this is non-linear then although it equals 0, it is not homogeneous.

I just wanted to see if I was thinking of this the correct way.

Your input would be very much appreciated!

P.S: Sorry to just add this in under this heading but I got stuck on this as well:

Is $\displaystyle \frac{\delta u}{\delta t}=\frac{\delta^2 u}{\delta x^2}+\frac{\delta^2 u}{\delta y^2}$ autonomous?

(i'm not sure if $\displaystyle \delta$ is the right symbol for a partial DE, but it looked the most like the one i've written down).

Once again, consulting the almighty book of knowledge I get that $\displaystyle \frac {dy}{dx}=(y)$. However it doesn't mention anything regarding partial DEs! I gather that if the equation does not contain any t, x and y's then it is autonomous, however it is difficult to see if the equation actually does.

When push comes to shove, I would say this is autonomous. Is this correct?
• Oct 6th 2008, 01:55 PM
nmatthies1
$\displaystyle \frac{d^2 x}{dt^2}+4\frac{dx}{dt}+9x=2cos3t$ is linear? I have it down as nonlinear.

Also, I classified $\displaystyle \frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)}$ as linear.
• Oct 6th 2008, 06:44 PM
mr fantastic
Quote:

Originally Posted by nmatthies1
$\displaystyle \frac{d^2 x}{dt^2}+4\frac{dx}{dt}+9x=2cos3t$ is linear? I have it down as nonlinear. Mr F says: Wrong. It's second order linear.

Also, I classified $\displaystyle \frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)}$ as linear. Mr F says: It is not linear. It is not homogenous.

..
• Oct 6th 2008, 07:52 PM
Chris L T521
Quote:

Originally Posted by Showcase_22
$\displaystyle \frac{d^2 x}{dt^2}+4\frac{dx}{dt}+9x=2cos3t$

Is this ordinary, linear differential equation homogeneous? I was thinking about this because 2cos3t can =0 at $\displaystyle \frac {\pi}{2}+n\pi$ so at these points could it be considered homogeneous?

I would first recommend that you solve the homogeneous counterpart and come up with the complementary solution $\displaystyle y_c$:

$\displaystyle \frac{d^2 x}{dt^2}+4\frac{dx}{dt}+9x=0$

Then assume a particular solution of the form $\displaystyle y_p=A\sin(3x)+B\cos(3x)$. Substitute this into the original DE and solve for the coefficients A and B.

The final solution will take the form $\displaystyle y=y_c+y_p$

Can you try to take the problem from here?

Quote:

$\displaystyle \frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)}$

This is an ordinary, non-linear differential equation. However, is it homogeneous?

My book defines a homogenous equation as "a differential equation that can be written as $\displaystyle \frac{dy}{dx}=F(\frac{y}{x})$". $\displaystyle \frac{y}{x}$ is present on the RHS, however $\displaystyle \frac{2-3x}{1-3y}$ cannot be written in this way. Therefore the equation is non-homogeneous.

Alternatively this can be rearranged to $\displaystyle \frac{dy}{dx}+\frac{y(2-3x)}{x(1-3y)}=0$. Since this is non-linear then although it equals 0, it is not homogeneous.

I just wanted to see if I was thinking of this the correct way.

Your input would be very much appreciated!
I think you're over-thinking this one. This is a little easier than it looks...

Try and see if you can "group" certain variables together...hint, hint...

--Chris
• Oct 6th 2008, 11:08 PM
Showcase_22
Quote:

Originally Posted by Chris L T521
I would first recommend that you solve the homogeneous counterpart and come up with the complementary solution $\displaystyle y_c$:

$\displaystyle \frac{d^2 x}{dt^2}+4\frac{dx}{dt}+9x=0$

Then assume a particular solution of the form $\displaystyle y_p=A\sin(3x)+B\cos(3x)$. Substitute this into the original DE and solve for the coefficients A and B.

The final solution will take the form $\displaystyle y=y_c+y_p$

Can you try to take the problem from here?

Okay then. If I solved it when it was equaal to zero and if I solved it when it was equal to cos3t (ie. I found some general solutions) the two results would be different?

The final result I would get would be, as you stated,$\displaystyle y=y_c+y_p$ but i'm finding it a little hard to see how it would be helpful.

However, I have a lecture in a few minutes so i'll have a go at this when I get back. Expect a later post where i've worked them both out!

Quote:

I think you're over-thinking this one. This is a little easier than it looks..

Try and see if you can "group" certain variables together...hint, hint...

--Chris
How's this:

$\displaystyle \frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)}$

$\displaystyle \frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)} \frac{\frac{1}{x}}{\frac{1}{x}}$

$\displaystyle \frac{dy}{dx}=\frac{-\frac{y}{x}(2-3x)}{(1-3y)}$

$\displaystyle \frac{dy}{dx}=(\frac{-y}{x} )\frac{2-3x}{1-3y}$

i'm not really sure what else to do to it. Is this what you meant or did you mean using partial fractions?
• Oct 7th 2008, 01:45 AM
nmatthies1
Ok... But isn't a linear DE one that you can write in the form an(t)d^ny/dt^n + ... + a1(t)dy/dt + a0(t)y = f(t) ? But then where is the function of time (t) in the term 9x of $\displaystyle \frac{d^2 x}{dt^2}+4\frac{dx}{dt}+9x=2cos3t$ ?

Also, in $\displaystyle \frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)}$ I see functions of x on the LHS and the RHS, so shouldn't it be linear?

I guess I'm in desperate need of an explanation ;)
• Oct 7th 2008, 02:10 AM
Showcase_22
Ahhh! You're confusing me!

Using the general form of writing a linear DE (that you wrote down) we get:

$\displaystyle a_2(t)\frac{d^2 y}{dt^2}+a_1(t) \frac{dy}{dt}+a_0(t)y=f(t)$

Comparing this with:

$\displaystyle \frac{d^2 x}{dt^2}+4\frac{dx}{dt}+9x=2cos3t$

gives:

$\displaystyle a_2(t)=1$,$\displaystyle a_1(t)=4$ and $\displaystyle a_0 (t)=9.$

This shows the equation is linear. The functions of t are constant (ie. constant functions). I'd get someone to second this though since i'm a bit shaky on this subject.

Quote:

Also, in $\displaystyle \frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)}$ I see functions of x on the LHS and the RHS, so shouldn't it be linear?
If we rearrange it we get:

$\displaystyle x(1-3y)\frac{dy}{dx}=-y(2-3x)$

$\displaystyle x\frac{dy}{dx}-3xy\frac{dy}{dx}=-2y+3xy$

and this does not follow the general pattern of a linear DE.

P.S:nmathies1 these are the questions i'm stuck on for the induction homework:http://www.mathhelpforum.com/math-he...sequences.html

I've reached a dead end on them. Have you done them?
• Oct 7th 2008, 02:31 AM
Showcase_22
Sorry to double post but I don't want to mix this with what I wrote before.

Quote:

Okay then. If I solved it when it was equaal to zero and if I solved it when it was equal to cos3t (ie. I found some general solutions) the two results would be different?
Here we go!!!

$\displaystyle \frac{d^2x}{dt^2}+a\frac{dx}{dt}+9x=0$

$\displaystyle \lambda^2+4\lambda+9=0$

$\displaystyle \lambda=-2\pm \sqrt {3}i$ (from the quadratic equation)

After a bit more working I end up with:

$\displaystyle x=e^{-2t}((A+B)cos \sqrt{3}t+(A-B)sin \sqrt{3}t)$

$\displaystyle x=e^{-2t}(Dcos \sqrt{3}t+Esin \sqrt{3}t)$

where D=A+B and E=A-B.

Quote:

Then assume a particular solution of the form $\displaystyle y_p=A\sin(3x)+B\cos(3x)$. Substitute this into the original DE and solve for the coefficients A and B.

I decided to use$\displaystyle x_p=A\sin(3t)+B\cos(3t)$ for the particular solution because it is closer to the variables I have.

$\displaystyle \frac{d^2 x}{dt^2}+4\frac{dx}{dt}+9x=2cos3t$

$\displaystyle x=Acos 3t +Bsin 3t$

$\displaystyle \frac{dx}{dt}=-3Asin3t+3Bcos3t$

$\displaystyle \frac{d^x}{dt^2}=-9Acos3t-9Bsin3t$

$\displaystyle -9Acos3t-9Bsin3t+4(-3Asin3t+3Bcos3t)+9(Acos 3t +Bsin 3t)=cos3t$

$\displaystyle -9Acos3t-9Bsin3t-12Asin3t+12Bcos3t+9Acos 3t +9Bsin 3t=cos3t$

$\displaystyle -9Acos3t+12Bcos3t+9Acos3t-9Bsin3t-12Asin3t+9Bsin3t=cos3t$

$\displaystyle 12Bcos3t-12Asin3t=cos3t$

Therefore: 12B=1 and -12A=0

A=0 and $\displaystyle B=\frac{1}{12}$

So the solution is $\displaystyle x=\frac{1}{12}sin3t$

Firstly, have I done this correctly?

Secondly, how does this show that the DE is non-homogeneous?
• Oct 7th 2008, 06:33 AM
nmatthies1
Ok I get where you're coming from ;) but am still quite confused... but following your logic the fourth DE on the assignment eg td²x/dt² + e^(-cos(t))dx/dt = t²x is not linear either then?

I'm under the impression that the lecturer's explanations/ definitions weren't quite clear, at least for me.

I think the rest of the assignment is quite clear though, altough I'm having trouble with B2 part a): solve the following Initial Value Problem: dy/dx = - 1/ (1-x²)
So, how do I integrate -1/(1-x²) ? There must be a way I'm not familiar with...

:)
• Oct 7th 2008, 06:49 AM
Showcase_22
This is quite fun really! Two warwick students versus the world!!!

Quote:

Ok I get where you're coming from but am still quite confused... but following your logic the fourth DE on the assignment eg td²x/dt² + e^(-cos(t))dx/dt = t²x is not linear either then?
Firstly, take Latex for a spin. It's just easier to see where you're coming from.

$\displaystyle t\frac{d^2x}{dt^2}+e^{-cost}\frac{dx}{dt}=t^2 x$

Let's write it:

$\displaystyle t\frac{d^2x}{dt^2}+e^{-cost}\frac{dx}{dt}-t^2 x=0$

We also know that:

$\displaystyle a_2(t)\frac{d^2 y}{dt^2}+a_1(t) \frac{dy}{dt}+a_0(t)y=f(t)$

Looking at what we have, there are functions of t in front of all the terms. Therefore it is linear.

Quote:

I'm under the impression that the lecturer's explanations/ definitions weren't quite clear, at least for me.
preaching to the choir dudette!!!

Quote:

I think the rest of the assignment is quite clear though, altough I'm having trouble with B2 part a): solve the following Initial Value Problem: dy/dx = - 1/ (1-x²)
So, how do I integrate -1/(1-x²) ? There must be a way I'm not familiar with...
I'm in a rush since I have to go to the vectors and matrices lecture now. take advantage of the fact that $\displaystyle 1-x^2$ is the difference of two squares and use partial fractions on it.

I can solve it later if you like.
• Oct 7th 2008, 06:55 AM
nmatthies1
Yeah, it really is fun ;)

Okay, so If I (finally) understood it right, it basically means that a DE is linear when it follows the general form ie $\displaystyle a_2(t)\frac{d^2 y}{dt^2}+a_1(t) \frac{dy}{dt}+a_0(t)y=f(t)$ , whether or not there are actually "'t"s in the coefficients? Is that right?

As for the integral, I don't think we have done partial fractions at school. So If you have time, it would be great if you could show me :)

• Oct 7th 2008, 08:31 AM
Showcase_22
Quote:

Okay, so If I (finally) understood it right, it basically means that a DE is linear when it follows the general form ie $\displaystyle a_2(t)\frac{d^2 y}{dt^2}+a_1(t) \frac{dy}{dt}+a_0(t)y=f(t)$ , whether or not there are actually "'t"s in the coefficients? Is that right?
Yup!! If the function of t just equals a number then it's a constant function. (Rofl)

Quote:

As for the integral, I don't think we have done partial fractions at school. So If you have time, it would be great if you could show me
This is what I would do:

$\displaystyle \int -\frac{1}{1-x^2} dx$

$\displaystyle -\int \frac{1+x-x}{(1+x)(1-x)}dx$

$\displaystyle -\int \frac{1+x}{(1+x)(1-x)}-\frac{x}{(1+x)(1-x)}dx$

$\displaystyle -\int \frac{1}{1-x}-\frac{x}{1-x^2} dx$

$\displaystyle \int \frac{x}{1-x^2}-\frac{1}{1-x}dx$

I think you can take it from there(Cool)

It goes into ln territory from there.

Quote:

The land of chocolate!! Speaking of which I need to go to Tesco and buy some.

How did you do in the tests? I passed all three with 38,37 and 37. I could have done better but I think my mind was somewhere else that day.

Also, I created a team. Feel free to join it!
• Oct 7th 2008, 10:05 AM
Chris L T521
Quote:

Originally Posted by Showcase_22
Okay then. If I solved it when it was equaal to zero and if I solved it when it was equal to cos3t (ie. I found some general solutions) the two results would be different?

The final result I would get would be, as you stated,$\displaystyle y=y_c+y_p$ but i'm finding it a little hard to see how it would be helpful.

However, I have a lecture in a few minutes so i'll have a go at this when I get back. Expect a later post where i've worked them both out!

How's this:

$\displaystyle \frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)}$

$\displaystyle \frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)} \frac{\frac{1}{x}}{\frac{1}{x}}$

$\displaystyle \frac{dy}{dx}=\frac{-\frac{y}{x}(2-3x)}{(1-3y)}$

$\displaystyle \frac{dy}{dx}=(\frac{-y}{x} )\frac{2-3x}{1-3y}$

i'm not really sure what else to do to it. Is this what you meant or did you mean using partial fractions?

Once I show you a couple manipulations, you'll be like (Doh), I should have seen that! :D

$\displaystyle \frac{\,dy}{\,dx}=\frac{-y(2-3x)}{x(1-3y)}\implies\frac{\,dy}{\,dx}=-\frac{y}{1-3y}\cdot\frac{2-3x}{x}\implies \left(\frac{3y-1}{y}\right)\,dy=\left(\frac{2-3x}{x}\right)\,dx$

I'm sure you can take it from here. :D

However, I think this will not simplify that well...thus leading to an implicit solution.

--Chris
• Oct 7th 2008, 10:13 AM
Showcase_22
(Doh) I should have seen that!:D lol

Okay then. Going back to the question, how does this tell me whether this is homogeneous or not?
• Oct 7th 2008, 10:24 AM
Chris L T521
Quote:

Originally Posted by Showcase_22
(Doh) I should have seen that!:D lol

Okay then. Going back to the question, how does this tell me whether this is homogeneous or not?

Here is one way to test and see if its homogeneous:

We (in a sense) see that $\displaystyle N(x,y)=\frac{1-3y}{y}$ and $\displaystyle M(x,y)=\frac{2-3x}{x}$ after getting our equation in the form of $\displaystyle M(x,y)\,dx+N(x,y)\,dy=0$

It is homogeneous if we can observe the following:

$\displaystyle M(tx,ty)=t^{\alpha}M(x,y)$ AND $\displaystyle N(tx,ty)=t^{\alpha}N(x,y)$.

If $\displaystyle t^{\alpha}$ from $\displaystyle M(x,y)$ and $\displaystyle N(x,y)$ are equal, then its homogeneous. Otherwise, its non-homogeneous.

I can give you a heads up now and tell you its not homogeneous. But I want you to show that it isn't. :D

--Chris
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