Homogeneous Differential Equations.

**Showcase_22** · October 4th 2008, 11:00 AM

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

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

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

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 $\frac{dy}{dx}=F(\frac{y}{x})$ ". $\frac{y}{x}$ is present on the RHS, however $\frac{2-3x}{1-3y}$ cannot be written in this way. Therefore the equation is non-homogeneous.

Alternatively this can be rearranged to $\frac{dy}{dx}+\frac{y(2-3x)}{x(1-3y)}=0$ . 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 $\frac{\delta u}{\delta t}=\frac{\delta^2 u}{\delta x^2}+\frac{\delta^2 u}{\delta y^2}$ autonomous?

(i'm not sure if $\delta$ 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 $\frac {dy}{dx}=(y)$ . 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?

**nmatthies1** · October 6th 2008, 01:55 PM

$ \frac{d^2 x}{dt^2}+4\frac{dx}{dt}+9x=2cos3t $ is linear? I have it down as nonlinear.

Also, I classified $\frac{dy}{dx}=\frac{-y(2-3x)}{x(1-3y)}$ as linear.

**mr fantastic** · October 6th 2008, 06:44 PM

Originally Posted by nmatthies1

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

..

**Chris L T521** · October 6th 2008, 07:52 PM

Originally Posted by Showcase_22

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

Is this ordinary, linear differential equation homogeneous? I was thinking about this because 2cos3t can =0 at $\frac {\pi}{2}+n\pi$ 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 $y_c$ :

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

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

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

Can you try to take the problem from here?

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

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 $\frac{dy}{dx}=F(\frac{y}{x})$ ". $\frac{y}{x}$ is present on the RHS, however $\frac{2-3x}{1-3y}$ cannot be written in this way. Therefore the equation is non-homogeneous.

Alternatively this can be rearranged to $\frac{dy}{dx}+\frac{y(2-3x)}{x(1-3y)}=0$ . 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

**Showcase_22** · October 6th 2008, 11:08 PM

Originally Posted by Chris L T521

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

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

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

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

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, $y=y_c+y_p$ 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!

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:

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

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

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

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

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

**nmatthies1** · October 7th 2008, 01:45 AM

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

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

**Showcase_22** · October 7th 2008, 02:10 AM

Ahhh! You're confusing me!

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

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

Comparing this with:

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

gives:

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

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.

Also, in $ \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:

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

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

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?

**Showcase_22** · October 7th 2008, 02:31 AM

Sorry to double post but I don't want to mix this with what I wrote before.

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!!!

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

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

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

After a bit more working I end up with:

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

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

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

Then assume a particular solution of the form $ 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 $x_p=A\sin(3t)+B\cos(3t)$ for the particular solution because it is closer to the variables I have.

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

$x=Acos 3t +Bsin 3t$

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

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

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

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

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

$12Bcos3t-12Asin3t=cos3t$

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

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

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

Firstly, have I done this correctly?

Secondly, how does this show that the DE is non-homogeneous?

**nmatthies1** · October 7th 2008, 06:33 AM

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...

**Showcase_22** · October 7th 2008, 06:49 AM

This is quite fun really! Two warwick students versus the world!!!

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.

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

Let's write it:

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

We also know that:

$ 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.

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

preaching to the choir dudette!!!

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 $1-x^2$ is the difference of two squares and use partial fractions on it.

I can solve it later if you like.

**nmatthies1** · October 7th 2008, 06:55 AM

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

Oh yeah, and to answer your question I'm from Belgium.

**Showcase_22** · October 7th 2008, 08:31 AM

Okay, so If I (finally) understood it right, it basically means that a DE is linear when it follows the general form ie $ 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.

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:

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

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

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

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

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

I think you can take it from there

It goes into ln territory from there.

Oh yeah, and to answer your question I'm from Belgium.

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!

**Chris L T521** · October 7th 2008, 10:05 AM

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, $y=y_c+y_p$ 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:

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

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

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

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

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

, I should have seen that!

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

I'm sure you can take it from here.

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

--Chris

**Showcase_22** · October 7th 2008, 10:13 AM

I should have seen that!

lol

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

**Chris L T521** · October 7th 2008, 10:24 AM

Originally Posted by Showcase_22

I should have seen that!

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

It is homogeneous if we can observe the following:

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

If $t^{\alpha}$ from $M(x,y)$ and $N(x,y)$ 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.

--Chris

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