Solving a Non-Linear System of Differential Equations

[iqjump123]

Homework Statement

assuming dy/dt = Dy, d^2y/dt^2 =D^2, etc:

determine the general and particular solutions to the following linear pair of differential equations:

2D^2y-Dy-4x=2t
2Dx-4Dy-3y=0

Homework Equations

The Attempt at a Solution

I have went through algebraic manipulation to come up with the first equation:
16D^4x+4D^3x-6D^2x-4x=2t.
It was close, but wasn't an equidimensional equation. Now I would have to solve this- but a 4th order equation that isn't linear?

Thanks in advance!

 

[tiny-tim]

hi iqjump123!

(try using the X² icon just above the Reply box )

I have went through algebraic manipulation to come up with the first equation:
16D^4x+4D^3x-6D^2x-4x=2t

how did you get that?

just differentiate the first equation once, and substitute for Dx from the second …

what do you get? ​

(and you can forget about the 2t while you're solving the general solution)

 

[iqjump123]

hi iqjump123!

(try using the X² icon just above the Reply box )


how did you get that?

just differentiate the first equation once, and substitute for Dx from the second …

what do you get? ​

(and you can forget about the 2t while you're solving the general solution)

Hello tinytim,

First of all, thanks for the help!
I am a little lost on how to approach your suggestion.
Differentiating the first equation will get me 2D^3y-D^2y=0, right?
I am lost on how I can substitute this value into the second equation, however.

(sorry for not using the sup and sub script method- I am typing the reply from my iPad and i have trouble typing into the forum box when i mess with the options)

Many thanks!

 

[tiny-tim]

hello iqjump123!

2D^2y-Dy-4x=2t
2Dx-4Dy-3y=0

Differentiating the first equation will get me 2D^3y-D^2y=0, right?

nooo …

differentiating the first equation gives you 2D³y - D²y - 4Dx = 2 …

carry on from there ​

 

[iqjump123]

hello iqjump123!



nooo …

differentiating the first equation gives you 2D³y - D²y - 4Dx = 2 …

carry on from there ​

Thanks tinytim!
I substituted the modified differential equation to the second equation and solved for y(t), giving me three real roots. This is different than what wolfram gave me, but I think my answer makes more sense.

My next procedure will be to substitute the answer to the first ode and solve for x(t)- then that will give me the set of general solutions. Is this correct?

Also, the solving of this set of DE was done by differentiating the first equation only. However, wouldn't it have to be that I will have to differentiate the second equation as well if I differentiate the first ode?

Thanks, as always.

Iqjump123

 

[tiny-tim]

hello iqjump123!

(i've been away for couple of weeks )

My next procedure will be to substitute the answer to the first ode and solve for x(t)- then that will give me the set of general solutions. Is this correct?

yes (but i think you mean y(t) )

and once you have a general y, then integrate (4Dy + 3y)/2 to get x, and finally use the 2 on the RHS to find a particular solution

(though the cubic equation doesn't seem to have any convenient roots … are you sure the original equations are correct?)

Also, the solving of this set of DE was done by differentiating the first equation only. However, wouldn't it have to be that I will have to differentiate the second equation as well if I differentiate the first ode?

no!

 

[iqjump123]

hello tiny tim! It has been a little bit but I wanted to revisit this thread. I really appreciate your help. As you explained, it seems that the 3rd order characteristic equation equating for y is getting me 3 roots that don't seem plausible for my problem. (one decimal real value, two complex numbers)

Unless anybody else can see an easier straight forward way to solve this problem, my guess is that there was a typo in the problem I was solving.

However, thanks to you I definitely got a grasp on how to tackle this system of ODE problem.

Thanks again.

iqjump123

 

[tiny-tim]

hello iqjump123! happy new year!

… As you explained, it seems that the 3rd order characteristic equation equating for y is getting me 3 roots that don't seem plausible for my problem. (one decimal real value, two complex numbers) …

in principle, you can solve that …

it'll be a combination of the form y = Ae^at + (Bcosct + Csinct)e^bt …

but as you say, there's probably a typo! ​