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Solution of 2 x'' y + 3 x' y' + x y'' = 0 where ' is d/dt

  • Thread starter Dukon
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  • #1
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Homework Statement


Given [itex]\dot{x} \equiv { \mathrm{d}x \over \mathrm{d}t }[/itex] and [itex]\ddot{x}[/itex] is [itex]{ \mathrm{d^2}x \over \mathrm{d}t^2 }[/itex]

what is solution for x(t) and y(t) which satsifies

[itex]2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0[/itex]

Homework Equations



[itex]2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0[/itex]

The Attempt at a Solution



clearly, guesses could be made for x(t) and y(t) but is there a most general solution which somehow includes all possible solutions?
 
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Answers and Replies

  • #2
LCKurtz
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(can the [itex] construction be used in the post titles also or only in the body?)
Don't put equations in the titles in the first place. It violates forum rules.
 
  • #3
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Homework Statement


Given [itex]\dot{x} \equiv { \mathrm{d}x \over \mathrm{d}t }[/itex] and [itex]\ddot{x}[/itex] is [itex]{ \mathrm{d^2}x \over \mathrm{d}t^2 }[/itex]

what is solution for x(t) and y(t) which satsifies

[itex]2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0[/itex]

Homework Equations



[itex]2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0[/itex]

The Attempt at a Solution



clearly, guesses could be made for x(t) and y(t) but is there a most general solution which somehow includes all possible solutions?
Let ##x = e^{rt}## and let ##y = e^{st}##.
See where that gets you.
 
  • #4
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I know but I said that already in "3. The Attempt at a Solution "
[itex]e^{rt}[/itex] is just a guess
I am looking for a more general more formal solution method
 
  • #5
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I know but I said that already in "3. The Attempt at a Solution "
[itex]e^{rt}[/itex] is just a guess
I am looking for a more general more formal solution method
This is more of an "educated guess" than just a guess. This is also a standard (and formal) technique that is used in solving differential equations.
 
  • #6
LCKurtz
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I know but I said that already in "3. The Attempt at a Solution "
[itex]e^{rt}[/itex] is just a guess
I am looking for a more general more formal solution method
It's kind of an odd problem with a second order DE and two unknown functions with no second equation nor boundary conditions. Not sure what to expect, but Mark's suggestion leads to two solutions of the form $$
(x(t),y(t)) = C(e^{rt},e^{-2rt})\text{ and } D(e^{rt},e^{-rt})$$each with a free parameter ##r## and arbitrary constants. Of course, your question involves whether or not that is all there are. That question is usually answered by existence and uniqueness theorems. But this is a nonlinear setting so all bets are off. Whether there are such theorems for this kind of setting, I don't know.
 
  • #7
haruspex
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is there a most general solution which somehow includes all possible solutions?
It's too easy to generate whole families of solutions. E.g. set x=tm, y=tn (not even bothering with constant multiples) gives a solution whenever neither n nor m is zero and n is either -2m or 1-m.
 
  • #8
pasmith
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Homework Statement


Given [itex]\dot{x} \equiv { \mathrm{d}x \over \mathrm{d}t }[/itex] and [itex]\ddot{x}[/itex] is [itex]{ \mathrm{d^2}x \over \mathrm{d}t^2 }[/itex]

what is solution for x(t) and y(t) which satsifies

[itex]2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0[/itex]

Homework Equations



[itex]2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0[/itex]

The Attempt at a Solution



clearly, guesses could be made for x(t) and y(t) but is there a most general solution which somehow includes all possible solutions?
Let [itex]x(t)[/itex] be arbitrary. You are then left with a second-order linear ODE for [itex]y(t)[/itex] with singular points wherever [itex]x(t) = 0[/itex].

EDIT: You have [tex]
2\ddot x y + 3 \dot x \dot y + x \ddot y = (\ddot x y + 2 \dot x \dot y + x \ddot y) + (\ddot x y + \dot x \dot y)
= \frac{d^2}{dt^2}(xy) + \frac{d}{dt}(\dot x y) = 0[/tex] so you can integrate once immediately.
 
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  • #9
haruspex
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Let [itex]x(t)[/itex] be arbitrary. You are then left with a second-order linear ODE for [itex]y(t)[/itex] with singular points wherever [itex]x(t) = 0[/itex].

EDIT: You have [tex]
2\ddot x y + 3 \dot x \dot y + x \ddot y = (\ddot x y + 2 \dot x \dot y + x \ddot y) + (\ddot x y + \dot x \dot y)
= \frac{d^2}{dt^2}(xy) + \frac{d}{dt}(\dot x y) = 0[/tex] so you can integrate once immediately.
taking that a bit further, we can set x to be any function of t and obtain ##y=\frac c{x^2}\int x.dt##
 
  • #10
Buzz Bloom
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Hi @Doug Brown:

I am not sure if this idea would be of general use, but I seem to remember learning about this technique as an undergraduate many decades ago.

Define two differential operators as follows:
Dx = (d/dt)(∂/∂y)
Dy = (d/dt)(∂/∂x)​
Then the equation can be written as:
[ 3 Dx2 + 2 Dx Dy +Dy2 ] xy = 0​
This can be factored as follows:
[ [ Dx + Dy ] × [ 2 Dx + Dy ] ] xy = 0​
This produces two equations:
[ Dx + Dy ] xy = 0
and
[ 2 Dx + Dy ] xy = 0​
I don't have a way of using a dot to indicate differentiation WRT t, so I will use an apostrophe instead.
These equations can be written as:
(1) x'y + xy' = 0
(2) 2x'y + xy' = 0
I have not checked this out, but I think that if a pair P1{x(t),y(t)} satisfies (1), and P[SUB2[/SUB]{x(t),y(t)} satisfies (2), then one general solution (of possibly many) of the original equation would be
A × P1{x(t),y(t)} + B × P2{x(t),y(t)}​
with A and B arbitrary constants.

Regards,
Buzz
 
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  • #11
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Hi @Doug Brown:

I am not sure if this idea would be of general use, but I seem to remember learning about this technique as an undergraduate many decades ago.

Define two differential operators as follows:
Dx = (d/dt)(∂/∂y)
Dy = (d/dt)(∂/∂x)
(Above) Is this really what you meant? Usually Dx is a way of writing the operator ##\frac{d}{dx}## and similarly for Dy
Buzz Bloom said:
Then the equation can be written as:
[ 3 Dx2 + 2 Dx Dy +Dy2 ] xy = 0​
This can be factored as follows:
[ [ Dx + Dy ] × [ 2 Dx + Dy ] ] xy = 0​
This produces two equations:
[ Dx + Dy ] xy = 0
and
[ 2 Dx + Dy ] xy = 0
This is what I did using the hint I gave back in post #3.
Buzz Bloom said:
I don't have a way of using a dot to indicate differentiation WRT t, so I will use an apostrophe instead.
These equations can be written as:
(1) x'y + xy' = 0
(2) 2x'y + xy' = 0
I have not checked this out, but I think that if a pair P1{x(t),y(t)} satisfies (1), and P[SUB2[/SUB]{x(t),y(t)} satisfies (2), then one general solution (of possibly many) of the original equation would be
A × P1{x(t),y(t)} + B × P2{x(t),y(t)}​
with A and B arbitrary constants.

Regards,
Buzz
 
  • #12
Buzz Bloom
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(Above) Is this really what you meant? Usually Dx is a way of writing the operator ddx\frac{d}{dx} and similarly for Dy
Hi Mark:

I am familiar with Dx being an alternative notation for d/dx. I don't remember what the notation was to represent (d/dt)(∂/∂y) when I learned differential equations back in the 1950s, so I borrowed the Dx notation for that.

This is what I did using the hint I gave back in post #3.
What I found was that (1) yields y = -1/x, and (2) y = -1/x2. One solution for (1) is x = tα, y = t. This solution seems to be valid for any value of α. Similarly for (2) x = tα, y = t-2α. I have not checked out combinations.

Regards,
Buzz
 
  • #13
haruspex
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Hi Mark:

I am familiar with Dx being an alternative notation for d/dx. I don't remember what the notation was to represent (d/dt)(∂/∂y) when I learned differential equations back in the 1950s, so I borrowed the Dx notation for that.


What I found was that (1) yields y = -1/x, and (2) y = -1/x2. One solution for (1) is x = tα, y = t. This solution seems to be valid for any value of α. Similarly for (2) x = tα, y = t-2α. I have not checked out combinations.

Regards,
Buzz
Seems to me nobody has grasped the significance of my post #8. It shows how you can get all possible solutions by plugging in any integrable function for x(t) that you like, then determining y(t). What more is there to say?
 
  • #14
Buzz Bloom
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Another aspect of the approach I suggested is that it allows one to directly see the possible relationships between the variables x and y in a form independent of the parameter t.

Regards,
Buzz
 

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