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

[Dukon]

Homework Statement

Given $\dot{x} \equiv { \mathrm{d}x \over \mathrm{d}t }$ and $\ddot{x}$ is ${ \mathrm{d^2}x \over \mathrm{d}t^2 }$

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

$2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0$

Homework Equations

$2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0$

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?

 

[LCKurtz]

(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.

 

[Mark44]

Homework Statement

Given $\dot{x} \equiv { \mathrm{d}x \over \mathrm{d}t }$ and $\ddot{x}$ is ${ \mathrm{d^2}x \over \mathrm{d}t^2 }$

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

$2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0$

Homework Equations

$2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0$

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.

 

[Dukon]

I know but I said that already in "3. The Attempt at a Solution "
$e^{rt}$ is just a guess
I am looking for a more general more formal solution method

 

[Mark44]

I know but I said that already in "3. The Attempt at a Solution "
$e^{rt}$ 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.

 

[LCKurtz]

I know but I said that already in "3. The Attempt at a Solution "
$e^{rt}$ 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.

 

[haruspex]

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=t^m, 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.

 

[pasmith]

Homework Statement

Given $\dot{x} \equiv { \mathrm{d}x \over \mathrm{d}t }$ and $\ddot{x}$ is ${ \mathrm{d^2}x \over \mathrm{d}t^2 }$

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

$2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0$

Homework Equations

$2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0$

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 $x(t)$ be arbitrary. You are then left with a second-order linear ODE for $y(t)$ with singular points wherever $x(t) = 0$.

EDIT: You have $$2\ddot x y + 3 \dot x \dot y + x \ddot y = (\ddot x y + 2 \dot x \dot y + x \ddot y) + (\ddot x y + \dot x \dot y) = \frac{d^2}{dt^2}(xy) + \frac{d}{dt}(\dot x y) = 0$$ so you can integrate once immediately.

 

[haruspex]

Let $x(t)$ be arbitrary. You are then left with a second-order linear ODE for $y(t)$ with singular points wherever $x(t) = 0$.

EDIT: You have $$2\ddot x y + 3 \dot x \dot y + x \ddot y = (\ddot x y + 2 \dot x \dot y + x \ddot y) + (\ddot x y + \dot x \dot y) = \frac{d^2}{dt^2}(xy) + \frac{d}{dt}(\dot x y) = 0$$ so you can integrate once immediately.

taking that a bit further, we can set x to be any function of t and obtain $y=\frac c{x^2}\int x.dt$

 

[Buzz Bloom]

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:

D_x = (d/dt)(∂/∂y)
D_y = (d/dt)(∂/∂x)​

Then the equation can be written as:

[ 3 D_x² + 2 D_x D_y +D_y² ] xy = 0​

This can be factored as follows:

[ [ D_x + D_y ] × [ 2 D_x + D_y ] ] xy = 0​

This produces two equations:

[ D_x + D_y ] xy = 0
and
[ 2 D_x + D_y ] 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

 

[Mark44]

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:

D_x = (d/dt)(∂/∂y)
D_y = (d/dt)(∂/∂x)
​

(Above) Is this really what you meant? Usually D_x is a way of writing the operator $\frac{d}{dx}$ and similarly for D_y
​

Then the equation can be written as:

[ 3 D_x² + 2 D_x D_y +D_y² ] xy = 0​

This can be factored as follows:

[ [ D_x + D_y ] × [ 2 D_x + D_y ] ] xy = 0​

This produces two equations:

[ D_x + D_y ] xy = 0
and
[ 2 D_x + D_y ] xy = 0
​

This is what I did using the hint I gave back in post #3.
​

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

 

[Buzz Bloom]

(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 D_x 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 D_x 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 = t^α, y = t^-α. This solution seems to be valid for any value of α. Similarly for (2) x = t^α, y = t^-2α. I have not checked out combinations.

Regards,
Buzz

 

[haruspex]

Hi Mark:

I am familiar with D_x 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 D_x notation for that.What I found was that (1) yields y = -1/x, and (2) y = -1/x². One solution for (1) is x = t^α, y = t^-α. This solution seems to be valid for any value of α. Similarly for (2) x = t^α, 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?

 

[Buzz Bloom]

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