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

• Dukon
In summary, the solution to the equation is x=C(e^{-rt},e^{-st}) where r is a free parameter and the constants C and D are arbitrary.

## 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?

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Doug Brown said:
(can the $construction be used in the post titles also or only in the body?) Don't put equations in the titles in the first place. It violates forum rules. Doug Brown said: ## Homework Statement Given [itex]\dot{x} \equiv { \mathrm{d}x \over \mathrm{d}t }$ 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.

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

Doug Brown said:
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.

Doug Brown said:
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.

Doug Brown said:
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.

Doug Brown said:

## 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.

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pasmith said:
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##

Hi @Doug Brown:

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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Buzz Bloom said:
Hi @Doug Brown:

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

Mark44 said:
(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.

Mark44 said:
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

Buzz Bloom said:
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?

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