Problem with Second Order Differential Equation

In summary, the conversation is about solving the equation (x'-t)x''-x'=0 by substituting x'=p and x''=p', and using integration to find the solution for x. The conversation includes discussions on recognizing the terms as derivatives of a certain function, rearranging the equation into Bernoulli form, and using substitution and separation of variables to find the solution. Eventually, the conversation leads to the solution x'= t+/-Sqrt[tc+t^2] and x=(t (t +/- Sqrt[2 c + t^2]) + 2 c Log[2 (t +/- Sqrt[2 c + t^2])])/2.
  • #1
scienceguy288
14
0
I am stuck and after several attempts, have made little progress.

Homework Statement



Solve: (x'-t)x''-x'=0

The Attempt at a Solution



I know that the dependent variable x is missing. Therefore, I substitute x'=p and x''=p', giving me, pp'-p't-p=0. However, here I get stuck. If I try to rearrange to give me a Bernoulli form, I get p-(p/p')=t and then get -p'/p+1/p=1/t, which cannot be solved by setting w=y^(1-n) as this would equal 1.

Can anyone give me a push in the right direction from this ( pp'-p't-p=0) point on?
 
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  • #2
pp' - (p't + p) = 0

The term in brackets is the derivative of something you can recognise? And so is the first term?
 
  • #3
epenguin said:
pp' - (p't + p) = 0

The term in brackets is the derivative of something you can recognise? And so is the first term?

Still stumped. I don't recognize that as anything I have seen before as far as I know...
 
  • #4
You don't recognise what when differentiated gives you pp'?

Then for the bracketed term, well to give you a product (p't) in it, the thing differentiated probably has to be a product. :wink:
 
  • #5
epenguin said:
You don't recognise what when differentiated gives you pp'?

Nope...

epenguin said:
Then for the bracketed term, well to give you a product (p't) in it, the thing differentiated probably has to be a product. :wink:

I see that [pt]'=p't+p, but I don't see how that can help solve this...
 
  • #6
[p*p]'=2*p*p'.
 
  • #7
Dick said:
[p*p]'=2*p*p'.

Right, so [(y^2)/2)]'=pp'. I still don't see how that is helpful...Sorry for my thickheadedness, but I am just not seeing it...
 
  • #8
pp' - (p't + p)=[p*p/2-pt]'=0. Are SURE you don't see it? Integrate both sides.
 
  • #9
Dick said:
pp' - (p't + p)=[p*p/2-pt]'=0. Are SURE you don't see it? Integrate both sides.

Okay, so the left will become p/2(p-2t)+c. The right is the integral of a derivative, so it will become simply (p^2)/2-pt. After simplification this just results in 0=0, though?
 
  • #10
Just what sort of 'simplification' are you referring to? p isn't a constant. It's a function.
 
  • #11
Well, Integrating both sides will give us

p/2(p-2t)+c=(p^2)/2-pt
(p^2)/2-2tp/2+c=(p^2)/2-pt
((p^2)/2-pt)+c=(p^2)/2-pt
0=0
 
  • #12
I meant integrate both sides of [p*p/2-pt]'=0. What are YOU doing?
 
  • #13
(p^2)/2-pt=constant?
 
  • #14
scienceguy288 said:
(p^2)/2-pt=constant?

Yes, now solve for p.
 
  • #15
Alright, so we get p=t+/-Sqrt[tc+t^2], integrating that in terms of t to get x=(t (t + Sqrt[2 c + t^2]) + 2 c Log[2 (t + Sqrt[2 c + t^2])])/2 and x=(t (t - Sqrt[2 c + t^2]) - 2 c Log[2 (t + Sqrt[2 c + t^2])])/2
 
  • #16
Thanks for helping me, I finally got the right answer. Your advise was truly indispensable!
 
  • #17
scienceguy288 said:
I am stuck and after several attempts, have made little progress.

Homework Statement



Solve: (x'-t)x''-x'=0

The Attempt at a Solution



I know that the dependent variable x is missing. Therefore, I substitute x'=p and x''=p', giving me, pp'-p't-p=0. However, here I get stuck. If I try to rearrange to give me a Bernoulli form, I get p-(p/p')=t and then get -p'/p+1/p=1/t, which cannot be solved by setting w=y^(1-n) as this would equal 1.

Can anyone give me a push in the right direction from this ( pp'-p't-p=0) point on?
I'm comming to this late and clearly Mark44 has given you sufficient help. But when I looked at that equation, I saw it as (p- t)p'= p or
[tex]\frac{dp}{dt}= \frac{p}{p-t}[/tex]

Now let y= p- t. Then dy/dt= dp/dt- 1 so that dp/dt= dy/dt+ 1 and p= y+ t. The equation becomes
[tex]\frac{dy}{dt}+ 1= \frac{y- t}{y}= 1- \frac{t}{y}[/tex]
which is separable:
[tex]ydy= -tdt[/tex]
so that [itex]y^2= -t^2+ C[/itex].

Now, go back to x: y= p- t= x'- t so that the equation is
[tex](x'- t)^2= C- t^2[/itex]

[itex]x'= t\pm\sqrt{C- t^2}[/itex]
 

FAQ: Problem with Second Order Differential Equation

What is a second order differential equation?

A second order differential equation is a mathematical equation that involves a second derivative of a function. It is commonly used to model physical systems and describe how a variable changes over time.

What are the types of solutions to a second order differential equation?

The types of solutions to a second order differential equation are general solution, particular solution, and boundary value solution. The general solution includes all possible solutions to the equation, while a particular solution is a specific solution that satisfies the given initial conditions. A boundary value solution is a particular solution that satisfies the given boundary conditions.

What is the difference between a linear and a non-linear second order differential equation?

A linear second order differential equation is one in which the dependent variable and its derivatives only appear to the first power. In contrast, a non-linear second order differential equation contains terms with powers higher than one, making it more complex to solve.

What is the method of undetermined coefficients?

The method of undetermined coefficients is a technique used to find a particular solution to a second order differential equation. It involves assuming a form for the particular solution and solving for the coefficients using the given initial or boundary conditions.

How are second order differential equations used in science?

Second order differential equations are used in various scientific fields, such as physics, engineering, and biology, to model and analyze dynamic systems. They can help predict the behavior of physical phenomena and aid in the development of technologies and solutions to real-world problems.

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