Problem with Second Order Differential Equation

[scienceguy288]

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?

 

[epenguin]

pp' - (p't + p) = 0

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

 

[scienceguy288]

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...

 

[epenguin]

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.

 

[scienceguy288]

You don't recognise what when differentiated gives you pp'?

Nope...

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

I see that [pt]'=p't+p, but I don't see how that can help solve this...

 

[Dick]

[p*p]'=2*p*p'.

 

[scienceguy288]

[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...

 

[Dick]

pp' - (p't + p)=[p*p/2-pt]'=0. Are SURE you don't see it? Integrate both sides.

 

[scienceguy288]

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?

 

[Dick]

Just what sort of 'simplification' are you referring to? p isn't a constant. It's a function.

 

[scienceguy288]

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

 

[Dick]

I meant integrate both sides of [p*p/2-pt]'=0. What are YOU doing?

 

[scienceguy288]

(p^2)/2-pt=constant?

 

[Dick]

(p^2)/2-pt=constant?

Yes, now solve for p.

 

[scienceguy288]

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

 

[scienceguy288]

Thanks for helping me, I finally got the right answer. Your advise was truly indispensable!

 

[HallsofIvy]

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 coming 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
$$\frac{dp}{dt}= \frac{p}{p-t}$$

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
$$\frac{dy}{dt}+ 1= \frac{y- t}{y}= 1- \frac{t}{y}$$
which is separable:
$$ydy= -tdt$$
so that $y^2= -t^2+ C$.

Now, go back to x: y= p- t= x'- t so that the equation is
[tex](x'- t)^2= C- t^2[/itex]<br /> <br /> $x'= t\pm\sqrt{C- t^2}$[/tex]