Differential equation problem

  • Thread starter Ed Aboud
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  • #1
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Homework Statement



Solve

[tex] y \frac{d^2y}{dt^2} + (\frac{dy}{dt})^2 = 1 [/tex]

Homework Equations





The Attempt at a Solution



[tex] \frac{dy}{dt} = v[/tex]

[tex] \frac{d^2y}{dt^2} = v \frac{dv}{dy}[/tex]

[tex] yv \frac{dv}{dy} + v^2 =1 [/tex]

[tex] \frac{dv}{dy} + \frac{v}{y} = \frac{1}{vy} [/tex]

[tex] I(y) = exp ( \int \frac{1}{y} dy)[/tex]

[tex] I(y) = y [/tex]

[tex] y \frac{dv}{dy} + v = \frac{1}{v}[/tex]

[tex] \frac{d}{dy} (vy) = \frac{1}{v}[/tex]

[tex] y= \frac{ln(v)}{v} + \frac{C}{v} [/tex]

Thanks.
 

Answers and Replies

  • #2
156
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Hi Ed,
[tex] \frac{d}{dy} (vy) = \frac{1}{v}[/tex]

[tex] y= \frac{ln(v)}{v} + \frac{C}{v} [/tex]
This is where the error shows up: you are integrating one side with respect to y and the other with respect to v. Notice that your fourth line
[tex] \frac{dv}{dy} + \frac{v}{y} = \frac{1}{vy} [/tex]
is not of the form you would normally apply the integrating factor method you have used: the right-hand side should be a function of y only.

All is not lost, however. From your third line
[tex] yv \frac{dv}{dy} + v^2 =1 [/tex]
you can instead use the method of separation of variables.
 
  • #3
Dick
Science Advisor
Homework Helper
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You are also taking rather of the long way around. The problem is much easier if you notice that your equation is
[tex]
\frac{d^2 (y^2/2)}{dt^2} = 1
[/tex]
 
  • #4
199
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Ok so heres my new attempt :

[tex] y v \frac{dv}{dy} + v^2 = 1[/tex]


[tex] y \frac{dv}{dy} + v = \frac{1}{v} [/tex]


[tex] y \frac{dv}{dy} = \frac{1 - v^2}{v} [/tex]


[tex] \int \frac{v}{1 - v^2} dv = \int \frac{1}{y} dy [/tex]

[tex] ln ( \frac{1}{ \sqrt{1 - v^2} }) = ln (y) + C [/tex]

[tex] y = exp(ln(\frac{1}{\sqrt{1 - v^2}}) + C ) [/tex]

[tex] y = \frac{1}{\sqrt{1 - v^2}} . e^C [/tex]

[tex] y \sqrt{1 - v^2} = e^C [/tex]

I don't think I am getting anywhere with it.

Thanks for the help.
 
  • #5
199
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Sorry I'm not sure where you got
[tex]

\frac{d^2 (y^2/2)}{dt^2} = 1

[/tex]

from.
 
  • #6
Dick
Science Advisor
Homework Helper
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Sorry I'm not sure where you got
[tex]

\frac{d^2 (y^2/2)}{dt^2} = 1

[/tex]

from.

I differentiated y^2/2 twice with respect to t. (y^2/2)'=y*y'. (y*y')'=y*y''+(y')^2. Your equation is y*y''+(y')^2=1.
 
  • #7
199
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I see it now, thanks very much for your help!!!
 
  • #8
199
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Sorry for double-posting but now I'm having doubts if I'm doing the next integral right.

[tex] \frac{d^2}{dt^2} (\frac{1}{2} y^2) = 1[/tex]

[tex] \alpha = \frac{1}{2} y^2[/tex]

[tex] \frac{d \alpha}{dt} = v[/tex]

[tex] \frac{dv}{dt} = 1[/tex]

[tex] v = t + C [/tex]

[tex] \frac{d \alpha}{dt} = t + C[/tex]

[tex] \alpha = \frac{1}{2} t^2 + Ct[/tex]

[tex] y^2 = t^2 + Ct [/tex]

[tex] y = \sqrt{t^2 + Ct} [/tex]

Correct?
 
  • #9
Dick
Science Advisor
Homework Helper
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Just integrate the first equation twice. The first time you get d(y^2/2)/dt=t+C. The second time you get y^2/2=t^2/2+Ct+D. So y^2=t^2+Ct+D. (I didn't multiply C and D by 2 since they are just constants anyway). So your answer is right. But you should have picked up another constant when you integrated dalpha/dt.
 
  • #10
156
0
Ok so heres my new attempt :

[tex] y v \frac{dv}{dy} + v^2 = 1[/tex]


[tex] y \frac{dv}{dy} + v = \frac{1}{v} [/tex]


[tex] y \frac{dv}{dy} = \frac{1 - v^2}{v} [/tex]


[tex] \int \frac{v}{1 - v^2} dv = \int \frac{1}{y} dy [/tex]

[tex] ln ( \frac{1}{ \sqrt{1 - v^2} }) = ln (y) + C [/tex]

[tex] y = exp(ln(\frac{1}{\sqrt{1 - v^2}}) + C ) [/tex]

[tex] y = \frac{1}{\sqrt{1 - v^2}} . e^C [/tex]

[tex] y \sqrt{1 - v^2} = e^C [/tex]

I don't think I am getting anywhere with it.

Thanks for the help.
This wasn't wrong (although technically the C is -C). You can solve your last equation for v, which is dy/dt, and then separate variables again to finish the job (taking care to remember there are two square roots). Just a general note (Dick will go down the other track with you), you should expect two arbitrary constants in your solution.
 
  • #11
199
0
Cool, got it now.

Thanks again!!
 

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