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Differential equation problem

  • Thread starter Ed Aboud
  • Start date
197
0
1. Homework Statement

Solve

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

2. Homework Equations



3. 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.
 
156
0
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.
 

Dick

Science Advisor
Homework Helper
26,258
618
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]
 
197
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.
 
197
0
Sorry I'm not sure where you got
[tex]

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

[/tex]

from.
 

Dick

Science Advisor
Homework Helper
26,258
618
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.
 
197
0
I see it now, thanks very much for your help!!!
 
197
0
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?
 

Dick

Science Advisor
Homework Helper
26,258
618
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.
 
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.
 
197
0
Cool, got it now.

Thanks again!!
 

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