Solving a Nonlinear Differential Equation: y'' + (y')^2 = 1

[Ed Aboud]

Homework Statement

Solve

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

Homework Equations

The Attempt at a Solution

\frac{dy}{dt} = v

\frac{d^2y}{dt^2} = v \frac{dv}{dy}

yv \frac{dv}{dy} + v^2 =1

\frac{dv}{dy} + \frac{v}{y} = \frac{1}{vy}

I(y) = exp ( \int \frac{1}{y} dy)

I(y) = y

y \frac{dv}{dy} + v = \frac{1}{v}

\frac{d}{dy} (vy) = \frac{1}{v}

y= \frac{ln(v)}{v} + \frac{C}{v}

Thanks.

 

[Unco]

Hi Ed,

\frac{d}{dy} (vy) = \frac{1}{v}

y= \frac{ln(v)}{v} + \frac{C}{v}

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
\frac{dv}{dy} + \frac{v}{y} = \frac{1}{vy}
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
yv \frac{dv}{dy} + v^2 =1
you can instead use the method of separation of variables.

 

[Dick]

You are also taking rather of the long way around. The problem is much easier if you notice that your equation is
<br /> \frac{d^2 (y^2/2)}{dt^2} = 1<br />

 

[Ed Aboud]

Ok so here's my new attempt :

y v \frac{dv}{dy} + v^2 = 1


y \frac{dv}{dy} + v = \frac{1}{v}


y \frac{dv}{dy} = \frac{1 - v^2}{v}


\int \frac{v}{1 - v^2} dv = \int \frac{1}{y} dy

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

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

y = \frac{1}{\sqrt{1 - v^2}} . e^C

y \sqrt{1 - v^2} = e^C

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

Thanks for the help.

 

[Ed Aboud]

Sorry I'm not sure where you got
<br /> <br /> \frac{d^2 (y^2/2)}{dt^2} = 1<br /> <br />

from.

 

[Dick]

Sorry I'm not sure where you got
<br /> <br /> \frac{d^2 (y^2/2)}{dt^2} = 1<br /> <br />

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.

 

[Ed Aboud]

I see it now, thanks very much for your help!

 

[Ed Aboud]

Sorry for double-posting but now I'm having doubts if I'm doing the next integral right.

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

\alpha = \frac{1}{2} y^2

\frac{d \alpha}{dt} = v

\frac{dv}{dt} = 1

v = t + C

\frac{d \alpha}{dt} = t + C

\alpha = \frac{1}{2} t^2 + Ct

y^2 = t^2 + Ct

y = \sqrt{t^2 + Ct}

Correct?

 

[Dick]

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.

 

[Unco]

Ok so here's my new attempt :

y v \frac{dv}{dy} + v^2 = 1y \frac{dv}{dy} + v = \frac{1}{v}y \frac{dv}{dy} = \frac{1 - v^2}{v}\int \frac{v}{1 - v^2} dv = \int \frac{1}{y} dy

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

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

y = \frac{1}{\sqrt{1 - v^2}} . e^C

y \sqrt{1 - v^2} = e^C

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.

 

[Ed Aboud]

Cool, got it now.

Thanks again!