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

 

[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
$$<br /> \frac{d^2 (y^2/2)}{dt^2} = 1<br />$$

from.

 

[Dick]

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

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 :

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

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

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

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

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

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

Thanks for the help.

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!