Second Order Seperable Differential Equations

[-JammyDodger-]

Hello and thanks to all who read this.

I'm trying to teach myself applied mathematics, specifically (simple) differential equations.

I've been solving them all fine, so far, but now I've come across second order DE's with a squared term, and I can't seem to get the right answer. I'll give an example:

d^2x/dt^2 = (dx/dt)^2

I've been using a substitution as follows:

u = dx/dt

Therefore:

du/dt = u^2

Then seperating:

du/u^2 = dt

Then integrating both sides:

-1/u = t + k

The inital conditions are if x=0, dx/dt (u) =1 and t=0

Therefore k = -1

Then:

-1/u = (t-1)
-1/(dx/dt) = (t-1)
-1/dx = (t-1)dt

Integrating again:

-x = (t^2)/2 - t + C

Initial conditions mean C = 0

Therefore, I get my answer to be:

x = t - (t^2)/2

The correct answer is: x = ln[1/(1-t)]

Could someone please show me where I'm going wrong? Is it just a simple algebraic mistake? Or is my method completely wrong? Thanks.

 

[Ben Niehoff]

-1/u = (t-1)
-1/(dx/dt) = (t-1)
-1/dx = (t-1)dt

Your mistake is here. Try going over this part again.

 

[-JammyDodger-]

Thanks for your reply, I think I see where I was going wrong now.

The -1/(dx/dt) becomes -dt/dx; is that right?

Funnily, I'm still not getting the right answer, I keep getting x=ln[(1-t)/1], whereas the correct answer is x=ln[1/(1-t)].

Any ideas where I'm going wrong?

 

[Ben Niehoff]

Could be a minus sign somewhere, considering that

\ln \left( \frac{1}{1-t} \right) = - \ln \left( \frac{1-t}{1} \right)

 

[-JammyDodger-]

I just can't seem to get the right answer. I get either x=ln[(1-t)] or x=ln[1/(t-1)]. The correct answer is x=ln[1/(1-t)]. Would it be possible for you to run through the solution? I don't think I'm going to get anywhere with it because I keep making the same mistake somewhere.

I get -1/u = t-1 ; I then multiply by -1 to change signs: 1/u = 1-t ; and u =dx/dt, therefore dt/dx = 1-t ; and when I separate I get: dt/(1-t) = dx. Is my error somewhere there?

For reference the problem is d^2x/dt^2 = (dx/dt)^2 ; with initial conditions being: x=0 and dx/dt = 1 when t=0.

Thanks for your replies.

 

[adriank]

It looks like what you have up to dt/(1 - t) = dx is ok. But when you integrate that, you should be getting x = -ln|1 - t|.

 

[coomast]

I just can't seem to get the right answer. I get either x=ln[(1-t)] or x=ln[1/(t-1)]. The correct answer is x=ln[1/(1-t)]. Would it be possible for you to run through the solution? I don't think I'm going to get anywhere with it because I keep making the same mistake somewhere.

I get -1/u = t-1 ; I then multiply by -1 to change signs: 1/u = 1-t ; and u =dx/dt, therefore dt/dx = 1-t ; and when I separate I get: dt/(1-t) = dx. Is my error somewhere there?

For reference the problem is d^2x/dt^2 = (dx/dt)^2 ; with initial conditions being: x=0 and dx/dt = 1 when t=0.

Thanks for your replies.

Hello JammyDodger, When solving these type of problems I do them slightly different. By this I mean that I leave the integration constants as they are and apply the initial and/or boundary conditions to the final solution and not earlier. I can't proof this but I suspect that in some cases you could eliminate solutions by applying the conditions at an early stage. So by leaving the integration constant, you get after the first integral:

\frac{dx}{dt}=u=\frac{-1}{t+C}

or thus:

dx=\frac{-dt}{t+C}

giving:

x=-ln(t+C)+K

which is the general solution. Now applying the condition:

at \qquad t=0 \qquad x=0

gives you:

K=ln(C)

and thus:

x=ln\left( \frac{C}{t+C}\right)

Applying the other one:

at \qquad t=0 \qquad \frac{dx}{dt}=1

gives:

C=-1

and thus the solution:

x=ln\left(\frac{1}{1-t}\right)

Hope this helps. I assume that you have an error somewhere in your calculation, check it again carefully and you should end up with the correct one.

coomast