- #1
-JammyDodger-
- 8
- 0
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.
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.