What is the Solution for x'(t) - x^2 = 1 given x(0) = 1?

[brollysan]

Homework Statement

x' - x^2 = 1, x(0) = 1
find x(t)

Homework Equations

The Attempt at a Solution

This was for an assignment due today, I kept trying but had to give up after 2 hours. I don't get the idea of seperable equations when there is just freaking x involved, what is it that I separate? And all the examples in my calculus made sure they did have a y and x, geez must be hard to separate those..

I just took a wild shot and arrived at: arctan x + 1

 

[Dick]

You separate x and t. x' must mean d(x(t))/dt. There certainly is an arctan in the problem. Take the examples in your calculus book and replace y and x by x and t.

 

[brollysan]

ok so I end up with

dx/dt = 1+ x^2

and then I invert it and get dt/dx = 1/(1+x^2)
and then I get dt = 1/(1+x^2) dx
and take the integral and get x(t) = arctanx + C
and then 1 = arctan 0 + C

So I end up with x(t) = arctanx + 1 ?

I tried drawing it in python using eulers, and apparently what I have arrived at here is wrong

 

[Dick]

If you integrate dt=1/(1+x^2) dx you get t+C=arctan(x). You don't get an x on both sides. Integral dt=t+C. Not x+C.

 

[brollysan]

Meaning that t+ C = arctan(x)

tan(t + C) = tan(arctan(x))
x(t) = tan(t + C) ?

So the solution is x(t) = tan(t + 0.25pi) ?

edit: It would have to be tan(t - 0.25pi) right?

 

[Dick]

Yes, x(t)=tan(t+pi/4). Just because the calculus book always uses y as the dependent variable and x as the independent variable doesn't mean they aren't allowed to throw a problem at you where t is the independent variable and x is the dependent. Just switch the letters around.

 

[Dick]

Meaning that t+ C = arctan(x)

tan(t + C) = tan(arctan(x))
x(t) = tan(t + C) ?

So the solution is x(t) = tan(t + 0.25pi) ?

edit: It would have to be tan(t - 0.25pi) right?

No, tan(0-pi/4)=(-1). Your initial condition said x(0)=+1, not -1.