# Thread: Solve the Differential Equation

1. ## Solve the Differential Equation

I have been having some trouble with the following question - i'm not sure how the substitution works and how to change back at the end.

Any help greatly appreciated!

Show that posing x = t +1/z changes the equation x' = x^2 - tx +1 to a linear equation. Then Solve.

2. Originally Posted by rmangan
I have been having some trouble with the following question - i'm not sure how the substitution works and how to change back at the end.

Any help greatly appreciated!

Show that posing x = t +1/z changes the equation x' = x^2 - tx +1 to a linear equation. Then Solve.
Have you made the change of variable? What did you get for the ODE in z?

You should have started from:

$\frac{d}{dt} \left(t+\frac{1}{z}\right) =\left(t+\frac{1}{z}\right)^2 -t\left(t+\frac{1}{z}\right)+1$

and simplified.

CB

3. The substitution is in fact x = t + 1/z. I have substituted in but came across a few difficulties. Firstly, i was unsure as to how to calculate (t + 1/z)'. I came up with t'. Is this correct?

Assuming it is, the ODE yielded t' - (1/z)t = 1 + 1/z^2.

I tried to solve this and ended up with t(z) = zln|z| - 1/2z.

Does this seem correct. My major problem was how to change it back to get a solution x(t).

THanks so much for your help.

4. Originally Posted by rmangan
The substitution is in fact x = t + 1/z. I have substituted in but came across a few difficulties. Firstly, i was unsure as to how to calculate (t + 1/z)'. I came up with t'. Is this correct?
$\frac{d}{dt}\left(t+\frac{1}{z}\right)=1+(-1)\frac{1}{z^2}z'$

CB

5. Ok, then I got z' +tz = -1.

I solved z' + tz = 0 first and got Ce^(-t^2/2)

I couldn't seem to find a particular solution so I varied the constant and ended up with the general solution z(t) = g(t)/g'(t) + c, where g(t) is the integral of e^(t^2/2)

Does this seem correct? The question asks me to express this in terms of the initial conditions (t(0), x(0) ). How do I translate it back to a solution for x(t)?

Thanks again for all your help.