I have been stuck with the theory side to this question.

If :
$(2t+x)x' = 2t^3 -4t -2x $
$x(1) = -1$
Then $x(t) = t^2 -2t$ is a continuously differentiable solution to the IVP valid for all t.

Find another continuously differentiable solution to the IVP , valid fou all t, and discuss what uniqueness theorem allows two solutions to exist.

My attempt:

Solving the differential equation I got:
$x(t) = \sqrt{t^4} - 2t $

Theory: Now this describes two curves which are both solutions to the differential equation passing through (1,-1) .
My confusion comes from why two solutions can exist and how to find them.
If I let $x' = \frac{2t^3 -4t-2x}{2t+x} $
And differentiate the right hand side with respect to x the derivative will be continuous at (1,-1) but discontinuous at $(\frac{1}{2} , -1)$ and infinitely many other points.
Can I then assume that many solutions can exist and if so what theorem/why?

  • $\begingroup$ $$(2t+x)x' = t^2 (-2 + 2 t) \ne 2t^3 -4t -x = t^2-2 t$$ Are you sure you wrote the problem and / or the solution $x(t)$ correctly? $\endgroup$ – Moo Dec 15 '17 at 20:54
  • $\begingroup$ @Moo fixed, sorry for the typo, $-2x$ instead of $-x$ in DE $\endgroup$ – Matthew Dec 15 '17 at 20:59
  • $\begingroup$ The square root sign is always positive. Presumably what you mean to say with $\sqrt{t^4}$ is $\pm t^2$ $\endgroup$ – Ross Millikan Dec 15 '17 at 21:07
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    $\begingroup$ @ocallam: Would $x(t) = -t^2 - 2t$ satisfy the IC? $\endgroup$ – Moo Dec 15 '17 at 21:12
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    $\begingroup$ @RossMillikan does the initial condition not restrict it to positive? Also if you know of any theorems/ other problems relevant to this it would be very beneficial $\endgroup$ – Matthew Dec 15 '17 at 21:21

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