# Differential Equations: Uniqueness and two solutions.

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

If :
$(2t+x)x' = 2t^3 -4t -2x$
With:
$x(1) = -1$
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?

• $$(2t+x)x' = t^2 (-2 + 2 t) \ne 2t^3 -4t -x = t^2-2 t$$ Are you sure you wrote the problem and / or the solution $x(t)$ correctly? – Moo Dec 15 '17 at 20:54
• @Moo fixed, sorry for the typo, $-2x$ instead of $-x$ in DE – Matthew Dec 15 '17 at 20:59
• The square root sign is always positive. Presumably what you mean to say with $\sqrt{t^4}$ is $\pm t^2$ – Ross Millikan Dec 15 '17 at 21:07
• @ocallam: Would $x(t) = -t^2 - 2t$ satisfy the IC? – Moo Dec 15 '17 at 21:12
• @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 – Matthew Dec 15 '17 at 21:21