I haven't taking Differential Equations for over 2 or 3 years and it escapes my memory how to determine when would an IVP (Initial Value Problem) would have

  • no solutions
  • more than one solution
  • precisely one solution

Can someone refresh my memory? For example, we have the problem

$tx'=2x$ with $t \epsilon [-1,1]$ and $x(t_0)=x_0$

After working out the problem I have the following solution:

$x_0=C t_0^2$ where C is arbitrary.

When does this have no solution? And etc.


1 Answer 1


Hint: Consider the case $t_0=0$. Then from the ODE $tx'=2x$ evaluated at $t_0$, we get $$0\cdot x'_0=2x_0 \iff 2x_0=0 \iff x_0=0.$$ Therefore, if $t_0=0$ and $x_0\neq 0$ then there are no solutions. If OTOH $t_0=x_0=0$, then $x(t)=c\,t^2$ is a solution for any $c\in\mathbb{R}$, so there are infinite solutions.

The $t_0\neq 0$ case is left as an exercise.


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