Compute the solution of the IVP
$$
\begin{cases}
x'(t) = – \frac{t}{x(t)},\\
\\
x(t_0)=x_0, x_0>0,
\end{cases}
$$
and determine the largest interval where the solution
exists and satisfies the IVP,. depending on $t_0$ and $x_0$.
I have worked out I have to use the method for separable equations but am unsure of how it works if anyone can help me out it would be deeply appreciated sorry this isn't as progressed as it really should be but I have hit a bit of a wall with this subject.
Best Answer
The method for separable equations gives the general solution for the ODE:
$$x(t)=\sqrt{-2t+C},$$
where $C \in \mathbb R.$
If $x(t_0)=x_0$, then
$$x_0=\sqrt{-2_0t+C},$$
Hence $C=x_0^2+2t_0$. Therefore the solution of the IVP is given by
$$x(t)=\sqrt{-2t+x_0^2+2t_0},$$
therefore we must have $-2t+x_0^2+2t_0 >0.$ This means $ t < \frac{x_0^2+2t_0}{2}.$
Consequence: largest interval where the solution exists and satisfies the IVP is
$$(- \infty, \frac{x_0^2+2t_0}{2}).$$