The problem is from Adam's Calculus (7th Ed). It is an initial value problem, and I solved it:
[itex]
\begin{cases}
y'=\frac{3+2x^{2}}{x^{2}} \\
y(-2)=1
\end{cases} \\
\implies y=-\frac{3}{x}+2x+\frac{7}{2}
[/itex]
I can see that the solution is not valid for x=0, but the book says that the solutions is only valid for the interval
[itex](-\infty,0)[/itex] because "that is the largest interval that contains the initial point -2 but not the point x=0.
I do not understand this. Why can x not be larger than 0?
[itex]
\begin{cases}
y'=\frac{3+2x^{2}}{x^{2}} \\
y(-2)=1
\end{cases} \\
\implies y=-\frac{3}{x}+2x+\frac{7}{2}
[/itex]
I can see that the solution is not valid for x=0, but the book says that the solutions is only valid for the interval
[itex](-\infty,0)[/itex] because "that is the largest interval that contains the initial point -2 but not the point x=0.
I do not understand this. Why can x not be larger than 0?
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