# Domain of a solution to an IVP

Consider the initial value problem (IVP) $$ty'+2y=4t^2$$ and $y(1)=2$. We find by the integrating factor method that the general solution is $$y=t^2+\frac{c}{t^2}$$ where $c$ is an arbitrary constant. Until here it is clear, now the author writes that the solution to the IVP is $$y=t^2+\frac{1}{t^2},\; t>0$$ My question is about writing $t>0$ I don't understand why $t$ should be $>0$. Moreover, he adds that the function $$y=t^2+\frac{1}{t^2},\;t<0$$ is not part of the solution of this IVP. Could someone explain why ? thank you for your help!

The Domain of the maximal solution is the largest interval include in domain of definition of the local solution which on contain the initial data (time). In your case $t_0 = 1$ is the initial data time and the domain of your local solution $$y(t) = t^2+\frac{1}{t^2}$$ is $D_u = (-\infty, 0)\cup(0, \infty)$.
Hence $(0, \infty)$ is the largest interval containing $t_0=1$ such and contained in $D_u.$ so the solution is $$y(t) = t^2+\frac{1}{t^2},~~~t\in 0, \infty)$$