# 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!

## 1 Answer

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 $$y(t) = t^2+\frac{1}{t^2},~~~t\in 0, \infty)$$

• Is this a definition ? and why the domain should be an interval ? – palio Sep 1 '17 at 16:46
• yes, It is a definition of maximal solution – Guy Fsone Sep 1 '17 at 17:01