### Author Topic: The restriction of the variable $t$?  (Read 999 times)

#### Wei Cui

• Full Member
•   • Posts: 16
• Karma: 11 ##### The restriction of the variable $t$?
« on: September 13, 2018, 02:50:10 PM »
Hello,

I have a question for the Example 4 on 10th edition textbook in page 37.

To satisfy the initial condition, $c$ should be $1$, thus the answer for the question is: $y=t^2+\frac{1}{t^2}$,  $t>0$. The textbook explains that the restriction of variable $t$ to the interval $0<t<\infty$ results from infinite discontinuity in the coefficient $p(t)$ at the origin, but how can we figure it out? How to determine the restriction of $t$?

#### Victor Ivrii ##### Re: The restriction of the variable $t$?
« Reply #1 on: September 13, 2018, 04:43:35 PM »
Solve the initial value problem
\begin{align}
&ty' + 2y = 4t^2, \\
&y(1) = 2.
\end{align}
We see from the equation that there is a problematic point $t=0$ because here bone cannot express $y'$. Solving equation, we get a solution which indeed breaks here and only here. Since the initial condition is at $t=1$, we pick up $(0,\infty)$.

If the initial condition was at $t=-2$ we would pick up $(-\infty, 0)$. We cannot impose an initial condition at $t=0$ or, at least, we need to modify it (the theory will come later).