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The restriction of the variable $t$?

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**Wei Cui**:

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 $t$?

Thanks in advance!

**Victor Ivrii**:

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).

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