Consider the initial value problem:

\[y'(t) = ty(t),\qquad y(0) = 1.\]

Suppose \(\gamma\) is a solution on \([0,1/2]\), then we could use the fundamental theorem of calculus to see that

\[\gamma(t) = 1 + \int_{0}^{t}s\gamma(s)\,ds\]

The idea in Peano is to find an approximate form of \((\ast)\) which we can solve: For each \(k\in\N\) set

\[\gamma_{k}(t) = 1 \quad\text{for }t\in\left[0,\frac{1}{2k}\right]\]

\((\ast)\)

and

\[\gamma_{k}(t) = 1+\int_{0}^{t-\frac{1}{2k}}s\gamma_{k}(s)\,ds\quad\text{for }t\in\left(\frac{1}{2k},\frac{1}{2}\right]\]

\(k=2\)

\(k=3\)

\(k=4\)

\(k=5\)

\(k=6\)

\(k=7\)

\(k=8\)

\(k=9\)

\(k=10\)

\(k=20\)

\(k=100\)

\(k=1000\)

\[\gamma_{k}(t) =\begin{cases}1 & t\in\left[0,\frac{1}{2k}\right]\\ \displaystyle{1+\int_{0}^{t-\frac{1}{2k}}s\gamma_{k}(s)\,ds} & t\in\left(\frac{1}{2k},\frac{1}{2}\right]\end{cases}\]

\(\gamma_{k}'(t)\)

\(t\gamma_{k}(t)\)

\(k=2\)

\(k=3\)

\(k=4\)

\(k=5\)

Recall, we're trying to solve:

\[y'(t) = ty(t)\]

\(k=6\)

\(k=7\)

\(k=8\)

\(k=9\)

\(k=10\)

\(k=20\)

\(k=100\)

\(k=1000\)

So, let's see how close are \(\gamma_{k}'(t)\) and \(t\gamma_{k}(t)\):

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By John Jasper

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