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\begin{document}


{\bf Question}

Write the following initial value problems in the form of an
integral equation and hence find the solution using Picard
iteration.
\begin{description}
\item[(a)] $y'=y \qquad y(0)=1$
\item[(b)] $y'=y^2 \qquad y(0)=1$
\item[(c)] $y'=2x(1+y) \qquad y(0)=0$
\end{description}



\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)] $y'=y \qquad y(0)=1 \hfill(1)$

The integral equation is $\ds y(x) - 1 = \int_0^x y(t) \, dt$
\hfill(2)

Iteration $\ds y_{n+1}(x) = 1 + \int_0^xy(t) \, dt, \hspace{.2in}
y_0 = 1 $

Hence, \begin{eqnarray*} y_1(x) &amp; = &amp; 1 + \int_0^x dt = 1 + x \\
y_2(x) &amp; = &amp; 1 + \int_0^x (1 + t) \, dt = 1 + x + \frac{1}{2}x^2
\\ y_3(x) &amp; = &amp; 1 + \int_0^x(1 + x + \frac{1}{2}x^2)\, dt = 1 + x +
\frac{1}{2}x^2 + \frac{1}{6}x^3 \\ \\ {\rm This\ suggests }
\\  y_n(x) &amp; = &amp; 1 + x + \frac{x^2}{2} + ... + \frac{x^n}{n!}
\end{eqnarray*}
Prove this by induction:

True for $n = 1.$

Suppose true for $n$ then \begin{eqnarray*} y_{n+1}(x) &amp; = &amp; 1 +
\int_0^x(1 + t + \frac{t^2}{2} + ... + \frac{t^n}{n!}) \, dt \\ &amp;
= &amp; 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...+
\frac{x^{n+1}}{(n+1)!} \end{eqnarray*}

So true for $n$ $\Rightarrow$ true for $n + 1.$

Since it was true for $n = 1$ then by the induction hypothesis it
is true for all $n\in \textbf{N}.$

As $n \rightarrow \infty \hspace{.2in} y_n(x) \rightarrow y(x) = 1
+ x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = e^x$

(You can check that $y = e^x$ satisfies (1))



\item[(b)]$y'=y^2 \qquad y(0)=1 \hfill(1)$

The integral equation is $\ds y(x) - 1 = \int_0^x y^2(t) \, dt$
\hfill(2)

Iteration $\ds y_{n+1}(x) = 1 + \int_0^xy^2(t) \, dt,
\hspace{.2in} y_0 = 1 $

Hence, \begin{eqnarray*} y_1(x) &amp; = &amp; 1 + \int_0^x dt = 1 + x \\
y_2(x) &amp; = &amp; 1 + \int_0^x (1 + t)^2 \, dt \\ &amp; = &amp; 1 + x + x^2 +
\frac{x^3}{3} \\ y_3(x) &amp; = &amp; 1 + \int_0^x(1 + t + t^2 +
\frac{t^3}{3})^2\, dt \\ &amp; = &amp; 1 + \int_0^x (1 + 2t + 3t^2
\frac{8}{3}t^3 + \frac{5}{3}t^4 + \frac{2}{3}t^5 + \frac{1}{9}t^6)
\, dt \\ &amp; = &amp; 1 + x + x^2 + x^3 + \frac{2}{3}x^4 + \frac{1}{3}
x^5 + \frac{1}{9}x^6 + \frac{x^7}{63} \\ {\rm this\ suggests }
\\  y_n(x) &amp; = &amp; (1 + x + x^2 + ... + x^n)p_n(x) \hspace{1in}(3)
\end{eqnarray*}
where $p_n(x)$ is some polynomial.

If $y_n(x)$ has the form (3) then:

\begin{eqnarray*} y_{n+1}(x) &amp; = &amp; 1 + \int_0^x(1 + t + t^2 + ...
+ t^n  + t^{n+1} p_n(t))^2 \, dt  \\ &amp; = &amp; 1 + \int_0^x(1 + 2t +
3t^2 + ... + nt^n  + t^{n+1} q_n(t)) \, dt \\ &amp; &amp;  \hspace{1in}
{\rm (where\ }q_n(t) {\rm \ is\ some\ polynomial)} \\ &amp; = &amp; 1 + x
+ x^2 + ... + x^{n+1} + x^{n+2} r_{n+1}(x) \\ &amp; &amp;  \hspace{1in}
{\rm (where\ }r_n(x) {\rm \ is\ some\ polynomial)}
\end{eqnarray*}

Hence it seems reasonable to suppose that as $n \rightarrow
\infty$

$ y_n(x) \rightarrow y(x) = 1 + x + x^2 + x^3 + ... = (1 -
x)^{-1}$

In fact one can prove this for $|x| &lt; 1$ by looking more carefully
at the remainder term.

One can check that $\ds y = \frac{1}{1 - x}$ is a solution of (1).


\item[(c)]$y'=2x(1+y) \qquad y(0)=0 \hfill(1)$

The integral equation is $\ds y(x) = \int_0^x 2t( 1 + y(t)) \, dt$
\hfill(2)

Iteration: \begin{eqnarray*} y_{n+1}(x) &amp; = &amp; \int_0^x t(1 +
y_n(t)) \, dt \hspace{.2in} y_0 = 0 \\y_1(x) &amp; = &amp; \int_0^x 2t \,
dt = x^2
\\ y_2(x)&amp; = &amp; \int_0^x 2t(1 + t^2) \, dt \\ &amp; = &amp; x^2 + \frac{x^4}{2}
\\ y_3(x) &amp; = &amp; \int_0^x2t(1  + t^2 + \frac{t^4}{2}) \, dt
\\ &amp; = &amp; x^2 + \frac{x^4}{2} + \frac{x^6}{6} \\ {\rm This\ suggests }
\\  y_n(x) &amp; = &amp; x^2 + \frac{x^4}{2!} + ... + \frac{x^{2n}}{n!} \hspace{.2in}(3)
\end{eqnarray*}
If we assume this is true then \begin{eqnarray*} y_{n+1}(x) &amp; = &amp;
\int_0^x 2t(1 + t^2 + \frac{t^4}{2!} + ... + \frac{t^{2n}}{n!}) \,
dt \\ &amp; = &amp; (2t + 2t^3 + t^5 + ... + \frac{2t^{2n+1}}{(n+1)!}) \,
dt \\ &amp; = &amp; x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + ... +
\frac{x^{2(n+1)}}{(n+1)!}\end{eqnarray*}

Hence it is true for  $n \Rightarrow$ true for $n+1.$

Also true for  $n = 1$ so by induction $\ds y_n(x) = x^2 +
\frac{x^4}{2!} + ... + \frac{x^{2n}}{n!}$

As $\ds n \rightarrow \infty \hspace{.2in} y_n(x) \rightarrow y(x)
= x^2 + \frac{x^4}{2!} + ...  = e^{x^2} - 1$

and one can check that $ y = e^{x^2} - 1$ is a solution of (1)


\end{description}


\end{document}
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