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\begin{document}


{\bf Question}

Find the Green's function for the following equations
and use it to solve the corresponding initial-value problems
\begin{description}
\item[(a)] $\ds y''+2y'-8y=2e^{3x}; \quad y(0)=1, \quad y'(0)=-\frac{2}{7}$
\item[(b)] $y''+2y'+y=e^{2x}; \quad y(0)=0, \quad y'(0)=0$
\item[(c)] $xy''-3y'=4x-6; \quad y(1)=0, \quad y'(1)=1$
\end{description}


\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)] $\ds y''+2y'-8y=2e^{3x}; \quad y(0)=1, \quad y'(0)=-\frac{2}{7}$

The homogeneous equation is: \begin{eqnarray*} y''+2y'-8y & = & 0
\\ {\rm A.E.\ \ } m^2 + 2m - 8 & = & 0 \\ (m+4)(m-2) & = & 0 \\ y_1 =
e^{-4x} & & y_2 = e^{2x}
\end{eqnarray*}

Looking at the Wronskian:

$W = \left| \begin{array}{cc} e^{-4x} & e^{2x} \\ -4e^{-4x} &
2e^{2x} \end{array} \right| = 6e^{-2x} $

$\ds G(x,t) = \frac{y_1(t)y_2(x) - y_1(x)y_2(t)}{W(t)} =
\frac{1}{6} \left(e^{2(x-t)} - e^{-4(x-t)}\right)$

Solution of $y'' + 2y' - 8y = 0$ satisfying $y(0) = 1, \, y'(0) =
-\frac{2}{7}$

$\ds \begin{array}{cccc} y(x) = Ae^{-4x} + Be^{2x} & y(0) = 1 &
\Rightarrow & A + B = 1  \\ y'(x) = -4Ae^{-4x} + 2Be^{2x} & y'(0)
= -\frac{2}{7} & \Rightarrow & -4A + 2B = -\frac{2}{7}
\end{array}$

Thus $\ds A = \frac{8}{21};\ \   B = \frac{13}{21}$

So the solution to $\ds y''+2y'-8y=2e^{3x}$ is:
\begin{eqnarray*} y & =& \int_0^x G(x,t)R(t) dt +
\frac{8}{21}e^{-4x} + \frac{13}{21}e^{2x} \\ & = & \frac{1}{3}
\int_0^x (e^{2x}e^t - e^{-4x} e^{7t}) \, dt + \frac{8}{21}e^{-4x}
+ \frac{13}{21}e^{2x} \\ & =& \frac{1}{3}\left[e^{2x}e^t -
\frac{1}{7}e^{3x}e^{7t}\right]_0^x +  \frac{8}{21}e^{-4x} +
\frac{13}{21}e^{2x} \\ & = & \frac{1}{3} \left( e^{3x} - e^{2x} -
\frac{1}{7}e^{3x} + \frac{1}{7}e^{-4x}\right)  \frac{8}{21}e^{-4x}
+ \frac{13}{21}e^{2x} \\ & = & \frac{2}{7}e^{3x} +
\frac{2}{7}e^{2x} + \frac{3}{7} e^{-4x} \end{eqnarray*}


\item[(b)]
$y''+2y'+y=e^{2x}; \quad y(0)=0, \quad y'(0)=0$

The homogeneous equation is: \begin{eqnarray*} y''+2y'+y & = & 0
\\ {\rm A.E.\ \ } m^2 + 2m + 1 & = & 0 \\ (m+1)^2 & = & 0 \\ m = -1
{\rm \ \ twice}\\ y_1 = e^{-x} & & y_2 = xe^{-x}
\end{eqnarray*}

Looking at the Wronskian:

$W = \left| \begin{array}{cc} e^{-x} & xe^{-x} \\ -e^{-x} &
(1-x)e^{-x} \end{array} \right| = e^{-2x} $

$\ds G(x,t) = \frac{y_1(t)y_2(x) - y_1(x)y_2(t)}{W(t)} = (x - t)
e^{-(x-t)}$

Solution of $y'' + 2y' + y = 0$ satisfying $y(0) = 0, \, y'(0) =
0$ is $y(x) = 0 $ (for all $x$)

Solution of $y'' + 2y' + y = e^{2x}$ is:
\begin{eqnarray*} y(x) & = & \int_0^x G(x,t) R(t) \, dt \\ & = &
\int_0^x(xe^{-x}e^{3t} - te^{-x}e^{3t}) \, dt \\ & = & \left[
\frac{1}{3}xe^{-x}e^{3t} \right]_0^x - \left[\frac{1}{3}t
e^{-x}e^{3t} \right]_0^x + \frac{1}{3} \int_0^xe^{-x}e^{3t} \, dt
\\ & = &  \left[\frac{1}{3}xe^{-x}e^{3t} - \frac{1}{3}te^{-x}e^{3t} +
\frac{1}{9}e^{-x}e^{3t}\right]_0^x \\ & =  & \frac{1}{3}xe^{2x} -
\frac{1}{3}xe^{2x} + \frac{1}{9}e^{2x} - \frac{1}{3}xe^{-x} -
\frac{1}{9}e^{-x} \\ & = & \frac{1}{9}(e^{2x} - (1 + 3x)e^{-x})
\end{eqnarray*}

\item[(c)] $xy''-3y'=4x-6; \quad y(1)=0, \quad y'(1)=1$

The homogeneous equation is: \begin{eqnarray*} xy''- 3y' & = & 0
\\ \Rightarrow x^2y'' - 3xy' & = & 0  \\ {\rm so\ Euler\ type\
equation}
\\{\rm Let\ }y=x^K.\\ K(K-1) - 3K & = & 0 \\ \Rightarrow K^2 - 4K & = & 0 \\ K(K - 4) &
= & 0 \\ K = 0 &{\rm or}& K = 4 \\ {\rm so \ \ \ }y_1 = x^4 & &
y_2 = 1 \end{eqnarray*}

Looking at the Wronskian:

$W = \left| \begin{array}{cc} x^4 & 1 \\ 4x^3 & 0
\end{array} \right| = -4x^3 $

$\ds G(x,t) = \frac{y_1(t)y_2(x) - y_1(x)y_2(t)}{W(t)} =
\left(\frac{t^4 - x^4}{-4t^3} \right) = \frac{1}{4}
\left(\frac{x^4}{t^3} - t\right)$

Solution of $\ds y'' - \frac{3}{x}y' = 0$ satisfying $y(1) = 0, \,
y'(1) = 1$ is

$\ds \begin{array}{cccl} y(x) = A + Bx^4 & y(1) = 0 & \Rightarrow
& A + B = 0  \\ y'(x) = 4Bx^{3} & y'(1) = 1 & \Rightarrow & B =
\frac{1}{4} \Rightarrow A = -\frac{1}{4}\end{array}$

Solution of $\ds y'' - \frac{3}{x}y' = 4 - \frac{6}{x}$ is: (N.B.
coefficient of $y''$ is 1)

\begin{eqnarray*}y & = & \int_1^xG(x,t)R(t) \, dt - \frac{1}{4} +
\frac{1}{4}x^4 \\ & = & \frac{1}{4} \int_1^x \left(
\frac{x^4}{t^3} - t\right)\left(4 - \frac{6}{t}\right) \, dt -
\frac{1}{4} + \frac{1}{4}x^4 \\ & = & \frac{1}{4} \int_1^x \left(
\frac{4x^4}{t^3} - 4t - \frac{6x^4}{t^4} + 6 \right)\, dt -
\frac{1}{4} + \frac{1}{4}x^4 \\ & = & \frac{1}{4} \left[
-\frac{2x^4}{t^2} - 2t^2 + \frac{2x^4}{t^3} + 6t\right]_1^x -
\frac{1}{4} + \frac{1}{4}x^4 \\ & = & -x^2 + 2x + \frac{1}{4}x^4 -
\frac{5}{4}
\end{eqnarray*}


\end{description}




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