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\begin{document}
\title{Solutions, 316-IV}
\maketitle

\section{Problem 2.3.18}
Solve the IVP:
\[ \frac{dy}{dx} + 4y -e^{-x} = 0 \ ,\  y(0) = \frac{4}{3}
\]
{\bf Solution:}\\
The integrating factor, $\mu(x)$, is 
\[ \mu(x) = e^{\int 4 dx} = e^{4x} \]
so that
\begin{eqnarray*}
e^{4x}\left(\frac{dy}{dx} + 4y\right) & = &  e^{-x}*e^{4x}=e^{3x} \\
\frac{d}{dx}\left(y e^{4x}    \right) & = &  e^{3x} \\
y e^{4x}& =& \int e^{3x}dx + C \\
y e^{4x} & =& \frac{1}{3}e^{3x} + C  \\
y(x) & =& \frac{1}{3}e^{-x} + Ce^{-4x}  \\
y(0) & = & \frac{1}{3} + C = \frac{4}{3} \rightarrow C = 1
\end{eqnarray*}
Then the solution to the IVP is:
\[ y(x) = \frac{1}{3}e^{-x} + e^{-4x} \ . \]
\newpage
\section{Problem 2.3.23} In Ex. 2 we have that $Ra^1$ decays
into $Ra^2$ at the rate of $40e^{-20t}$kg/sec and the decay 
constant for $Ra^2$ is $k = 5$/sec. I.e. the equation describing the
total amount of $Ra^2$ present, $y(t)$ is:
\[ \frac{dy}{dt} + 5y = 40e^{-20t} \ . \]
Find $y(t)$ for $t \ge 0$ if $y(0) = 10$kg. \\
{\bf Solution:}
The integrating factor, $\mu(t)$, is
\[ \mu(t) = e^{\int 5 dt} = e^{5t} \]
so that
\begin{eqnarray*}
e^{5t}\left(\frac{dy}{dt} + 5y\right) & = &  40e^{-20t}*e^{5t}=40e^{-15x} \\
\frac{d}{dt}\left(y e^{5t}    \right) & = &  40e^{-15t} \\
y e^{5t}& =& 40\int e^{-15t}dt + C \\
ye^{5t} & =&-\frac{40}{15}e^{-15t} + C  \\
y(t) & =&-\frac{40}{15}e^{-20t} + Ce^{-5t}  \\
y(0) & = &-\frac{8}{3} + C = 10 \rightarrow C = \frac{38}{3} = 12.67
\end{eqnarray*}
Then the solution to the IVP is:
\[ y(t) =-\frac{8}{3}e^{-20t} + \frac{38}{3}e^{-5t} \ . \]
\newpage
\section{Problem 2.6.30}
Solve the differential equation:
\[ (x+y-1)dx + (y-x-5)dy = 0 \ . \]
{\bf Solution:} We solve this problem in 3 steps:
\begin{enumerate}
\item Change to $(u,v)$ variables such that the constants are removed:
$u = x-x_0\ ,\ v = y-y_0$ so that
\[x+y-1 = u + v = (x-x_0)+(y-y_0)=x+y-(x_0+y_0)\]
 and
\[y-x-5 =-u + v =-(x-x_0)+(y-y_0)=-x+y +(x_0-y_0)\]
 so that we get the system
\[ \begin{array}{rcr}
   x_0+y_0 & = & 1 \\
   x_0-y_0 & = & -5 \end{array} \]
with solution: $x_0 = -2$, $y_0 = 3$, i.e. 
$u = x+2$, $v = y-3$. In terms of $u,v$ the equation becomes:
\[ (u+v)du + (v-u)dv = 0 \ . \]

\item We now write with $u$ and $v$ as independent and dependent variables,
respectively. We have:
\[ \frac{dv}{du} = \frac{u+v}{u-v} \ .
\] 
This equation can be solved by the substitution
\[ v = uz \ \Rightarrow \frac{dv}{du} = z + u \frac{dz}{du} \ . \]
Substituting:
\[ \frac{dv}{du} =  z + u \frac{dz}{du} = \frac{u+v}{u-v} 
= \frac{1+v/u}{1-v/u}
= \frac{1+z}{1-z}
\ ,
\]
or
\begin{eqnarray*} 
z + u \frac{dz}{du} & = & \frac{1+z}{1-z}  \\
 \frac{dz}{du} & = & \frac{1}{u} \left(\frac{1+z}{1-z} -z\right) \\
 \frac{dz}{du} & = & \frac{1}{u} \frac{1+z-z+z^2}{1-z} \\
\int \frac{1-z}{1+z^2}dz & = & \int \frac{1}{u} du + C\\
\tan^{-1}{z} - \frac{1}{2}\log{(1+z^2)} & = & \log\left|u\right| + C
\end{eqnarray*} 
\item Recal now that $z = v/u$ so that
\begin{eqnarray*} 
 \tan^{-1}{(v/u)} - \frac{1}{2}\log{(1+(v/u)^2)} & = & \log\left|u\right| + C \\
 \tan^{-1}{(v/u)} - \frac{1}{2}\log{\frac{u^2+v^2}{u^2}} & = & \log\left|u\right| + C \\
 \tan^{-1}{(v/u)} - \frac{1}{2}\log{(u^2+v^2)}+ \frac{1}{2}\log{(u^2)} & = & \log\left|u\right| + C \\
 \tan^{-1}{(v/u)} - \frac{1}{2}\log{(u^2+v^2)}+ \log\left|u\right| & = & \log\left|u\right| + C \\
 \tan^{-1}{(v/u)} & = &\frac{1}{2}\log{(u^2+v^2)} + C \ .
\end{eqnarray*} 
We finally substitute back $x,y$:
\[ \tan^{-1}{\left(\frac{y-3}{x+2}\right)}  = \frac{1}{2}\log{((x+2)^2+(y-3)^2)} + C \ .\]
\end{enumerate}
\end{document}
