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

\section{Problem 4.3.12}
Verify independence, give general solution and solve IVP for:
\[y''-y = 0\ ;\ y_1(x) = \cosh{x} \ ,\  y_2(x) = \sinh{x} \]
\[y(0) = 1\ ,\ y'(0) = -1 \ . \]
{\bf Solution:}\\
We compute the Wronskian:
\[
\left| \begin{array}{rr}
y_1(x) & y_2(x) \\
y_1^\prime(x) & y_2^\prime(x) \end{array} \right|
=
\left| \begin{array}{rr}
\cosh{x} & \sinh{x} \\
\sinh{x} & \cosh{x} \end{array} \right| = \cosh^2{x} - \sinh^2{x} = 1
\]
Also, both functions satisfy the ODE since
$(\cosh{x})'' = \cosh{x}$, $(\sinh{x})'' = \sinh{x}$
so that $y_1(x)$ and $y_2(x)$ form a fundamental set of solutions.
The general solution is therefore
\[ y_g(x) = A \cosh{x} + B \sinh{x} \]
and
\[ y_g^\prime(x) = A \sinh{x} + B \cosh{x} \]
so 
\begin{eqnarray*} 
y_g(0) = A \cosh{0} + B \sinh{0} & \Rightarrow & A = 1\\
y_g^\prime(0) = A \sinh{0} + B \cosh{0}& \Rightarrow & B = -1
\end{eqnarray*}
The solution to the IVP is
\[ y(t) = \cosh{x} - \sinh{x} \ . \]
\newpage
\section{Problem 4.3.19}
Use Abel's identity to determine the Wronskian of two solutions of
\[ xy'' + (x-1) y' + 3 y = 0 \ . \]
{\bf Solution:}\\
First write equation into normal form $ y''+p(x)y' + q(x) y = 0$:
\[ y'' + \frac{x-1}{x} y' +\frac{3}{x} y = 0 \]
so that 
\[ p(x) = \frac{x-1}{x} \]
and
\[ \int p(x) dx = \int \sfrac{x-1}{x} dx = x-\log{|x|} \ .\]
Then
\[ W\left[y_1,y_2\right] (x) = C e^{-\int p(x) dx } = Cxe^{-x} \ .\]
\newpage
\section{Problem 4.3.23}
Show that the substitution $y(x) = v(x) u(x) $ where
\[ u(x) = e^{-\sfrac{1}{2}\int p(x) dx } \ , \]
transforms the differential equation
\[ y''+p(x)y' + q(x) y = 0 \]
into an equation of the form:
\[ u '' + f(x) u = 0 \ . \]
{\bf Solution:} \\
\begin{eqnarray*}
y & = & v(x) u(x) \\
y'& = & v(x) u'(x) + v'(x) u (x) \\
y''& = & v(x) u''(x) + 2 v'(x)u'(x)+ v''(x) u (x) 
\end{eqnarray*}
and substituting for $u(x)$
\begin{eqnarray*}
q(x) y & = & q(x) v(x) e^{-\sfrac{1}{2}\int p(x) dx} \\
p(x) y'& = &
p(x) \left(-\sfrac{1}{2}v(x) p(x) + v'(x)\right) e^{-\sfrac{1}{2}\int p(x) dx} \\
y''& = &
\left(-\sfrac{1}{2}v'(x) p(x) -\sfrac{1}{2}v(x) p'(x)+ v''(x)+
 \sfrac{1}{4}v(x) p^2(x) - \sfrac{1}{2}v'(x) p(x)\right) e^{-\sfrac{1}{2}\int p(x) dx}
\end{eqnarray*}
and combining we see that the term involving $v'(x)$ cancels and we get
\[ \left( v''(x) + \left[ q(x) -\sfrac{1}{4}p^2(x)-\sfrac{1}{2}p'(x)
\right] v(x) \right)e^{-\sfrac{1}{2}\int p(x) dx} = 0 \ .\]
This is of the desired form with
\[ f(x) = q(x) -\sfrac{1}{4}p^2(x)-\sfrac{1}{2}p'(x) \ . \]
\section{Problem 4.4.5}
Find a second solution for
\[ tx''-(t+1)x'+x = 0 \ , \  t>0 \]
given a solution $x_1(1) = exp(t)$.\\
{\bf Solution:} \\
The equation in canonical form is
\[ x''-\frac{t+1}{t}x'+\frac{1}{t}x = 0 \ , \  t>0 \]
so that 
\[ p(t) = -\frac{t+1}{t} \]
and 
\[ \int p(x) dx =- t - \log{|t|} \Rightarrow e^{-\int p(t) dt} = te^t \]
A second solution can be found in the form
\[ x_2(t) = e^t \int \frac{e^{-\int p(t) dt}}{e^{2t}} dt 
          = e^t \int \frac{te^t}{e^{2t}} dt \]
\[ x_2(t) =  e^t \int te^{-t} dt =
-e^t (t+1) e^{-t} = -(t+1) \ .\]

\newpage
\section{Problem 4.4.13b}
Consider Hermite's eq.
\[ y'' - 2 x y' + \lambda y = 0 \ ,\]
with $\lambda$ a parameter. Give a second solution if we have
$y_1(x) = f(x) = 3x - 2x^3$ for $\lambda = 6$. \\
{\bf Solution:} \\
We have that 
\[ p(x) = - 2 x \Rightarrow e^{-\int p(x) dx} = e^{x^2} \]
and 
\[ y_2(t) = (3x - 2x^3) \int \frac{e^{-\int p(t) dt}}{(3x - 2x^3)^2} dx
          = (3x - 2x^3) \int \frac{e^{x^2}}{(3x - 2x^3)^2} dx \ .
\]
\section{Problem 4.4.14b}
Find a second solution of Legendre's equation
\[ (1-x^2) y'' - 2 x y' + \lambda(\lambda+1)y = 0  \ , \ -1<x<1 \ ,\]
given
\[ y_1(x) = f(x) = 3x^2-1 \ ,\  \lambda = 2 \ . \]
{\bf Solution:} \\
The equation in canonical form is
\[ y'' - \frac{2 x}{(1-x^2)}  y'+ \frac{\lambda(\lambda+1)}{(1-x^2)}  y =0 \]
so that 
\[ p(x) = - \frac{2 x}{(1-x^2)}
\Rightarrow e^{-\int p(x) dx} = (1-x^2)
\]
and 
\[ y_2(t) = (3x - 2x^2) \int \frac{e^{-\int p(t) dt}}{(3x - 2x^2)^2} dx
          = (3x^2 - 1) \int \frac{1-x^2}{(3x^2 - 1)^2} dx \ .
\]
\end{document}
