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\begin{center}
\textbf{Math 2233.21569}\\[0pt]SOLUTIONS TO SECOND EXAM \\[0pt]April 1, 2021
\end{center}

1. (5 pts) Explain in words and formulas how you would construct the general
solution of $y^{\prime\prime}+p\left(  x\right)  y^{\prime}+q\left(  x\right)
y=g(x)$, given that $y_{1}(x)$ is a solution of $y^{\prime\prime}+p\left(
x\right)  y^{\prime}+q\left(  x\right)  y=0$. (That is, describe the general
procedure, writing down the relevant formulas. It is \textbf{not} necessary to
carry out any calculations.)

\begin{itemize}
\item Step 1: Use Reduction of Order to find a second, independent, solution
of the homogenous equation:%
\[
y_{2}=y_{1}\left(  x\right)  \int\frac{1}{\left(  y_{1}\left(  x\right)
\right)  ^{2}}\exp\left(  -\int^{x}p\left(  s\right)  ds\right)  dx
\]


\item Step 2: Use Variations of Parameters to find a particular solution
$y_{p}\left(  x\right)  $ of the inhomogeneous equation%
\[
y_{p}\left(  x\right)  =-y_{1}\left(  x\right)  \int\frac{y_{2}\left(
x\right)  g\left(  x\right)  }{W\left[  y_{1},y_{2}\right]  }dx\quad+\quad
y_{2}\left(  x\right)  \int\frac{y_{1}\left(  x\right)  g\left(  x\right)
}{W\left[  y_{1},y_{2}\right]  }dx
\]


\item Step 3: The general solution of the inhomogeneous equation can now be
constructed from $y_{1}$, $y_{2}$ and $y_{p}:$%
\[
y\left(  x\right)  =y_{p}\left(  x\right)  +c_{1}y_{1}\left(  x\right)
+c_{2}y_{2}\left(  x\right)
\]
with $c_{1}$ and $c_{2}$ arbitrary constants.
\end{itemize}

2. (10 pts) Given that $y_{1}(x)=x^{2}$ is one solution of $x^{2}%
y^{\prime\prime}-3xy^{\prime}+4y=0$, use Reduction of Order to determine the
general solution of this differential equation.

\begin{itemize}
\item Putting the ODE in standard form, one has $y^{\prime\prime}-\frac{3}%
{x}y^{\prime}+\frac{4}{x^{2}}y=0$ and so $p\left(  x\right)  =-\frac{3}{x}$.
We now apply the Reduction of Order formula to calculate a second independent
solution%
\begin{align*}
y_{2}\left(  x\right)   &  =y_{1}\left(  x\right)  \int\frac{1}{\left(
y_{1}\left(  x\right)  \right)  ^{2}}\exp\left[  -\int p\left(  x\right)
dx\right]  dx\\
&  =x^{2}\int\frac{1}{x^{4}}\exp\left[  -\int\frac{-3}{x}dx\right]  dx\\
&  =x^{2}\int\frac{1}{x^{4}}\exp\left[  +3\ln\left\vert x\right\vert \right]
dx=x^{2}\int\frac{1}{x^{4}}x^{3}dx\\
&  =x^{2}\int\frac{1}{x}\ dx=x^{2}\\
&  =x^{2}\ln\left\vert x\right\vert
\end{align*}
The general solution is thus%
\[
y\left(  x\right)  =c_{1}y_{1}\left(  x\right)  +c_{2}y_{2}\left(  x\right)
=c_{1}x^{2}+c_{2}x^{2}\ln\left\vert x\right\vert
\]

\end{itemize}

\newpage

3. Determine the general solution of the following differential equations.

(a) (5 pts) $y^{\prime\prime}-5y^{\prime}+6y=0$

\begin{itemize}
\item This a constant coefficient ODE. \ Its characteristic equation is
\[
0=\lambda^{2}-5\lambda+6=\left(  \lambda-2\right)  \left(  \lambda-3\right)
\quad\Rightarrow\quad\lambda=2,3\quad\Rightarrow\quad y_{1}\left(  x\right)
=e^{2x}\ ,\ y_{2}\left(  x\right)  =e^{3x}%
\]
and so%
\[
y\left(  x0\right)  =c_{1}e^{2x}+c_{2}e^{3x}%
\]

\end{itemize}

(b) (5 pt) $x^{2}y^{\prime\prime}+5xy^{\prime}-12y=0$

\begin{itemize}
\item This is an Euler type ODE. \ Its auxiliary equation is%
\[
0=r^{2}+\left(  5-1\right)  r-12=\left(  \lambda+6\right)  \left(
\lambda-2\right)  \quad\Rightarrow\quad\lambda=2,-6\quad\Rightarrow\quad
y_{1}\left(  x\right)  =x^{2}\ \ ,\ y_{2}=x^{-6}%
\]
and so%
\[
y\left(  x\right)  =c_{1}x^{2}+c_{2}x^{-6}%
\]

\end{itemize}

(c) (5 pts) $x^{2}y^{\prime\prime}-xy^{\prime}+y=0$

\begin{itemize}
\item This is an Euler type ODE. Its auxiliary equation is%
\[
0=r^{2}+\left(  -1-1\right)  r+1=\left(  r-1\right)  ^{2}\quad\Rightarrow\quad
r=1\quad\Rightarrow\quad y_{1}\left(  x\right)  =x^{1}\ ,\ y_{2}\left(
x\right)  =x^{1}\ln\left\vert x\right\vert
\]
and so%
\[
y\left(  x\right)  =c_{1}x+c_{2}x\ln\left\vert x\right\vert
\]

\end{itemize}

(d) (5 pts) $y^{\prime\prime}+10y^{\prime}+25y=0.$

\begin{itemize}
\item This is a constant coefficient ODE. \ Its characteristic equation is%
\[
0=\lambda^{2}+10\lambda+25=\left(  \lambda+5\right)  ^{2}\quad\Rightarrow
\quad\lambda=-5\quad\Rightarrow\quad y_{1}\left(  x\right)  =e^{-5x}%
\ ,\ y_{2}\left(  x\right)  =xe^{-5x}%
\]
and so%
\[
y\left(  x\right)  =c_{1}e^{-5x}+c_{2}xe^{-5x}%
\]

\end{itemize}

(e) (5 pts) $x^{2}y^{\prime\prime}+5xy^{\prime}+5y=0$

\begin{itemize}
\item This is an Euler type ODE. \ Its auxiliary equation is%
\[
0=r^{2}+\left(  5-1\right)  r+5=r^{2}+4r+5\quad\Rightarrow r=\frac{-4\pm
\sqrt{16-20}}{2}=-2\pm i\quad
\]
Since we have two complex roots $\lambda=\alpha\pm i\beta$ with $\alpha
=-2\ ,\ \beta=1$ , the general solution will be%
\[
y\left(  x\right)  =c_{1}x^{-2}\cos\left(  x \right)
+c_{2}x^{-2}\sin\left(   x \right)
\]

\end{itemize}

(f) (5 pts) $y^{\prime\prime}-2y^{\prime}+10y=0$

\begin{itemize}
\item This is a constant coefficients type ODE. \ Its characteristic equation
is%
\[
0=\lambda^{2}-2\lambda+13\quad\Rightarrow\quad\lambda=\frac{2\pm\sqrt{4-40}%
}{2}=1\pm3i
\]
Since we have two complex roots $\lambda=\alpha\pm i\beta$ with $\alpha=1$ ,
$\beta=3$, the general solution will be%
\[
y\left(  x\right)  =c_{1}e^{x}\cos\left(  3x\right)  +c_{2}e^{x}\sin\left(
3x\right)
\]

\end{itemize}

\newpage

4. Given that $y_{1}(x)=e^{-x}$ and $y_{2}(x)=e^{2x}$ are solutions of
$y^{\prime\prime}-y^{\prime}-2y=0$ :

(a) (10 pts) Use the Method of Variation of Parameters to find a particular
solution of $y^{\prime\prime}-y^{\prime}-2y=6e^{x}.$

\begin{itemize}
\item The differential equation is in standard form with $g\left(  x\right)
=6e^{x}$. \ We have%
\[
W\left[  y_{1},y_{2}\right]  =y_{1}y_{2}^{\prime}=y_{1}^{\prime}y_{2}=\left(
e^{-x}\right)  \left(  2e^{2x}\right)  -\left(  -e^{-x}\right)  \left(
e^{2x}\right)  =3e^{x}%
\]
Applying the Variation of Parameters formula%
\begin{align*}
y_{p}\left(  x\right)   &  =-y_{1}\int\frac{y_{2}g}{W\left[  y_{1}%
,y_{2}\right]  }dx+y_{2}\int\frac{y_{1}g}{W\left[  y_{1},y_{2}\right]  }dx\\
&  =-e^{-x}\int\frac{e^{2x}\left(  6e^{x}\right)  }{3e^{x}}dx+e^{2x}\int%
\frac{e^{-x}\left(  6e^{x}\right)  }{3e^{x}}dx\\
&  =-e^{-x}\int2e^{2x}dx+e^{2x}\int2e^{-x}dx\\
&  =-e^{-x}\left(  e^{2x}\right)  +e^{2x}\left(  -2e^{-x}\right) \\
&  =-3e^{x}%
\end{align*}

\end{itemize}

(b) (10 pts) Find the solution of the differential equation in part (a)
satisfying $y(0)=-1$, $y^{\prime}(0)=-8.$

\begin{itemize}
\item The general solution to the nonhomogeneous ODE will be%
\[
y\left(  x\right)  =y_{p}\left(  x\right)  +c_{1}y_{1}\left(  x\right)
+c_{2}y_{2}\left(  x\right)  =-3e^{x}+c_{1}e^{-x}+c_{2}e^{2x}%
\]
Imposing the initial conditions on the general solution we find%
\begin{align*}
-1  &  =y\left(  0\right)  =-3+c_{1}+c_{2}\\
-8  &  =y^{\prime}\left(  0\right)  =\left.  \left(  -3e^{x}-c_{1}%
e^{-x}+2c_{2}e^{2x}\right)  \right\vert _{x=0}=-3-c_{1}+2c_{2}%
\end{align*}


Solving these two linear equations for $c_{1}$ and $c_{2}$ yields%
\[
c_{1}=3\quad,\quad c_{2}=-1
\]
and so our solution is%
\[
y\left(  x\right)  =-3e^{x}+3e^{-x}-e^{2x}%
\]

\end{itemize}

\newpage

5. Invert the following Laplace Transforms (i.e find the function $f\left(
t\right)  $ with the given Laplace transform).

(a) (10 pts) $\displaystyle\mathcal{L}\left[  f\right]  \left(  s\right)
=\frac{s+2}{s^{2}-2s-3}$

\begin{itemize}
\item
\[
\mathcal{L}\left[  f\right]  \left(  s\right)  =\frac{s+2}{s^{2}-2s-3}%
=\frac{s+2}{\left(  s-3\right)  \left(  s+1\right)  }%
\]
Setting up a Partial Fractions Expansion of the function on the right%
\[
\frac{s+2}{\left(  s-3\right)  \left(  s+1\right)  }=\frac{A}{s-3}+\frac
{B}{s+1}\quad\Rightarrow\quad s+2=\left(  s+1\right)  A+\left(  s-3\right)  B
\]
Using special values of $s$%
\begin{align*}
s  &  =3\quad\Rightarrow\quad5=\left(  4\right)  A+\left(  0\right)
B\quad\Rightarrow\quad A=\frac{5}{4}\\
s  &  =-1\quad\Rightarrow\quad1=\left(  0\right)  A+\left(  -4\right)
B\quad\Rightarrow\quad B=-\frac{1}{4}%
\end{align*}
and so%
\[
\mathcal{L}\left[  f\right]  \left(  s\right)  =\frac{5}{4}\frac{1}{s-3}%
-\frac{1}{4}\frac{1}{s+1}=\frac{5}{4}\mathcal{L}\left[  e^{3x}\right]
-\frac{1}{4}\mathcal{L}\left[  e^{-x}\right]  =\mathcal{L}\left[  \frac{5}%
{4}e^{3x}-\frac{1}{4}e^{-x}\right]
\]
and so%
\[
f\left(  x\right)  =\frac{5}{4}e^{3x}-\frac{1}{4}e^{-x}%
\]

\end{itemize}

(b) (10 pts) $\displaystyle\mathcal{L}\left[  f\right]  \left(  s\right)
=\frac{s}{s^{2}-2s+10}$ (Hint: try completing the square in the dominator)

\begin{itemize}
\item We have%
\[
s^{2}-2s+10=s^{2}-2s+1-1+10=\left(  s-1\right)  ^{2}+9=\left(  s-1\right)
^{2}+3^{2}%
\]
The denominator is thus a sum of squares of the form $\left(  s-a\right)
^{2}+b^{2}$ with $a=1$, $b=3$. We'll thus try to express $\mathcal{L}\left[
f\right]  \left(  s\right)  $ as a combination of the Laplace transforms of
$\mathcal{L}\left[  e^{x}\cos\left(  3x\right)  \right]  =\frac{s-1}{\left(
s-1\right)  ^{2}+3^{2}}$ and $\mathcal{L}\left[  e^{x}\sin\left(  3x\right)
\right]  =\frac{3}{\left(  s-1\right)  ^{2}+3^{2}}:$%
\begin{align*}
\mathcal{L}\left[  f\right]  \left(  s\right)   &  =\frac{s}{s^{2}%
-2x-10}=A\frac{s-1}{\left(  s-1\right)  ^{2}+3^{2}}+B\frac{3}{\left(
s-1\right)  ^{2}+3^{2}}\\
\quad &  \Rightarrow\quad s=\left(  s-1\right)  A+3B
\end{align*}
But now%
\begin{align*}
s &  =1\quad\Rightarrow\quad1=3B\quad\Rightarrow\quad B=\frac{1}{3}\\
s &  =0\quad\Rightarrow\quad0=-A+3B=-A+1\quad\Rightarrow\quad A=1
\end{align*}
Thus,%
\begin{align*}
\mathcal{L}\left[  f\right]  \left(  s\right)   &  =\frac{s-1}{\left(
s-1\right)  ^{2}+3^{2}}+\frac{1}{3}\frac{3}{\left(  s-1\right)  ^{2}+3^{2}}\\
&  =\mathcal{L}\left[  e^{x}\cos\left(  3x\right)  \right]  +\frac{1}%
{3}\mathcal{L}\left[  e^{x}\sin\left(  3x\right)  \right]  =\mathcal{L}\left[
e^{x}\cos\left(  3x\right)  +\frac{1}{3}e^{x}\sin\left(  3x\right)  \right]
\end{align*}
and so%
\[
f\left(  x\right)  =e^{x}\cos\left(  3x\right)  +\frac{1}{3}e^{x}\sin\left(
3x\right)
\]

\end{itemize}

\newpage

6. (15 pts) Solve the following initial value problems \textbf{using the
Laplace transform method}.%
\[
y^{\prime\prime}+5y^{\prime}+6y=0\qquad;\qquad y\left(  0\right)
=2\quad,\quad y^{\prime}\left(  0\right)  =-6
\]


\begin{itemize}
\item Taking the Laplace transform of the differential equation yields%
\[
\left(  s^{2}\mathcal{L}\left[  y\right]  -sy\left(  0\right)  -y^{\prime
}\left(  0\right)  \right)  +5\left(  s\mathcal{L}\left[  y\right]  -y\left(
0\right)  \right)  +6\mathcal{L}\left[  y\right]  =0
\]
or%
\[
\left(  s^{2}+5s+6\right)  \mathcal{L}\left[  y\right]  -s\left(  2\right)
+6-\left(  5\right)  \left(  2\right)  =0
\]
or%
\[
\mathcal{L}\left[  y\right]  =\frac{2s+4}{s^{2}+5s+6}=\frac{2\left(
s+2\right)  }{\left(  s+2\right)  \left(  s+3\right)  }=\frac{2}%
{s+3}=2\mathcal{L}\left[  e^{-3x}\right]  =\mathcal{L}\left[  2e^{-3x}\right]
\]
and so the solution to the initial value problem is%
\[
y\left(  x\right)  =2e^{-3x}%
\]

\end{itemize}


\end{document}
