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\begin{document}

{\bf Question}

Use the method of variation of parameters to find a particular
solution to the following equations:

{\bf NOTE:} you may wish to use earlier methods to check the
answers but your solutions must use variation of parameters to get
the solution.

\begin{enumerate}

\item $y''-5y'+6y=2e^x\qquad (*)$

\item $y''+2y'+y=3e^{-x}$

\item $y''+ y=\tan x \quad 0&lt;x&lt;\pi/2$

\item $y'' -2y' +y = x^{3/2}\;e^x\qquad (*)$

\end{enumerate}



\vspace{0.25in}

{\bf Answer}

\begin{enumerate}

\item
Find $y_c$ from $y_c''-5y_c'+6y_c=0$\\ Auxiliary equation is
$m^2-5m+6=0$ with solutions $m=2$, $m=3$.\\ Hence
$y_c=Ae^{2x}+Be^{3x}$ and therefore take $y_1=e^{2x}$,
$y_2=e^{3x}$\\ Wronskian is $$ W(x)=\left|\begin{array}{cc} e^{2x}
&amp; e^{3x}\\ 2e^{2x} &amp; 3e^{3x} \end{array} \right| = e^{5x} $$ $\ds
v_1(x)=-\int\frac{y_2(x)\left(2e^x\right)}{W(x)}\;dx =
-\int\frac{2e^{4x}}{e^{5x}}\;dx = -\int 2e^{-x}\;dx=2e^{-x}$\\
$\ds v_2(x)=\int\frac{y_1(x)\left(2e^x\right)}{W(x)}\;dx =
\int\frac{2e^{3x}}{e^{5x}}\;dx = \int 2e^{-2x}\;dx=-e^{-2x}$\\
$\ds y_{pi}=v_1 y_1+v_2 y_2= 2e^{-x}e^{2x}+\left(-e^{-2x}
\right)e^{3x}=2e^x-e^x=e^x$\\ General solution is $\ds
y=Ae^{2x}+Be^{3x}+e^x$

\item
Find $y_c$ from $y_c''+2y_c'+y_c=0$\\ Auxiliary equation is
$m^2+2m+1=0$ with solutions $m=-1$ (repeated root).\\ Hence
$y_c=Ae^{-x}+Bxe^{-x}$ and therefore take $y_1=e^{-x}$,
$y_2=xe^{-x}$\\ Wronskian is $$ W(x)=\left|\begin{array}{cc}
e^{-x} &amp; xe^{-x}\\ -e^{-x} &amp; e^{-x}-xe^{-x} \end{array} \right| =
e^{-2x} $$ $\ds
v_1(x)=-\int\frac{y_2(x)\left(3e^{-x}\right)}{W(x)}\;dx =
-\int\frac{xe^{-x}3e^{-x}}{e^{-2x}}\;dx = -\int
3x\;dx=-\frac{3}{2}x^2$\\ $\ds
v_2(x)=\int\frac{y_1(x)\left(3e^{-x}\right)}{W(x)}\;dx =
\int\frac{e^{-x}3e^{-x}}{e^{-2x}}\;dx = \int 3\;dx=3x$\\ $\ds
y_{pi}=v_1 y_1+v_2 y_2= -\frac{3}{2}x^2e^{-x}+3x^2e^{-x}=
\frac{3}{2}x^2e^{-x}$\\ General solution is $\ds
y=Ae^{-x}+Bxe^{-x}+\frac{3}{2}x^2e^{-x}$\\

\item
Find $y_c$ from $y_c''+y_c=0$\\ Auxiliary equation is $m^2+1=0$
with solutions $m=i$, $m=-i$.\\ Hence $y_c=A\sin x + B\cos x$ and
therefore take $y_1=\sin x$, $y_2=\cos x$\\ Wronskian is $$
W(x)=\left|\begin{array}{cc} \sin x &amp; \cos x\\ \cos x &amp; -\sin x
\end{array} \right| = -\sin^2 x - \cos^2 x = -1 $$ $\ds
v_1(x)=-\int\frac{y_2(x)\frac{\sin x}{\cos x}}{W(x)}\;dx =
-\int\frac{\sin x}{-1}\;dx = -\cos x$\\ $\ds
v_2(x)=\int\frac{y_1(x)\frac{\sin x}{\cos x}}{W(x)}\;dx =
\int\frac{\frac{\sin^2 x}{\cos x}}{-1}\;dx = \sin x - \ln(\sec x +
\tan x)$\\ (Note: You can evaluate the integral as follows or
other ways including using computer packages such as Maple)\\
${\bf \Bigg\{} \quad \ds -\int\frac{\sin^2 x}{\cos x}\;dx = -\int
\frac{1-\cos^2x}{\cos x}\;dx =\int \cos x \; dx - \int
\frac{1}{\cos x}\;dx= \sin x -\int \frac{\cos x}{\cos^2 x}\;dx$
$\ds =\sin x -\int\frac{\cos x}{1-\sin^2 x}\; dx = \sin x
-\int\frac{\cos x}{2(1-\sin x)}\;dx +\int\frac{\cos x}{2(1+\sin
x)}\;dx$ $\ds =\sin x + \frac{1}{2}\ln\left(\frac{1-\sin x}{1+\sin
x}\right) =\sin x + \frac{1}{2}\ln\left(\frac{(1-\sin
x)^2}{1-\sin^2 x}\right) \qquad{\bf \Bigg\}}$\\ $\ds y_{pi}=v_1
y_1+v_2 y_2= -\sin x \cos x +\cos x \sin x -\cos x\;\ln(\sec x
+\tan x)$\\ $\ds y_{pi}= -\cos x\;\ln(\sec x +\tan x)$\\ General
solution is $\ds y=A\sin x+B\cos x - \cos x\;\ln(\sec x +\tan
x)$\\

\newpage
\item
Find $y_c$ from $y_c''-2y_c'+y_c=0$\\ Auxiliary equation is
$m^2-2m+1=0$ with solutions $m=1$ (repeated root).\\ Hence
$y_c=Ae^{x}+Bxe^{x}$ and therefore take $y_1=e^{x}$,
$y_2=xe^{x}$\\ Wronskian is $$ W(x)=\left|\begin{array}{cc} e^{x}
&amp; xe^{x}\\ e^{x} &amp; e^{x}+xe^{x} \end{array} \right| = e^{2x} $$
$\ds v_1(x)=-\int\frac{y_2(x)\left(x^{3/2}e^{x}\right)}{W(x)}\;dx
= -\int\frac{xe^{x}x^{3/2}e^{x}}{e^{2x}}\;dx = -\int x^{5/2}\;dx=
-\frac{2}{7}x^{7/2}$\\ $\ds
v_2(x)=\int\frac{y_1(x)\left(x^{3/2}e^{x}\right)}{W(x)}\;dx =
\int\frac{e^{x}x^{3/2}e^{x}}{e^{2x}}\;dx = -\int x^{3/2}\;dx
=\frac{2}{5}x^{5/2}$\\ $\ds y_{pi}=v_1 y_1+v_2 y_2=
-\frac{2}{7}x^{7/2}e^x + \frac{2}{5}x^{7/2}e^x
=\frac{4}{35}x^{7/2}e^x$\\ General solution is $\ds
y=Ae^{x}+Bxe^{x}+\frac{4}{35}x^{7/2}e^x$\\

\end{enumerate}

\end{document}
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