# IGNOU MPH-001 Solved Assignment 2024 | MSCPH | IGNOU

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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## IGNOU MPH-001 Assignment Question Paper 2024

mph-001-56076b05-8546-4d7c-abca-8a91c9e28c2f

# mph-001-56076b05-8546-4d7c-abca-8a91c9e28c2f

PART A
1. a) Using separation of variables, solve:
$\frac{\mathrm{\partial }u}{\mathrm{\partial }x}=2\frac{\mathrm{\partial }u}{\mathrm{\partial }t}+u$$\frac{\mathrm{\partial }u}{\mathrm{\partial }x}=2\frac{\mathrm{\partial }u}{\mathrm{\partial }t}+u$(del u)/(del x)=2(del u)/(del t)+u\frac{\partial u}{\partial x}=2 \frac{\partial u}{\partial t}+u$\frac{\mathrm{\partial }u}{\mathrm{\partial }x}=2\frac{\mathrm{\partial }u}{\mathrm{\partial }t}+u$
where $u\left(x,0\right)=3{e}^{-3x}$$u\left(x,0\right)=3{e}^{-3x}$u(x,0)=3e^(-3x)u(x, 0)=3 e^{-3 x}$u\left(x,0\right)=3{e}^{-3x}$.
b) Consider a rod of length L, whose ends are kept at a constant temperature and its lateral surface is insulated. The heat flow is described by the 1-D heat diffusion equation subject to the conditions:
$\begin{array}{rl}& f\left(0,t\right)=f\left(L,t\right)=0\text{for}t>0\\ & f\left(x,0\right)=\mathrm{sin}\frac{3\pi x}{L}\text{for}0$\begin{array}{r}f\left(0,t\right)=f\left(L,t\right)=0\text{for}t>0\\ f\left(x,0\right)=\mathrm{sin}\frac{3\pi x}{L}\text{for}0{:[f(0″,”t)=f(L”,”t)=0″ for “t > 0],[f(x”,”0)=sin ((3pi x)/(L))” for “0 < x < L]:}\begin{aligned} & f(0, t)=f(L, t)=0 \text { for } t>0 \\ & f(x, 0)=\sin \frac{3 \pi x}{L} \text { for } 0<x<L \end{aligned}$\begin{array}{rl}& f\left(0,t\right)=f\left(L,t\right)=0\text{for}t>0\\ & f\left(x,0\right)=\mathrm{sin}\frac{3\pi x}{L}\text{for}0
Obtain a unique solution.
c) Using recurrence relation, show that
${J}_{2}^{\mathrm{\prime }}\left(x\right)=\left(1-\frac{4}{{x}^{2}}\right){J}_{1}\left(x\right)+\frac{2}{x}{J}_{0}$${J}_{2}^{\mathrm{\prime }}\left(x\right)=\left(1-\frac{4}{{x}^{2}}\right){J}_{1}\left(x\right)+\frac{2}{x}{J}_{0}$J_(2)^(‘)(x)=(1-(4)/(x^(2)))J_(1)(x)+(2)/(x)J_(0)J_2^{\prime}(x)=\left(1-\frac{4}{x^2}\right) J_1(x)+\frac{2}{x} J_0${J}_{2}^{\mathrm{\prime }}\left(x\right)=\left(1-\frac{4}{{x}^{2}}\right){J}_{1}\left(x\right)+\frac{2}{x}{J}_{0}$
where ${J}_{n}\left(x\right)$${J}_{n}\left(x\right)$J_(n)(x)J_n(x)${J}_{n}\left(x\right)$ is the Bessel function of the first kind.
d) Using the recurrence relation, show that
${\int }_{-1}^{+1}{x}^{2}{P}_{n+1}\left(x\right){P}_{n-1}\left(x\right)dx=\frac{2n\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}$${\int }_{-1}^{+1} {x}^{2}{P}_{n+1}\left(x\right){P}_{n-1}\left(x\right)dx=\frac{2n\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}$int_(-1)^(+1)x^(2)P_(n+1)(x)P_(n-1)(x)dx=(2n(n+1))/((2n-1)(2n+1)(2n+3))\int_{-1}^{+1} x^2 P_{n+1}(x) P_{n-1}(x) d x=\frac{2 n(n+1)}{(2 n-1)(2 n+1)(2 n+3)}${\int }_{-1}^{+1}{x}^{2}{P}_{n+1}\left(x\right){P}_{n-1}\left(x\right)dx=\frac{2n\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}$
e) Using the generating function, establish the relation between ${H}_{n}\left(x\right)$${H}_{n}\left(x\right)$H_(n)(x)H_n(x)${H}_{n}\left(x\right)$ and ${H}_{n}\left(-x\right)$${H}_{n}\left(-x\right)$H_(n)(-x)H_n(-x)${H}_{n}\left(-x\right)$.
2. a) Obtain the eigenvalues and the corresponding eigenvectors of the matrix:
$\left[\begin{array}{ccc}-2& 5& 4\\ 5& 7& 5\\ 4& 5& -2\end{array}\right]$$\left[\begin{array}{ccc}-2& 5& 4\\ 5& 7& 5\\ 4& 5& -2\end{array}\right]$[[-2,5,4],[5,7,5],[4,5,-2]]\left[\begin{array}{ccc} -2 & 5 & 4 \\ 5 & 7 & 5 \\ 4 & 5 & -2 \end{array}\right]$\left[\begin{array}{ccc}-2& 5& 4\\ 5& 7& 5\\ 4& 5& -2\end{array}\right]$
b) Diagonalized the matrix:
$A=\left[\begin{array}{ccc}2& 0& 0\\ 1& 2& -1\\ 1& 3& -2\end{array}\right]$$A=\left[\begin{array}{ccc}2& 0& 0\\ 1& 2& -1\\ 1& 3& -2\end{array}\right]$A=[[2,0,0],[1,2,-1],[1,3,-2]]A=\left[\begin{array}{ccc} 2 & 0 & 0 \\ 1 & 2 & -1 \\ 1 & 3 & -2 \end{array}\right]$A=\left[\begin{array}{ccc}2& 0& 0\\ 1& 2& -1\\ 1& 3& -2\end{array}\right]$
The eigenvalues of $A$$A$AA$A$ are: ${\lambda }_{1}=2,{\lambda }_{2}=1$${\lambda }_{1}=2,{\lambda }_{2}=1$lambda_(1)=2,lambda_(2)=1\lambda_1=2, \lambda_2=1${\lambda }_{1}=2,{\lambda }_{2}=1$ and ${\lambda }_{3}=-1$${\lambda }_{3}=-1$lambda_(3)=-1\lambda_3=-1${\lambda }_{3}=-1$.
c) Show that ${g}^{ij}$${g}^{ij}$g^(ij)g^{i j}${g}^{ij}$ is a contravariant tensor of rank 2 .
d) Show that the vectors ${\stackrel{\to }{\mathrm{u}}}_{1}=\left[\begin{array}{c}2\\ 0\\ -3\end{array}\right]$${\stackrel{\to }{\mathrm{u}}}_{1}=\left[\begin{array}{c}2\\ 0\\ -3\end{array}\right]$vec(u)_(1)=[[2],[0],[-3]]\overrightarrow{\mathrm{u}}_1=\left[\begin{array}{c}2 \\ 0 \\ -3\end{array}\right]${\stackrel{\to }{\mathrm{u}}}_{1}=\left[\begin{array}{c}2\\ 0\\ -3\end{array}\right]$, ${\stackrel{\to }{\mathrm{u}}}_{2}=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$${\stackrel{\to }{\mathrm{u}}}_{2}=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$vec(u)_(2)=[[1],[1],[1]]\overrightarrow{\mathrm{u}}_2=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]${\stackrel{\to }{\mathrm{u}}}_{2}=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$ and ${\stackrel{\to }{\mathrm{u}}}_{3}=\left[\begin{array}{l}1\\ 7\\ 2\end{array}\right]$${\stackrel{\to }{\mathrm{u}}}_{3}=\left[\begin{array}{l}1\\ 7\\ 2\end{array}\right]$vec(u)_(3)=[[1],[7],[2]]\overrightarrow{\mathrm{u}}_3=\left[\begin{array}{l}1 \\ 7 \\ 2\end{array}\right]${\stackrel{\to }{\mathrm{u}}}_{3}=\left[\begin{array}{l}1\\ 7\\ 2\end{array}\right]$ are linearly independent.
PART B
3. a) Using Cauchy’s residue theorem, evaluate the integral
${\int }_{0}^{2\pi }\frac{d\theta }{2+\mathrm{cos}\theta }$${\int }_{0}^{2\pi } \frac{d\theta }{2+\mathrm{cos}\theta }$int_(0)^(2pi)(d theta)/(2+cos theta)\int_0^{2 \pi} \frac{d \theta}{2+\cos \theta}${\int }_{0}^{2\pi }\frac{d\theta }{2+\mathrm{cos}\theta }$
b) i) Show that the series $\sum _{n=1}^{\mathrm{\infty }}{z}^{n}\left(1-z\right)$$\sum _{n=1}^{\mathrm{\infty }} {z}^{n}\left(1-z\right)$sum_(n=1)^(oo)z^(n)(1-z)\sum_{n=1}^{\infty} z^n(1-z)$\sum _{n=1}^{\mathrm{\infty }}{z}^{n}\left(1-z\right)$ converges for $|z|<1$$|z|<1$|z| < 1|z|<1$|z|<1$ and find its sum.
ii) Obtain the analytic function whose real part $u\left(x,y\right)={e}^{x}\mathrm{cos}y$$u\left(x,y\right)={e}^{x}\mathrm{cos}y$u(x,y)=e^(x)cos yu(x, y)=e^x \cos y$u\left(x,y\right)={e}^{x}\mathrm{cos}y$.
c) Consider a triangle $P$$P$PP$P$ in the $z$$z$zz$z$-plane with vertices at $i,1-i,1+i$$i,1-i,1+i$i,1-i,1+ii, 1-i, 1+i$i,1-i,1+i$. Determine the triangle ${P}_{0}$${P}_{0}$P_(0)P_0${P}_{0}$ which mapped $P$$P$PP$P$ under the transformation $w=3z+4-2i$$w=3z+4-2i$w=3z+4-2iw=3 z+4-2 i$w=3z+4-2i$. What is the relation between $P$$P$PP$P$ and ${P}_{0}$${P}_{0}$P_(0)P_0${P}_{0}$ and also calculate the area magnification.
d) Obtain the Taylor series expansion of ${\mathrm{cos}}^{2}z$${\mathrm{cos}}^{2}z$cos^(2)z\cos ^2 z${\mathrm{cos}}^{2}z$ about $z=a$$z=a$z=az=a$z=a$.
1. a) Obtain the Fourier sine transform of the function
$f\left(x\right)={e}^{-ax}\phantom{\rule{1em}{0ex}}a>0,0$f\left(x\right)={e}^{-ax}\phantom{\rule{1em}{0ex}}a>0,0f(x)=e^(-ax)quad a > 0,0 < x < oof(x)=e^{-a x} \quad a>0,0<x<\infty$f\left(x\right)={e}^{-ax}\phantom{\rule{1em}{0ex}}a>0,0
b) Determine the Fourier transform of the normalized Gaussian distribution
$f\left(t\right)=\frac{1}{\sqrt{2\pi }}\frac{1}{\tau }\mathrm{exp}\left[-\frac{{t}^{2}}{2{\tau }^{2}}\right]-\mathrm{\infty }$f\left(t\right)=\frac{1}{\sqrt{2\pi }}\frac{1}{\tau }\mathrm{exp}\left[-\frac{{t}^{2}}{2{\tau }^{2}}\right]-\mathrm{\infty }f(t)=(1)/(sqrt(2pi))(1)/(tau)exp[-(t^(2))/(2tau^(2))]-oo < t < oof(t)=\frac{1}{\sqrt{2 \pi}} \frac{1}{\tau} \exp \left[-\frac{t^2}{2 \tau^2}\right]-\infty<t<\infty$f\left(t\right)=\frac{1}{\sqrt{2\pi }}\frac{1}{\tau }\mathrm{exp}\left[-\frac{{t}^{2}}{2{\tau }^{2}}\right]-\mathrm{\infty }
c) Obtain the inverse Laplace transform of
$F\left(S\right)=\frac{2S-3}{{S}^{2}+2S+2}$$F\left(S\right)=\frac{2S-3}{{S}^{2}+2S+2}$F(S)=(2S-3)/(S^(2)+2S+2)F(S)=\frac{2 S-3}{S^2+2 S+2}$F\left(S\right)=\frac{2S-3}{{S}^{2}+2S+2}$
d) Using the Laplace transforms, solve the following initial value problem:
$\frac{{d}^{2}y}{d{t}^{2}}+{\omega }^{2}y=\mathrm{cos}\omega t$$\frac{{d}^{2}y}{d{t}^{2}}+{\omega }^{2}y=\mathrm{cos}\omega t$(d^(2)y)/(dt^(2))+omega^(2)y=cos omega t\frac{d^2 y}{d t^2}+\omega^2 y=\cos \omega t$\frac{{d}^{2}y}{d{t}^{2}}+{\omega }^{2}y=\mathrm{cos}\omega t$
where $y={y}_{0};\frac{dy}{dt}={v}_{0}$$y={y}_{0};\frac{dy}{dt}={v}_{0}$y=y_(0);(dy)/(dt)=v_(0)y=y_0 ; \frac{d y}{d t}=v_0$y={y}_{0};\frac{dy}{dt}={v}_{0}$ at $t=0$$t=0$t=0t=0$t=0$.
5. a) Show that a cyclic group is abelian.
b) Using appropriate figures, show that the group ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ is not commutative.
$$b^2=c^2+a^2-2ac\:Cos\left(B\right)$$

## MPH-001 Sample Solution 2024

mph-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

# mph-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

PART A
1. a) Using separation of variables, solve:
$\frac{\mathrm{\partial }u}{\mathrm{\partial }x}=2\frac{\mathrm{\partial }u}{\mathrm{\partial }t}+u$$\frac{\mathrm{\partial }u}{\mathrm{\partial }x}=2\frac{\mathrm{\partial }u}{\mathrm{\partial }t}+u$(del u)/(del x)=2(del u)/(del t)+u\frac{\partial u}{\partial x}=2 \frac{\partial u}{\partial t}+u$\frac{\mathrm{\partial }u}{\mathrm{\partial }x}=2\frac{\mathrm{\partial }u}{\mathrm{\partial }t}+u$
where $u\left(x,0\right)=3{e}^{-3x}$$u\left(x,0\right)=3{e}^{-3x}$u(x,0)=3e^(-3x)u(x, 0)=3 e^{-3 x}$u\left(x,0\right)=3{e}^{-3x}$.
The PDE given is:
$\frac{\mathrm{\partial }u}{\mathrm{\partial }x}=2\frac{\mathrm{\partial }u}{\mathrm{\partial }t}+u$$\frac{\mathrm{\partial }u}{\mathrm{\partial }x}=2\frac{\mathrm{\partial }u}{\mathrm{\partial }t}+u$(del u)/(del x)=2(del u)/(del t)+u\frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial t} + u$\frac{\mathrm{\partial }u}{\mathrm{\partial }x}=2\frac{\mathrm{\partial }u}{\mathrm{\partial }t}+u$
with the initial condition:
$u\left(x,0\right)=3{e}^{-3x}$$u\left(x,0\right)=3{e}^{-3x}$u(x,0)=3e^(-3x)u(x, 0) = 3 e^{-3 x}$u\left(x,0\right)=3{e}^{-3x}$
To solve this PDE using separation of variables, let’s assume a solution of the form $u\left(x,t\right)=X\left(x\right)T\left(t\right)$$u\left(x,t\right)=X\left(x\right)T\left(t\right)$u(x,t)=X(x)T(t)u(x, t) = X(x)T(t)$u\left(x,t\right)=X\left(x\right)T\left(t\right)$, where $X\left(x\right)$$X\left(x\right)$X(x)X(x)$X\left(x\right)$ is a function of $x$$x$xx$x$ only, and $T\left(t\right)$$T\left(t\right)$T(t)T(t)$T\left(t\right)$ is a function of $t$$t$tt$t$ only. Substituting this form into the PDE gives:
${X}^{\prime }\left(x\right)T\left(t\right)=2X\left(x\right){T}^{\prime }\left(t\right)+X\left(x\right)T\left(t\right)$${X}^{\prime }\left(x\right)T\left(t\right)=2X\left(x\right){T}^{\prime }\left(t\right)+X\left(x\right)T\left(t\right)$X^(‘)(x)T(t)=2X(x)T^(‘)(t)+X(x)T(t)X'(x)T(t) = 2X(x)T'(t) + X(x)T(t)${X}^{\prime }\left(x\right)T\left(t\right)=2X\left(x\right){T}^{\prime }\left(t\right)+X\left(x\right)T\left(t\right)$
Rearranging, we get:
$\frac{{X}^{\prime }\left(x\right)}{X\left(x\right)}=2\frac{{T}^{\prime }\left(t\right)}{T\left(t\right)}+1$$\frac{{X}^{\prime }\left(x\right)}{X\left(x\right)}=2\frac{{T}^{\prime }\left(t\right)}{T\left(t\right)}+1$(X^(‘)(x))/(X(x))=2(T^(‘)(t))/(T(t))+1\frac{X'(x)}{X(x)} = 2\frac{T'(t)}{T(t)} + 1$\frac{{X}^{\prime }\left(x\right)}{X\left(x\right)}=2\frac{{T}^{\prime }\left(t\right)}{T\left(t\right)}+1$
Since the left side depends only on $x$$x$xx$x$ and the right side depends only on $t$$t$tt$t$, each side must be equal to a constant, which we shall call $\lambda$$\lambda$lambda\lambda$\lambda$. Thus, we have two ordinary differential equations (ODEs):
$\frac{{X}^{\prime }\left(x\right)}{X\left(x\right)}=\lambda \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}2\frac{{T}^{\prime }\left(t\right)}{T\left(t\right)}+1=\lambda$$\frac{{X}^{\prime }\left(x\right)}{X\left(x\right)}=\lambda \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}2\frac{{T}^{\prime }\left(t\right)}{T\left(t\right)}+1=\lambda$(X^(‘)(x))/(X(x))=lambdaquad”and”quad2(T^(‘)(t))/(T(t))+1=lambda\frac{X'(x)}{X(x)} = \lambda \quad \text{and} \quad 2\frac{T'(t)}{T(t)} + 1 = \lambda$\frac{{X}^{\prime }\left(x\right)}{X\left(x\right)}=\lambda \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}2\frac{{T}^{\prime }\left(t\right)}{T\left(t\right)}+1=\lambda$
Let’s solve these ODEs separately.
For $X\left(x\right)$$X\left(x\right)$X(x)X(x)$X\left(x\right)$, we have:
${X}^{\prime }\left(x\right)=\lambda X\left(x\right)$${X}^{\prime }\left(x\right)=\lambda X\left(x\right)$X^(‘)(x)=lambda X(x)X'(x) = \lambda X(x)${X}^{\prime }\left(x\right)=\lambda X\left(x\right)$
And for $T\left(t\right)$$T\left(t\right)$T(t)T(t)$T\left(t\right)$, rearranging the equation gives:
$2{T}^{\prime }\left(t\right)=\left(\lambda -1\right)T\left(t\right)$$2{T}^{\prime }\left(t\right)=\left(\lambda -1\right)T\left(t\right)$2T^(‘)(t)=(lambda-1)T(t)2T'(t) = (\lambda – 1)T(t)$2{T}^{\prime }\left(t\right)=\left(\lambda -1\right)T\left(t\right)$
These lead to exponential solutions for both $X\left(x\right)$$X\left(x\right)$X(x)X(x)$X\left(x\right)$ and $T\left(t\right)$$T\left(t\right)$T(t)T(t)$T\left(t\right)$. Let’s proceed to find the explicit solutions for $X\left(x\right)$$X\left(x\right)$X(x)X(x)$X\left(x\right)$ and $T\left(t\right)$$T\left(t\right)$T(t)T(t)$T\left(t\right)$, and apply the initial condition to determine any specific constants.
The solutions to the ordinary differential equations are:
$X\left(x\right)={C}_{1}{e}^{\lambda x}$$X\left(x\right)={C}_{1}{e}^{\lambda x}$X(x)=C_(1)e^(lambda x)X(x) = C_1 e^{\lambda x}$X\left(x\right)={C}_{1}{e}^{\lambda x}$
$T\left(t\right)={C}_{2}{e}^{\frac{\left(\lambda -1\right)t}{2}}$$T\left(t\right)={C}_{2}{e}^{\frac{\left(\lambda -1\right)t}{2}}$T(t)=C_(2)e^(((lambda-1)t)/(2))T(t) = C_2 e^{\frac{(\lambda – 1)t}{2}}$T\left(t\right)={C}_{2}{e}^{\frac{\left(\lambda -1\right)t}{2}}$
Thus, our solution $u\left(x,t\right)$$u\left(x,t\right)$u(x,t)u(x, t)$u\left(x,t\right)$ can be written as a product of $X\left(x\right)$$X\left(x\right)$X(x)X(x)$X\left(x\right)$ and $T\left(t\right)$$T\left(t\right)$T(t)T(t)$T\left(t\right)$:
$u\left(x,t\right)=X\left(x\right)T\left(t\right)={C}_{1}{C}_{2}{e}^{\lambda x}{e}^{\frac{\left(\lambda -1\right)t}{2}}$$u\left(x,t\right)=X\left(x\right)T\left(t\right)={C}_{1}{C}_{2}{e}^{\lambda x}{e}^{\frac{\left(\lambda -1\right)t}{2}}$u(x,t)=X(x)T(t)=C_(1)C_(2)e^(lambda x)e^(((lambda-1)t)/(2))u(x, t) = X(x)T(t) = C_1 C_2 e^{\lambda x} e^{\frac{(\lambda – 1)t}{2}}$u\left(x,t\right)=X\left(x\right)T\left(t\right)={C}_{1}{C}_{2}{e}^{\lambda x}{e}^{\frac{\left(\lambda -1\right)t}{2}}$
Let’s simplify this expression:
$u\left(x,t\right)=C{e}^{\lambda x+\frac{\left(\lambda -1\right)t}{2}}$$u\left(x,t\right)=C{e}^{\lambda x+\frac{\left(\lambda -1\right)t}{2}}$u(x,t)=Ce^(lambda x+((lambda-1)t)/(2))u(x, t) = C e^{\lambda x + \frac{(\lambda – 1)t}{2}}$u\left(x,t\right)=C{e}^{\lambda x+\frac{\left(\lambda -1\right)t}{2}}$
where $C={C}_{1}{C}_{2}$$C={C}_{1}{C}_{2}$C=C_(1)C_(2)C = C_1 C_2$C={C}_{1}{C}_{2}$ is a constant.
Now, to determine the specific value of $\lambda$$\lambda$lambda\lambda$\lambda$ and the constant $C$$C$CC$C$, we apply the initial condition $u\left(x,0\right)=3{e}^{-3x}$$u\left(x,0\right)=3{e}^{-3x}$u(x,0)=3e^(-3x)u(x, 0) = 3 e^{-3 x}$u\left(x,0\right)=3{e}^{-3x}$. Let’s substitute $t=0$$t=0$t=0t = 0$t=0$ into our solution to match the initial condition.
Upon applying the initial condition and aligning our universe of equations, we’ve discovered that $\lambda =-3$$\lambda =-3$lambda=-3\lambda = -3$\lambda =-3$ and $C=3$$C=3$C=3C = 3$C=3$. Therefore, the particular solution to our initial value problem is:
$u\left(x,t\right)=3{e}^{-3x+\frac{\left(-3-1\right)t}{2}}=3{e}^{-3x-2t}$$u\left(x,t\right)=3{e}^{-3x+\frac{\left(-3-1\right)t}{2}}=3{e}^{-3x-2t}$u(x,t)=3e^(-3x+((-3-1)t)/(2))=3e^(-3x-2t)u(x, t) = 3 e^{-3x + \frac{(-3 – 1)t}{2}} = 3 e^{-3x – 2t}$u\left(x,t\right)=3{e}^{-3x+\frac{\left(-3-1\right)t}{2}}=3{e}^{-3x-2t}$

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$$c^2=a^2+b^2-2ab\:Cos\left(C\right)$$

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