IGNOU MPH-001 Solved Assignment 2024 cover page for MScPH course

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:
∂ u ∂ x = 2 ∂ u ∂ t + u ∂ u ∂ x = 2 ∂ u ∂ t + u (del u)/(del x)=2(del u)/(del t)+u\frac{\partial u}{\partial x}=2 \frac{\partial u}{\partial t}+u∂u∂x=2∂u∂t+u
where u ( x , 0 ) = 3 e − 3 x u ( x , 0 ) = 3 e − 3 x u(x,0)=3e^(-3x)u(x, 0)=3 e^{-3 x}u(x,0)=3e−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:
f ( 0 , t ) = f ( L , t ) = 0 for t > 0 f ( x , 0 ) = sin ⁡ 3 π x L for 0 < x < L f ( 0 , t ) = f ( L , t ) = 0 for t > 0 f ( x , 0 ) = sin ⁡ 3 π x L for 0 < x < L {:[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}f(0,t)=f(L,t)=0 for t>0f(x,0)=sin⁡3πxL for 0<x<L
Obtain a unique solution.
c) Using recurrence relation, show that
J 2 ′ ( x ) = ( 1 − 4 x 2 ) J 1 ( x ) + 2 x J 0 J 2 ′ ( x ) = 1 − 4 x 2 J 1 ( x ) + 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_0J2′(x)=(1−4x2)J1(x)+2xJ0
where J n ( x ) J n ( x ) J_(n)(x)J_n(x)Jn(x) is the Bessel function of the first kind.
d) Using the recurrence relation, show that
∫ − 1 + 1 x 2 P n + 1 ( x ) P n − 1 ( x ) d x = 2 n ( n + 1 ) ( 2 n − 1 ) ( 2 n + 1 ) ( 2 n + 3 ) ∫ − 1 + 1   x 2 P n + 1 ( x ) P n − 1 ( x ) d x = 2 n ( n + 1 ) ( 2 n − 1 ) ( 2 n + 1 ) ( 2 n + 3 ) 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)}∫−1+1x2Pn+1(x)Pn−1(x)dx=2n(n+1)(2n−1)(2n+1)(2n+3)
e) Using the generating function, establish the relation between H n ( x ) H n ( x ) H_(n)(x)H_n(x)Hn(x) and H n ( − x ) H n ( − x ) H_(n)(-x)H_n(-x)Hn(−x).
2. a) Obtain the eigenvalues and the corresponding eigenvectors of the matrix:
[ − 2 5 4 5 7 5 4 5 − 2 ] − 2 5 4 5 7 5 4 5 − 2 [[-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][−25457545−2]
b) Diagonalized the matrix:
A = [ 2 0 0 1 2 − 1 1 3 − 2 ] A = 2 0 0 1 2 − 1 1 3 − 2 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=[20012−113−2]
The eigenvalues of A A AAA are: λ 1 = 2 , λ 2 = 1 λ 1 = 2 , λ 2 = 1 lambda_(1)=2,lambda_(2)=1\lambda_1=2, \lambda_2=1λ1=2,λ2=1 and λ 3 = − 1 λ 3 = − 1 lambda_(3)=-1\lambda_3=-1λ3=−1.
c) Show that g i j g i j g^(ij)g^{i j}gij is a contravariant tensor of rank 2 .
d) Show that the vectors u → 1 = [ 2 0 − 3 ] u → 1 = 2 0 − 3 vec(u)_(1)=[[2],[0],[-3]]\overrightarrow{\mathrm{u}}_1=\left[\begin{array}{c}2 \\ 0 \\ -3\end{array}\right]u→1=[20−3], u → 2 = [ 1 1 1 ] u → 2 = 1 1 1 vec(u)_(2)=[[1],[1],[1]]\overrightarrow{\mathrm{u}}_2=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]u→2=[111] and u → 3 = [ 1 7 2 ] u → 3 = 1 7 2 vec(u)_(3)=[[1],[7],[2]]\overrightarrow{\mathrm{u}}_3=\left[\begin{array}{l}1 \\ 7 \\ 2\end{array}\right]u→3=[172] are linearly independent.
PART B
3. a) Using Cauchy’s residue theorem, evaluate the integral
∫ 0 2 π d θ 2 + cos ⁡ θ ∫ 0 2 π   d θ 2 + cos ⁡ θ int_(0)^(2pi)(d theta)/(2+cos theta)\int_0^{2 \pi} \frac{d \theta}{2+\cos \theta}∫02πdθ2+cos⁡θ
b) i) Show that the series ∑ n = 1 ∞ z n ( 1 − z ) ∑ n = 1 ∞   z n ( 1 − z ) sum_(n=1)^(oo)z^(n)(1-z)\sum_{n=1}^{\infty} z^n(1-z)∑n=1∞zn(1−z) 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 ( x , y ) = e x cos ⁡ y u ( x , y ) = e x cos ⁡ y u(x,y)=e^(x)cos yu(x, y)=e^x \cos yu(x,y)=excos⁡y.
c) Consider a triangle P P PPP in the z z zzz-plane with vertices at i , 1 − i , 1 + i i , 1 − i , 1 + i i,1-i,1+ii, 1-i, 1+ii,1−i,1+i. Determine the triangle P 0 P 0 P_(0)P_0P0 which mapped P P PPP under the transformation w = 3 z + 4 − 2 i w = 3 z + 4 − 2 i w=3z+4-2iw=3 z+4-2 iw=3z+4−2i. What is the relation between P P PPP and P 0 P 0 P_(0)P_0P0 and also calculate the area magnification.
d) Obtain the Taylor series expansion of cos 2 ⁡ z cos 2 ⁡ z cos^(2)z\cos ^2 zcos2⁡z about z = a z = a z=az=az=a.
  1. a) Obtain the Fourier sine transform of the function
f ( x ) = e − a x a > 0 , 0 < x < ∞ f ( x ) = e − a x a > 0 , 0 < x < ∞ f(x)=e^(-ax)quad a > 0,0 < x < oof(x)=e^{-a x} \quad a>0,0<x<\inftyf(x)=e−axa>0,0<x<∞
b) Determine the Fourier transform of the normalized Gaussian distribution
f ( t ) = 1 2 π 1 τ exp ⁡ [ − t 2 2 τ 2 ] − ∞ < t < ∞ f ( t ) = 1 2 π 1 τ exp ⁡ − t 2 2 τ 2 − ∞ < t < ∞ 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<\inftyf(t)=12π1τexp⁡[−t22τ2]−∞<t<∞
c) Obtain the inverse Laplace transform of
F ( S ) = 2 S − 3 S 2 + 2 S + 2 F ( S ) = 2 S − 3 S 2 + 2 S + 2 F(S)=(2S-3)/(S^(2)+2S+2)F(S)=\frac{2 S-3}{S^2+2 S+2}F(S)=2S−3S2+2S+2
d) Using the Laplace transforms, solve the following initial value problem:
d 2 y d t 2 + ω 2 y = cos ⁡ ω t d 2 y d t 2 + ω 2 y = cos ⁡ ω t (d^(2)y)/(dt^(2))+omega^(2)y=cos omega t\frac{d^2 y}{d t^2}+\omega^2 y=\cos \omega td2ydt2+ω2y=cos⁡ωt
where y = y 0 ; d y d t = v 0 y = y 0 ; d y d t = v 0 y=y_(0);(dy)/(dt)=v_(0)y=y_0 ; \frac{d y}{d t}=v_0y=y0;dydt=v0 at t = 0 t = 0 t=0t=0t=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_3S3 is not commutative.
\(b=c\:cos\:A+a\:cos\:C\)

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:
∂ u ∂ x = 2 ∂ u ∂ t + u ∂ u ∂ x = 2 ∂ u ∂ t + u (del u)/(del x)=2(del u)/(del t)+u\frac{\partial u}{\partial x}=2 \frac{\partial u}{\partial t}+u∂u∂x=2∂u∂t+u
where u ( x , 0 ) = 3 e − 3 x u ( x , 0 ) = 3 e − 3 x u(x,0)=3e^(-3x)u(x, 0)=3 e^{-3 x}u(x,0)=3e−3x.
Answer:
The PDE given is:
∂ u ∂ x = 2 ∂ u ∂ t + u ∂ u ∂ x = 2 ∂ u ∂ t + u (del u)/(del x)=2(del u)/(del t)+u\frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial t} + u∂u∂x=2∂u∂t+u
with the initial condition:
u ( x , 0 ) = 3 e − 3 x u ( x , 0 ) = 3 e − 3 x u(x,0)=3e^(-3x)u(x, 0) = 3 e^{-3 x}u(x,0)=3e−3x
To solve this PDE using separation of variables, let’s assume a solution of the form u ( x , t ) = X ( x ) T ( t ) u ( x , t ) = X ( x ) T ( t ) u(x,t)=X(x)T(t)u(x, t) = X(x)T(t)u(x,t)=X(x)T(t), where X ( x ) X ( x ) X(x)X(x)X(x) is a function of x x xxx only, and T ( t ) T ( t ) T(t)T(t)T(t) is a function of t t ttt only. Substituting this form into the PDE gives:
X ′ ( x ) T ( t ) = 2 X ( x ) T ′ ( t ) + X ( x ) T ( t ) X ′ ( x ) T ( t ) = 2 X ( x ) T ′ ( t ) + X ( x ) T ( t ) 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′(x)T(t)=2X(x)T′(t)+X(x)T(t)
Rearranging, we get:
X ′ ( x ) X ( x ) = 2 T ′ ( t ) T ( t ) + 1 X ′ ( x ) X ( x ) = 2 T ′ ( t ) T ( t ) + 1 (X^(‘)(x))/(X(x))=2(T^(‘)(t))/(T(t))+1\frac{X'(x)}{X(x)} = 2\frac{T'(t)}{T(t)} + 1X′(x)X(x)=2T′(t)T(t)+1
Since the left side depends only on x x xxx and the right side depends only on t t ttt, each side must be equal to a constant, which we shall call λ λ lambda\lambdaλ. Thus, we have two ordinary differential equations (ODEs):
X ′ ( x ) X ( x ) = λ and 2 T ′ ( t ) T ( t ) + 1 = λ X ′ ( x ) X ( x ) = λ and 2 T ′ ( t ) T ( t ) + 1 = λ (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 = \lambdaX′(x)X(x)=λand2T′(t)T(t)+1=λ
Let’s solve these ODEs separately.
For X ( x ) X ( x ) X(x)X(x)X(x), we have:
X ′ ( x ) = λ X ( x ) X ′ ( x ) = λ X ( x ) X^(‘)(x)=lambda X(x)X'(x) = \lambda X(x)X′(x)=λX(x)
And for T ( t ) T ( t ) T(t)T(t)T(t), rearranging the equation gives:
2 T ′ ( t ) = ( λ − 1 ) T ( t ) 2 T ′ ( t ) = ( λ − 1 ) T ( t ) 2T^(‘)(t)=(lambda-1)T(t)2T'(t) = (\lambda – 1)T(t)2T′(t)=(λ−1)T(t)
These lead to exponential solutions for both X ( x ) X ( x ) X(x)X(x)X(x) and T ( t ) T ( t ) T(t)T(t)T(t). Let’s proceed to find the explicit solutions for X ( x ) X ( x ) X(x)X(x)X(x) and T ( t ) T ( t ) T(t)T(t)T(t), and apply the initial condition to determine any specific constants.
The solutions to the ordinary differential equations are:
X ( x ) = C 1 e λ x X ( x ) = C 1 e λ x X(x)=C_(1)e^(lambda x)X(x) = C_1 e^{\lambda x}X(x)=C1eλx
T ( t ) = C 2 e ( λ − 1 ) t 2 T ( t ) = C 2 e ( λ − 1 ) t 2 T(t)=C_(2)e^(((lambda-1)t)/(2))T(t) = C_2 e^{\frac{(\lambda – 1)t}{2}}T(t)=C2e(λ−1)t2
Thus, our solution u ( x , t ) u ( x , t ) u(x,t)u(x, t)u(x,t) can be written as a product of X ( x ) X ( x ) X(x)X(x)X(x) and T ( t ) T ( t ) T(t)T(t)T(t):
u ( x , t ) = X ( x ) T ( t ) = C 1 C 2 e λ x e ( λ − 1 ) t 2 u ( x , t ) = X ( x ) T ( t ) = C 1 C 2 e λ x e ( λ − 1 ) 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(x,t)=X(x)T(t)=C1C2eλxe(λ−1)t2
Let’s simplify this expression:
u ( x , t ) = C e λ x + ( λ − 1 ) t 2 u ( x , t ) = C e λ x + ( λ − 1 ) 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(x,t)=Ceλx+(λ−1)t2
where C = C 1 C 2 C = C 1 C 2 C=C_(1)C_(2)C = C_1 C_2C=C1C2 is a constant.
Now, to determine the specific value of λ λ lambda\lambdaλ and the constant C C CCC, we apply the initial condition u ( x , 0 ) = 3 e − 3 x u ( x , 0 ) = 3 e − 3 x u(x,0)=3e^(-3x)u(x, 0) = 3 e^{-3 x}u(x,0)=3e−3x. Let’s substitute t = 0 t = 0 t=0t = 0t=0 into our solution to match the initial condition.
Upon applying the initial condition and aligning our universe of equations, we’ve discovered that λ = − 3 λ = − 3 lambda=-3\lambda = -3λ=−3 and C = 3 C = 3 C=3C = 3C=3. Therefore, the particular solution to our initial value problem is:
u ( x , t ) = 3 e − 3 x + ( − 3 − 1 ) t 2 = 3 e − 3 x − 2 t u ( x , t ) = 3 e − 3 x + ( − 3 − 1 ) t 2 = 3 e − 3 x − 2 t 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(x,t)=3e−3x+(−3−1)t2=3e−3x−2t
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