# IGNOU MMT-007 Solved Assignment 2024 | M.Sc. MACS

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

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## IGNOU MMT-007 Assignment Question Paper 2024

mmt-007-solved-assignment-2024-qp-23dc1574-c435-4730-a3b0-c1df719cc7c8

# mmt-007-solved-assignment-2024-qp-23dc1574-c435-4730-a3b0-c1df719cc7c8

1. a) Show that $f\left(x,y\right)=xy$$f\left(x,y\right)=xy$f(x,y)=xyf(x, y)=x y$f\left(x,y\right)=xy$
i) satisfies a Lipschitz condition on any rectangle $a\le x\le b$$a\le x\le b$a <= x <= ba \leq x \leq b$a\le x\le b$ and $c\le y\le d$$c\le y\le d$c <= y <= dc \leq y \leq d$c\le y\le d$;
ii) satisfies a Lipschitz condition on any strip $a\le x\le b$$a\le x\le b$a <= x <= ba \leq x \leq b$a\le x\le b$ and $-\mathrm{\infty }$-\mathrm{\infty }-oo < y < oo-\infty<y<\infty$-\mathrm{\infty };
iii) does not satisfy a Lipschitz condition on the entire plane.
b) Use Frobenious method to find the series solution about $x=0$$x=0$x=0x=0$x=0$ of the equation
$x\left(1-x\right)\frac{{d}^{2}y}{d{x}^{2}}-\left(1+3x\right)\frac{dy}{dx}-y=0.$$x\left(1-x\right)\frac{{d}^{2}y}{d{x}^{2}}-\left(1+3x\right)\frac{dy}{dx}-y=0.$x(1-x)(d^(2)y)/(dx^(2))-(1+3x)(dy)/(dx)-y=0.x(1-x) \frac{d^2 y}{d x^2}-(1+3 x) \frac{d y}{d x}-y=0 .$x\left(1-x\right)\frac{{d}^{2}y}{d{x}^{2}}-\left(1+3x\right)\frac{dy}{dx}-y=0.$
1. a) For the following differential equation locate and classify its singular points on the $x$$x$xx$x$-axis
i) ${x}^{3}\left(x-1\right){y}^{\mathrm{\prime }\mathrm{\prime }}-2\left(x-1\right){y}^{\mathrm{\prime }}+3xy=0$${x}^{3}\left(x-1\right){y}^{\mathrm{\prime }\mathrm{\prime }}-2\left(x-1\right){y}^{\mathrm{\prime }}+3xy=0$x^(3)(x-1)y^(”)-2(x-1)y^(‘)+3xy=0x^3(x-1) y^{\prime \prime}-2(x-1) y^{\prime}+3 x y=0${x}^{3}\left(x-1\right){y}^{\mathrm{\prime }\mathrm{\prime }}-2\left(x-1\right){y}^{\mathrm{\prime }}+3xy=0$
ii) $\left(3x+1\right)x{y}^{\mathrm{\prime }\mathrm{\prime }}-\left(x+1\right){y}^{\mathrm{\prime }}+2y=0$$\left(3x+1\right)x{y}^{\mathrm{\prime }\mathrm{\prime }}-\left(x+1\right){y}^{\mathrm{\prime }}+2y=0$(3x+1)xy^(”)-(x+1)y^(‘)+2y=0(3 x+1) x y^{\prime \prime}-(x+1) y^{\prime}+2 y=0$\left(3x+1\right)x{y}^{\mathrm{\prime }\mathrm{\prime }}-\left(x+1\right){y}^{\mathrm{\prime }}+2y=0$
b) Show that ${\mathrm{L}}_{\mathrm{n}+1}\left(\mathrm{x}\right)=\left(2\mathrm{n}+1-\mathrm{x}\right){\mathrm{L}}_{\mathrm{n}}\left(\mathrm{x}\right)-{\mathrm{n}}^{2}{\text{}\mathrm{L}}_{\mathrm{n}-1}\left(\mathrm{x}\right)$${\mathrm{L}}_{\mathrm{n}+1}\left(\mathrm{x}\right)=\left(2\mathrm{n}+1-\mathrm{x}\right){\mathrm{L}}_{\mathrm{n}}\left(\mathrm{x}\right)-{\mathrm{n}}^{2}{\text{}\mathrm{L}}_{\mathrm{n}-1}\left(\mathrm{x}\right)$L_(n+1)(x)=(2n+1-x)L_(n)(x)-n^(2)L_(n-1)(x)\mathrm{L}_{\mathrm{n}+1}(\mathrm{x})=(2 \mathrm{n}+1-\mathrm{x}) \mathrm{L}_{\mathrm{n}}(\mathrm{x})-\mathrm{n}^2 \mathrm{~L}_{\mathrm{n}-1}(\mathrm{x})${\mathrm{L}}_{\mathrm{n}+1}\left(\mathrm{x}\right)=\left(2\mathrm{n}+1-\mathrm{x}\right){\mathrm{L}}_{\mathrm{n}}\left(\mathrm{x}\right)-{\mathrm{n}}^{2}{\text{}\mathrm{L}}_{\mathrm{n}-1}\left(\mathrm{x}\right)$.
c) 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)}$${\int }_{-1}^{1} {x}^{2}{P}_{n-1}\left(x\right){P}_{n+1}\left(x\right)dx=\frac{2n\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}$int_(-1)^(1)x^(2)P_(n-1)(x)P_(n+1)(x)dx=(2n(n+1))/((2n-1)(2n+1)(2n+3))\int_{-1}^1 x^2 P_{n-1}(x) P_{n+1}(x) d x=\frac{2 n(n+1)}{(2 n-1)(2 n+1)(2 n+3)}${\int }_{-1}^{1}{x}^{2}{P}_{n-1}\left(x\right){P}_{n+1}\left(x\right)dx=\frac{2n\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}$.
d) Construct Green’s function for the differential equation
$x{y}^{\mathrm{\prime }\mathrm{\prime }}+{y}^{\mathrm{\prime }}=0,\phantom{\rule{1em}{0ex}}0$x{y}^{\mathrm{\prime }\mathrm{\prime }}+{y}^{\mathrm{\prime }}=0,\phantom{\rule{1em}{0ex}}0xy^(”)+y^(‘)=0,quad0 < x < ℓx y^{\prime \prime}+y^{\prime}=0, \quad 0<x<\ell$x{y}^{\mathrm{\prime }\mathrm{\prime }}+{y}^{\mathrm{\prime }}=0,\phantom{\rule{1em}{0ex}}0
under the conditions that $\mathrm{y}\left(0\right)$$\mathrm{y}\left(0\right)$y(0)\mathrm{y}(0)$\mathrm{y}\left(0\right)$ is bounded and $\mathrm{y}\left(\ell \right)=0$$\mathrm{y}\left(\ell \right)=0$y(ℓ)=0\mathrm{y}(\ell)=0$\mathrm{y}\left(\ell \right)=0$.
3. a) Show that between every successive pair of zeros of ${J}_{0}\left(x\right)$${J}_{0}\left(x\right)$J_(0)(x)J_0(x)${J}_{0}\left(x\right)$ there exists a zero of ${J}_{1}\left(x\right)$${J}_{1}\left(x\right)$J_(1)(x)J_1(x)${J}_{1}\left(x\right)$.
b) Using the transformation $y={x}^{1/2}u,2{x}^{3/2}=3z$$y={x}^{1/2}u,2{x}^{3/2}=3z$y=x^(1//2)u,2x^(3//2)=3zy=x^{1 / 2} u, 2 x^{3 / 2}=3 z$y={x}^{1/2}u,2{x}^{3/2}=3z$ find the solution of the equation ${y}^{\mathrm{\prime }\mathrm{\prime }}+x\phantom{\rule{1em}{0ex}}y=0$${y}^{\mathrm{\prime }\mathrm{\prime }}+x\phantom{\rule{1em}{0ex}}y=0$y^(”)+x quad y=0y^{\prime \prime}+x \quad y=0${y}^{\mathrm{\prime }\mathrm{\prime }}+x\phantom{\rule{1em}{0ex}}y=0$ in terms of Bessel’s functions.
c) Show that ${\int }_{0}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{a}\mathrm{x}}{\mathrm{J}}_{0}\left(\mathrm{b}\mathrm{x}\right)\mathrm{d}\mathrm{x}=\frac{1}{\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}},\mathrm{a}>0\text{}\mathrm{b}>0$${\int }_{0}^{\mathrm{\infty }} {\mathrm{e}}^{-\mathrm{a}\mathrm{x}}{\mathrm{J}}_{0}\left(\mathrm{b}\mathrm{x}\right)\mathrm{d}\mathrm{x}=\frac{1}{\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}},\mathrm{a}>0\text{}\mathrm{b}>0$int_(0)^(oo)e^(-ax)J_(0)(bx)dx=(1)/(sqrt(a^(2)+b^(2))),a > 0b > 0\int_0^{\infty} \mathrm{e}^{-\mathrm{ax}} \mathrm{J}_0(\mathrm{bx}) \mathrm{dx}=\frac{1}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}, \mathrm{a}>0 \mathrm{~b}>0${\int }_{0}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{a}\mathrm{x}}{\mathrm{J}}_{0}\left(\mathrm{b}\mathrm{x}\right)\mathrm{d}\mathrm{x}=\frac{1}{\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}},\mathrm{a}>0\text{}\mathrm{b}>0$.
4. a) Find the Laplace transform of $\frac{\mathrm{cos}\sqrt{t}}{\sqrt{t}}$$\frac{\mathrm{cos}\sqrt{t}}{\sqrt{t}}$(cos sqrtt)/(sqrtt)\frac{\cos \sqrt{t}}{\sqrt{t}}$\frac{\mathrm{cos}\sqrt{t}}{\sqrt{t}}$.
b) If ${\mathrm{k}}_{\mathrm{m}}$${\mathrm{k}}_{\mathrm{m}}$k_(m)\mathrm{k}_{\mathrm{m}}${\mathrm{k}}_{\mathrm{m}}$ and ${\mathrm{k}}_{\mathrm{n}}$${\mathrm{k}}_{\mathrm{n}}$k_(n)\mathrm{k}_{\mathrm{n}}${\mathrm{k}}_{\mathrm{n}}$ are distinct roots of Bessel function ${\mathrm{J}}_{\mathrm{p}}\left(\mathrm{k}\mathrm{b}\right)=0$${\mathrm{J}}_{\mathrm{p}}\left(\mathrm{k}\mathrm{b}\right)=0$J_(p)(kb)=0\mathrm{J}_{\mathrm{p}}(\mathrm{kb})=0${\mathrm{J}}_{\mathrm{p}}\left(\mathrm{k}\mathrm{b}\right)=0$ with $\mathrm{p}\ge 0,\text{}\mathrm{b}>0$$\mathrm{p}\ge 0,\text{}\mathrm{b}>0$p >= 0,b > 0\mathrm{p} \geq 0, \mathrm{~b}>0$\mathrm{p}\ge 0,\text{}\mathrm{b}>0$ then show that
${\int }_{0}^{b}x{J}_{p}\left({k}_{m}x\right){J}_{p}\left({k}_{n}x\right)dx=\left\{\begin{array}{lll}0& \text{if}& m\ne n\\ \frac{{b}^{2}}{2}\left[{J}_{p+1}\left({k}_{n}b\right)\right]& \text{if}& m=n\end{array}.$${\int }_{0}^{b} x{J}_{p}\left({k}_{m}x\right){J}_{p}\left({k}_{n}x\right)dx=\left\{\begin{array}{l}0 \text{if} m\ne n\\ \frac{{b}^{2}}{2}\left[{J}_{p+1}\left({k}_{n}b\right)\right] \text{if} m=n\end{array}.\right\$int_(0)^(b)xJ_(p)(k_(m)x)J_(p)(k_(n)x)dx={[0,” if “,m!=n],[(b^(2))/(2)[J_(p+1)(k_(n)b)],” if “,m=n].:}\int_0^b x J_p\left(k_m x\right) J_p\left(k_n x\right) d x=\left\{\begin{array}{lll} 0 & \text { if } & m \neq n \\ \frac{b^2}{2}\left[J_{p+1}\left(k_n b\right)\right] & \text { if } & m=n \end{array} .\right.${\int }_{0}^{b}x{J}_{p}\left({k}_{m}x\right){J}_{p}\left({k}_{n}x\right)dx=\left\{\begin{array}{lll}0& \text{if}& m\ne n\\ \frac{{b}^{2}}{2}\left[{J}_{p+1}\left({k}_{n}b\right)\right]& \text{if}& m=n\end{array}.$
1. a) Solve the following IBVP using the Laplace transform technique:
$\begin{array}{rl}& {\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}},\phantom{\rule{1em}{0ex}}0<\mathrm{x}<1,\phantom{\rule{1em}{0ex}}\mathrm{t}>0.\\ & \mathrm{u}\left(0,\mathrm{t}\right)=1,\mathrm{u}\left(1,\mathrm{t}\right)=1,\phantom{\rule{1em}{0ex}}\mathrm{t}>0\\ & \mathrm{u}\left(\mathrm{x},0\right)=1+\mathrm{sin}\pi \mathrm{x},\phantom{\rule{1em}{0ex}}0<\mathrm{x}<1.\end{array}$$\begin{array}{r}{\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}},\phantom{\rule{1em}{0ex}}0<\mathrm{x}<1,\phantom{\rule{1em}{0ex}}\mathrm{t}>0.\\ \mathrm{u}\left(0,\mathrm{t}\right)=1,\mathrm{u}\left(1,\mathrm{t}\right)=1,\phantom{\rule{1em}{0ex}}\mathrm{t}>0\\ \mathrm{u}\left(\mathrm{x},0\right)=1+\mathrm{sin}\pi \mathrm{x},\phantom{\rule{1em}{0ex}}0<\mathrm{x}<1.\end{array}${:[u_(t)=u_(xx)”,”quad0 < x < 1″,”quadt > 0.],[u(0″,”t)=1″,”u(1″,”t)=1″,”quadt > 0],[u(x”,”0)=1+sin pix”,”quad0 < x < 1.]:}\begin{aligned} & \mathrm{u}_{\mathrm{t}}=\mathrm{u}_{\mathrm{xx}}, \quad 0<\mathrm{x}<1, \quad \mathrm{t}>0 . \\ & \mathrm{u}(0, \mathrm{t})=1, \mathrm{u}(1, \mathrm{t})=1, \quad \mathrm{t}>0 \\ & \mathrm{u}(\mathrm{x}, 0)=1+\sin \pi \mathrm{x}, \quad 0<\mathrm{x}<1 . \end{aligned}$\begin{array}{rl}& {\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}},\phantom{\rule{1em}{0ex}}0<\mathrm{x}<1,\phantom{\rule{1em}{0ex}}\mathrm{t}>0.\\ & \mathrm{u}\left(0,\mathrm{t}\right)=1,\mathrm{u}\left(1,\mathrm{t}\right)=1,\phantom{\rule{1em}{0ex}}\mathrm{t}>0\\ & \mathrm{u}\left(\mathrm{x},0\right)=1+\mathrm{sin}\pi \mathrm{x},\phantom{\rule{1em}{0ex}}0<\mathrm{x}<1.\end{array}$
b) If the Fourier cosine transform of $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ is ${\alpha }^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{a}\alpha }$${\alpha }^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{a}\alpha }$alpha^(n)e^(-aalpha)\alpha^{\mathrm{n}} \mathrm{e}^{-\mathrm{a} \alpha}${\alpha }^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{a}\alpha }$, then show that
$f\left(x\right)=\frac{2}{\pi }\frac{n!\mathrm{cos}\left(n+1\right)\theta }{{\left({a}^{2}+{x}^{2}\right)}^{\frac{n+1}{2}}}.$$f\left(x\right)=\frac{2}{\pi }\frac{n!\mathrm{cos}\left(n+1\right)\theta }{{\left({a}^{2}+{x}^{2}\right)}^{\frac{n+1}{2}}}.$f(x)=(2)/(pi)(n!cos(n+1)theta)/((a^(2)+x^(2))^((n+1)/(2))).f(x)=\frac{2}{\pi} \frac{n ! \cos (n+1) \theta}{\left(a^2+x^2\right)^{\frac{n+1}{2}}} .$f\left(x\right)=\frac{2}{\pi }\frac{n!\mathrm{cos}\left(n+1\right)\theta }{{\left({a}^{2}+{x}^{2}\right)}^{\frac{n+1}{2}}}.$
1. a) Find the displacement $u\left(x,t\right)$$u\left(x,t\right)$u(x,t)u(x, t)$u\left(x,t\right)$ of an infinite string using the method of Fourier transform given that the string is initially at rest and that the initial displacement is $\mathrm{f}\left(\mathrm{x}\right),-\mathrm{\infty }<\mathrm{x}<\mathrm{\infty }$$\mathrm{f}\left(\mathrm{x}\right),-\mathrm{\infty }<\mathrm{x}<\mathrm{\infty }$f(x),-oo < x < oo\mathrm{f}(\mathrm{x}),-\infty<\mathrm{x}<\infty$\mathrm{f}\left(\mathrm{x}\right),-\mathrm{\infty }<\mathrm{x}<\mathrm{\infty }$.
b) Using Fourier integral representation show that
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(\alpha \mathrm{x}\right)+\alpha \mathrm{sin}\left(\alpha \mathrm{x}\right)}{1+{\alpha }^{2}}\text{}\mathrm{d}\alpha =\left\{\begin{array}{ccc}0& \text{if}& \mathrm{x}<0\\ \pi /2& \text{if}& \mathrm{x}=0.\\ \pi {\mathrm{e}}^{-\mathrm{x}}& \text{if}& \mathrm{x}>0\end{array}$${\int }_{0}^{\mathrm{\infty }} \frac{\mathrm{cos}\left(\alpha \mathrm{x}\right)+\alpha \mathrm{sin}\left(\alpha \mathrm{x}\right)}{1+{\alpha }^{2}}\text{}\mathrm{d}\alpha =\left\{\begin{array}{ccc}0& \text{if}& \mathrm{x}<0\\ \pi /2& \text{if}& \mathrm{x}=0.\\ \pi {\mathrm{e}}^{-\mathrm{x}}& \text{if}& \mathrm{x}>0\end{array}\right\$int_(0)^(oo)(cos(alphax)+alpha sin(alphax))/(1+alpha^(2))dalpha={[0,” if “,x < 0],[pi//2,” if “,x=0.],[pie^(-x),” if “,x > 0]:}\int_0^{\infty} \frac{\cos (\alpha \mathrm{x})+\alpha \sin (\alpha \mathrm{x})}{1+\alpha^2} \mathrm{~d} \alpha=\left\{\begin{array}{ccc} 0 & \text { if } & \mathrm{x}<0 \\ \pi / 2 & \text { if } & \mathrm{x}=0 . \\ \pi \mathrm{e}^{-\mathrm{x}} & \text { if } & \mathrm{x}>0 \end{array}\right.${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(\alpha \mathrm{x}\right)+\alpha \mathrm{sin}\left(\alpha \mathrm{x}\right)}{1+{\alpha }^{2}}\text{}\mathrm{d}\alpha =\left\{\begin{array}{ccc}0& \text{if}& \mathrm{x}<0\\ \pi /2& \text{if}& \mathrm{x}=0.\\ \pi {\mathrm{e}}^{-\mathrm{x}}& \text{if}& \mathrm{x}>0\end{array}$
1. a) Using Runge-Kutta second order method with
(i) $\mathrm{h}=0.1$$\mathrm{h}=0.1$h=0.1\mathrm{h}=0.1$\mathrm{h}=0.1$, (ii) $\mathrm{h}=0.2$$\mathrm{h}=0.2$h=0.2\mathrm{h}=0.2$\mathrm{h}=0.2$, solve the initial value problem
${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{y}}^{2}\mathrm{sin}\mathrm{x},\phantom{\rule{1em}{0ex}}\mathrm{y}\left(0\right)=1.$${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{y}}^{2}\mathrm{sin}\mathrm{x},\phantom{\rule{1em}{0ex}}\mathrm{y}\left(0\right)=1.$y^(‘)=y^(2)sin x,quady(0)=1.\mathrm{y}^{\prime}=\mathrm{y}^2 \sin \mathrm{x}, \quad \mathrm{y}(0)=1 .${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{y}}^{2}\mathrm{sin}\mathrm{x},\phantom{\rule{1em}{0ex}}\mathrm{y}\left(0\right)=1.$
Upto $\mathrm{x}=0.4$$\mathrm{x}=0.4$x=0.4\mathrm{x}=0.4$\mathrm{x}=0.4$. If the exact solution is $\mathrm{y}=\mathrm{sec}\mathrm{x}$$\mathrm{y}=\mathrm{sec}\mathrm{x}$y=sec x\mathrm{y}=\sec \mathrm{x}$\mathrm{y}=\mathrm{sec}\mathrm{x}$, obtain the error.
b) Solve the heat conduction equation ${\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}}$${\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}}$u_(t)=u_(xx)\mathrm{u}_{\mathrm{t}}=\mathrm{u}_{\mathrm{xx}}${\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}}$ in the region $\mathrm{R}\left(0\le \mathrm{x}\le 1,\mathrm{t}>0\right)$$\mathrm{R}\left(0\le \mathrm{x}\le 1,\mathrm{t}>0\right)$R(0 <= x <= 1,t > 0)\mathrm{R}(0 \leq \mathrm{x} \leq 1, \mathrm{t}>0)$\mathrm{R}\left(0\le \mathrm{x}\le 1,\mathrm{t}>0\right)$ with the initial and boundary conditions $\mathrm{u}\left(\mathrm{x},0\right)=0,\mathrm{u}\left(0,\mathrm{t}\right)=0,\mathrm{u}\left(1,\mathrm{t}\right)=\mathrm{t}$$\mathrm{u}\left(\mathrm{x},0\right)=0,\mathrm{u}\left(0,\mathrm{t}\right)=0,\mathrm{u}\left(1,\mathrm{t}\right)=\mathrm{t}$u(x,0)=0,u(0,t)=0,u(1,t)=t\mathrm{u}(\mathrm{x}, 0)=0, \mathrm{u}(0, \mathrm{t})=0, \mathrm{u}(1, \mathrm{t})=\mathrm{t}$\mathrm{u}\left(\mathrm{x},0\right)=0,\mathrm{u}\left(0,\mathrm{t}\right)=0,\mathrm{u}\left(1,\mathrm{t}\right)=\mathrm{t}$ using Crank-Nicolson method with $\mathrm{h}=0.25$$\mathrm{h}=0.25$h=0.25\mathrm{h}=0.25$\mathrm{h}=0.25$ and $\lambda =1$$\lambda =1$lambda=1\lambda=1$\lambda =1$ upto two time steps.
8. a) Using second order finite Difference method, solve the boundary value problem
${\mathrm{y}}^{\mathrm{\prime }\mathrm{\prime }}+5{\mathrm{y}}^{\mathrm{\prime }}+4\mathrm{y}=1,\mathrm{y}\left(0\right)=0,\mathrm{y}\left(1\right)=0,\text{}\mathrm{h}=1/4\text{.}$${\mathrm{y}}^{\mathrm{\prime }\mathrm{\prime }}+5{\mathrm{y}}^{\mathrm{\prime }}+4\mathrm{y}=1,\mathrm{y}\left(0\right)=0,\mathrm{y}\left(1\right)=0,\text{}\mathrm{h}=1/4\text{.}$y^(”)+5y^(‘)+4y=1,y(0)=0,y(1)=0,h=1//4″. “\mathrm{y}^{\prime \prime}+5 \mathrm{y}^{\prime}+4 \mathrm{y}=1, \mathrm{y}(0)=0, \mathrm{y}(1)=0, \mathrm{~h}=1 / 4 \text {. }${\mathrm{y}}^{\mathrm{\prime }\mathrm{\prime }}+5{\mathrm{y}}^{\mathrm{\prime }}+4\mathrm{y}=1,\mathrm{y}\left(0\right)=0,\mathrm{y}\left(1\right)=0,\text{}\mathrm{h}=1/4\text{.}$
b) Solve the wave equation ${\mathrm{u}}_{\mathrm{t}\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}}$${\mathrm{u}}_{\mathrm{t}\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}}$u_(tt)=u_(xx)\mathrm{u}_{\mathrm{tt}}=\mathrm{u}_{\mathrm{xx}}${\mathrm{u}}_{\mathrm{t}\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}}$ with the initial and boundary conditions
$\mathrm{u}\left(\mathrm{x},0\right)=0,{\mathrm{u}}_{\mathrm{t}}\left(\mathrm{x},0\right)=0,\mathrm{u}\left(0,\mathrm{t}\right)=0,\mathrm{u}\left(1,\mathrm{t}\right)=100\mathrm{sin}\pi \mathrm{t}\text{.}$$\mathrm{u}\left(\mathrm{x},0\right)=0,{\mathrm{u}}_{\mathrm{t}}\left(\mathrm{x},0\right)=0,\mathrm{u}\left(0,\mathrm{t}\right)=0,\mathrm{u}\left(1,\mathrm{t}\right)=100\mathrm{sin}\pi \mathrm{t}\text{.}$u(x,0)=0,u_(t)(x,0)=0,u(0,t)=0,u(1,t)=100 sin pit”. “\mathrm{u}(\mathrm{x}, 0)=0, \mathrm{u}_{\mathrm{t}}(\mathrm{x}, 0)=0, \mathrm{u}(0, \mathrm{t})=0, \mathrm{u}(1, \mathrm{t})=100 \sin \pi \mathrm{t} \text {. }$\mathrm{u}\left(\mathrm{x},0\right)=0,{\mathrm{u}}_{\mathrm{t}}\left(\mathrm{x},0\right)=0,\mathrm{u}\left(0,\mathrm{t}\right)=0,\mathrm{u}\left(1,\mathrm{t}\right)=100\mathrm{sin}\pi \mathrm{t}\text{.}$
with $\mathrm{h}=\mathrm{k}=0.25$$\mathrm{h}=\mathrm{k}=0.25$h=k=0.25\mathrm{h}=\mathrm{k}=0.25$\mathrm{h}=\mathrm{k}=0.25$, using the explicit method upto four time levels.
9. a) Find an approximate value of $y\left(1.0\right)$$y\left(1.0\right)$y(1.0)y(1.0)$y\left(1.0\right)$ for the initial value problem
${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{x}}^{3}-{\mathrm{y}}^{3},\mathrm{y}\left(0\right)=1$${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{x}}^{3}-{\mathrm{y}}^{3},\mathrm{y}\left(0\right)=1$y^(‘)=x^(3)-y^(3),y(0)=1\mathrm{y}^{\prime}=\mathrm{x}^3-\mathrm{y}^3, \mathrm{y}(0)=1${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{x}}^{3}-{\mathrm{y}}^{3},\mathrm{y}\left(0\right)=1$
using the multiple method
${y}_{n+1}={y}_{n}+\frac{h}{3}\left[7{f}_{n}-2{f}_{n-1}+{f}_{n-2}\right]$${y}_{n+1}={y}_{n}+\frac{h}{3}\left[7{f}_{n}-2{f}_{n-1}+{f}_{n-2}\right]$y_(n+1)=y_(n)+(h)/(3)[7f_(n)-2f_(n-1)+f_(n-2)]y_{n+1}=y_n+\frac{h}{3}\left[7 f_n-2 f_{n-1}+f_{n-2}\right]${y}_{n+1}={y}_{n}+\frac{h}{3}\left[7{f}_{n}-2{f}_{n-1}+{f}_{n-2}\right]$
with step length $h=0.2$$h=0.2$h=0.2h=0.2$h=0.2$. Calculate the starting values using Runge-Kutta second order method with the same $\mathrm{h}$$\mathrm{h}$h\mathrm{h}$\mathrm{h}$.
b) Using standard five point formula, solve the Laplace equation ${\mathrm{\nabla }}^{2}u=0$${\mathrm{\nabla }}^{2}u=0$grad^(2)u=0\nabla^2 u=0${\mathrm{\nabla }}^{2}u=0$ in $R$$R$RR$R$ where $R$$R$RR$R$ is the square $0\le \mathrm{x}\le 1,0\le \mathrm{y}\le 1$$0\le \mathrm{x}\le 1,0\le \mathrm{y}\le 1$0 <= x <= 1,0 <= y <= 10 \leq \mathrm{x} \leq 1,0 \leq \mathrm{y} \leq 1$0\le \mathrm{x}\le 1,0\le \mathrm{y}\le 1$ subject to the boundary conditions $\mathrm{u}\left(\mathrm{x},\mathrm{y}\right)={\mathrm{x}}^{2}-{\mathrm{y}}^{2}$$\mathrm{u}\left(\mathrm{x},\mathrm{y}\right)={\mathrm{x}}^{2}-{\mathrm{y}}^{2}$u(x,y)=x^(2)-y^(2)\mathrm{u}(\mathrm{x}, \mathrm{y})=\mathrm{x}^2-\mathrm{y}^2$\mathrm{u}\left(\mathrm{x},\mathrm{y}\right)={\mathrm{x}}^{2}-{\mathrm{y}}^{2}$ on $\mathrm{x}=0,\mathrm{y}=0,\mathrm{y}=1$$\mathrm{x}=0,\mathrm{y}=0,\mathrm{y}=1$x=0,y=0,y=1\mathrm{x}=0, \mathrm{y}=0, \mathrm{y}=1$\mathrm{x}=0,\mathrm{y}=0,\mathrm{y}=1$ and $3\mathrm{u}+2\frac{\mathrm{\partial }\mathrm{u}}{\mathrm{\partial }\mathrm{x}}={\mathrm{x}}^{2}+{\mathrm{y}}^{2}$$3\mathrm{u}+2\frac{\mathrm{\partial }\mathrm{u}}{\mathrm{\partial }\mathrm{x}}={\mathrm{x}}^{2}+{\mathrm{y}}^{2}$3u+2(delu)/(delx)=x^(2)+y^(2)3 \mathrm{u}+2 \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\mathrm{x}^2+\mathrm{y}^2$3\mathrm{u}+2\frac{\mathrm{\partial }\mathrm{u}}{\mathrm{\partial }\mathrm{x}}={\mathrm{x}}^{2}+{\mathrm{y}}^{2}$ on $\mathrm{x}=1$$\mathrm{x}=1$x=1\mathrm{x}=1$\mathrm{x}=1$. Assume $\mathrm{h}=\mathrm{k}=1/2$$\mathrm{h}=\mathrm{k}=1/2$h=k=1//2\mathrm{h}=\mathrm{k}=1 / 2$\mathrm{h}=\mathrm{k}=1/2$.
10. a) Find an approximate value of $y\left(1.0\right)$$y\left(1.0\right)$y(1.0)y(1.0)$y\left(1.0\right)$ for the initial value problem
${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}-2\mathrm{y},\phantom{\rule{1em}{0ex}}\mathrm{y}\left(0\right)=1$${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}-2\mathrm{y},\phantom{\rule{1em}{0ex}}\mathrm{y}\left(0\right)=1$y^(‘)=x-2y,quady(0)=1\mathrm{y}^{\prime}=\mathrm{x}-2 \mathrm{y}, \quad \mathrm{y}(0)=1${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}-2\mathrm{y},\phantom{\rule{1em}{0ex}}\mathrm{y}\left(0\right)=1$
using Milne-Simpson’s method
${y}_{n+1}={y}_{n-1}+\frac{h}{3}\left[{f}_{n+1}+4{f}_{n}+{f}_{n-1}\right]$${y}_{n+1}={y}_{n-1}+\frac{h}{3}\left[{f}_{n+1}+4{f}_{n}+{f}_{n-1}\right]$y_(n+1)=y_(n-1)+(h)/(3)[f_(n+1)+4f_(n)+f_(n-1)]y_{n+1}=y_{n-1}+\frac{h}{3}\left[f_{n+1}+4 f_n+f_{n-1}\right]${y}_{n+1}={y}_{n-1}+\frac{h}{3}\left[{f}_{n+1}+4{f}_{n}+{f}_{n-1}\right]$
with the step length $h=0.2$$h=0.2$h=0.2h=0.2$h=0.2$. Calculate the starting value using Runge-Kutta fourth order method with the same $h$$h$hh$h$.
b) Using fourth order Taylor series method with $h=0.2$$h=0.2$h=0.2h=0.2$h=0.2$, solve the initial value problem
${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}+\mathrm{cos}\mathrm{y},\mathrm{y}\left(0\right)=0$${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}+\mathrm{cos}\mathrm{y},\mathrm{y}\left(0\right)=0$y^(‘)=x+cos y,y(0)=0\mathrm{y}^{\prime}=\mathrm{x}+\cos \mathrm{y}, \mathrm{y}(0)=0${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}+\mathrm{cos}\mathrm{y},\mathrm{y}\left(0\right)=0$
upto $\mathrm{x}=1$$\mathrm{x}=1$x=1\mathrm{x}=1$\mathrm{x}=1$.
$$\sec ^2 \theta=1+\tan ^2 \theta$$

## MMT-007 Sample Solution 2024

mmt-007-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

# mmt-007-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

1. a) Show that $f\left(x,y\right)=xy$$f\left(x,y\right)=xy$f(x,y)=xyf(x, y)=x y$f\left(x,y\right)=xy$
i) satisfies a Lipschitz condition on any rectangle $a\le x\le b$$a\le x\le b$a <= x <= ba \leq x \leq b$a\le x\le b$ and $c\le y\le d$$c\le y\le d$c <= y <= dc \leq y \leq d$c\le y\le d$;
ii) satisfies a Lipschitz condition on any strip $a\le x\le b$$a\le x\le b$a <= x <= ba \leq x \leq b$a\le x\le b$ and $-\mathrm{\infty }$-\mathrm{\infty }-oo < y < oo-\infty<y<\infty$-\mathrm{\infty };
iii) does not satisfy a Lipschitz condition on the entire plane.
To show that the function $f\left(x,y\right)=xy$$f\left(x,y\right)=xy$f(x,y)=xyf(x, y) = xy$f\left(x,y\right)=xy$ satisfies a Lipschitz condition on certain domains, we need to find a constant $L$$L$LL$L$ such that for all points $\left({x}_{1},{y}_{1}\right)$$\left({x}_{1},{y}_{1}\right)$(x_(1),y_(1))(x_1, y_1)$\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$$\left({x}_{2},{y}_{2}\right)$(x_(2),y_(2))(x_2, y_2)$\left({x}_{2},{y}_{2}\right)$ in the domain,
$|f\left({x}_{1},{y}_{1}\right)-f\left({x}_{2},{y}_{2}\right)|\le L\sqrt{\left({x}_{1}-{x}_{2}{\right)}^{2}+\left({y}_{1}-{y}_{2}{\right)}^{2}}$$|f\left({x}_{1},{y}_{1}\right)-f\left({x}_{2},{y}_{2}\right)|\le L\sqrt{\left({x}_{1}-{x}_{2}{\right)}^{2}+\left({y}_{1}-{y}_{2}{\right)}^{2}}$|f(x_(1),y_(1))-f(x_(2),y_(2))| <= Lsqrt((x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2))|f(x_1, y_1) – f(x_2, y_2)| \leq L\sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2}$|f\left({x}_{1},{y}_{1}\right)-f\left({x}_{2},{y}_{2}\right)|\le L\sqrt{\left({x}_{1}-{x}_{2}{\right)}^{2}+\left({y}_{1}-{y}_{2}{\right)}^{2}}$
i) On a rectangle $a\le x\le b$$a\le x\le b$a <= x <= ba \leq x \leq b$a\le x\le b$ and $c\le y\le d$$c\le y\le d$c <= y <= dc \leq y \leq d$c\le y\le d$:
Consider any two points $\left({x}_{1},{y}_{1}\right)$$\left({x}_{1},{y}_{1}\right)$(x_(1),y_(1))(x_1, y_1)$\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$$\left({x}_{2},{y}_{2}\right)$(x_(2),y_(2))(x_2, y_2)$\left({x}_{2},{y}_{2}\right)$ in the rectangle. Then,
$\begin{array}{rl}|f\left({x}_{1},{y}_{1}\right)-f\left({x}_{2},{y}_{2}\right)|& =|{x}_{1}{y}_{1}-{x}_{2}{y}_{2}|\\ & =|{x}_{1}{y}_{1}-{x}_{1}{y}_{2}+{x}_{1}{y}_{2}-{x}_{2}{y}_{2}|\\ & \le |{x}_{1}||{y}_{1}-{y}_{2}|+|{y}_{2}||{x}_{1}-{x}_{2}|\\ & \le max\left\{|{x}_{1}|,|{y}_{2}|\right\}\left(|{x}_{1}-{x}_{2}|+|{y}_{1}-{y}_{2}|\right)\end{array}$$\begin{array}{r}|f\left({x}_{1},{y}_{1}\right)-f\left({x}_{2},{y}_{2}\right)|=|{x}_{1}{y}_{1}-{x}_{2}{y}_{2}|\\ =|{x}_{1}{y}_{1}-{x}_{1}{y}_{2}+{x}_{1}{y}_{2}-{x}_{2}{y}_{2}|\\ \le |{x}_{1}||{y}_{1}-{y}_{2}|+|{y}_{2}||{x}_{1}-{x}_{2}|\\ \le max\left\{|{x}_{1}|,|{y}_{2}|\right\}\left(|{x}_{1}-{x}_{2}|+|{y}_{1}-{y}_{2}|\right)\end{array}${:[|f(x_(1)”,”y_(1))-f(x_(2)”,”y_(2))|=|x_(1)y_(1)-x_(2)y_(2)|],[=|x_(1)y_(1)-x_(1)y_(2)+x_(1)y_(2)-x_(2)y_(2)|],[ <= |x_(1)||y_(1)-y_(2)|+|y_(2)||x_(1)-x_(2)|],[ <= max{|x_(1)|”,”|y_(2)|}(|x_(1)-x_(2)|+|y_(1)-y_(2)|)]:}\begin{align*} |f(x_1, y_1) – f(x_2, y_2)| &= |x_1y_1 – x_2y_2| \\ &= |x_1y_1 – x_1y_2 + x_1y_2 – x_2y_2| \\ &\leq |x_1||y_1 – y_2| + |y_2||x_1 – x_2| \\ &\leq \max\{|x_1|, |y_2|\}(|x_1 – x_2| + |y_1 – y_2|) \end{align*}$\begin{array}{rl}|f\left({x}_{1},{y}_{1}\right)-f\left({x}_{2},{y}_{2}\right)|& =|{x}_{1}{y}_{1}-{x}_{2}{y}_{2}|\\ & =|{x}_{1}{y}_{1}-{x}_{1}{y}_{2}+{x}_{1}{y}_{2}-{x}_{2}{y}_{2}|\\ & \le |{x}_{1}||{y}_{1}-{y}_{2}|+|{y}_{2}||{x}_{1}-{x}_{2}|\\ & \le max\left\{|{x}_{1}|,|{y}_{2}|\right\}\left(|{x}_{1}-{x}_{2}|+|{y}_{1}-{y}_{2}|\right)\end{array}$
Since $a\le x\le b$$a\le x\le b$a <= x <= ba \leq x \leq b$a\le x\le b$ and $c\le y\le d$$c\le y\le d$c <= y <= dc \leq y \leq d$c\le y\le d$, we can take $L=max\left\{|a|,|b|,|c|,|d|\right\}$$L=max\left\{|a|,|b|,|c|,|d|\right\}$L=max{|a|,|b|,|c|,|d|}L = \max\{|a|, |b|, |c|, |d|\}$L=max\left\{|a|,|b|,|c|,|d|\right\}$. Then,
$|f\left({x}_{1},{y}_{1}\right)-f\left({x}_{2},{y}_{2}\right)|\le L\left(|{x}_{1}-{x}_{2}|+|{y}_{1}-{y}_{2}|\right)\le L\sqrt{2}\sqrt{\left({x}_{1}-{x}_{2}{\right)}^{2}+\left({y}_{1}-{y}_{2}{\right)}^{2}}$$|f\left({x}_{1},{y}_{1}\right)-f\left({x}_{2},{y}_{2}\right)|\le L\left(|{x}_{1}-{x}_{2}|+|{y}_{1}-{y}_{2}|\right)\le L\sqrt{2}\sqrt{\left({x}_{1}-{x}_{2}{\right)}^{2}+\left({y}_{1}-{y}_{2}{\right)}^{2}}$|f(x_(1),y_(1))-f(x_(2),y_(2))| <= L(|x_(1)-x_(2)|+|y_(1)-y_(2)|) <= Lsqrt2sqrt((x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2))|f(x_1, y_1) – f(x_2, y_2)| \leq L(|x_1 – x_2| + |y_1 – y_2|) \leq L\sqrt{2}\sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2}$|f\left({x}_{1},{y}_{1}\right)-f\left({x}_{2},{y}_{2}\right)|\le L\left(|{x}_{1}-{x}_{2}|+|{y}_{1}-{y}_{2}|\right)\le L\sqrt{2}\sqrt{\left({x}_{1}-{x}_{2}{\right)}^{2}+\left({y}_{1}-{y}_{2}{\right)}^{2}}$
So, $f\left(x,y\right)=xy$$f\left(x,y\right)=xy$f(x,y)=xyf(x, y) = xy$f\left(x,y\right)=xy$ satisfies a Lipschitz condition on any rectangle $a\le x\le b$$a\le x\le b$a <= x <= ba \leq x \leq b