Details For MMT-007 Solved Assignment

IGNOU MMT-007 Assignment Question Paper 2024

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

1. a) Show that $f(x,y)=xy$$f(x,y)=xy$f(x,y)=xyf(x, y)=x y$f(x,y)=xy$
   i) satisfies a Lipschitz condition on any rectangle $a\le x\le b$$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 }<y<\mathrm{\infty }$$-\mathrm{\infty }<y<\mathrm{\infty }$-oo < y < oo-infty<y<infty$-\mathrm{\infty }<y<\mathrm{\infty }$;
   iii) does not satisfy a Lipschitz condition on the entire plane.
   b) Use Frobenious method to find the series solution about $x=0$$x=0$x=0x=0$x=0$ of the equation

$$x(1-x)\frac{d^{2}y}{d{x}^2}-(1+3x)\frac{dy}{dx}-y=0.$$$$x(1-x)\frac{d^{2}y}{d{x}^2}-(1+3x)\frac{dy}{dx}-y=0.$$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 .$$x(1-x)\frac{d^{2}y}{d{x}^2}-(1+3x)\frac{dy}{dx}-y=0.$$

2. a) For the following differential equation locate and classify its singular points on the $x$$x$xx$x$-axis
   i) $x^3(x-1)y^{\mathrm{\prime }\mathrm{\prime }}-2(x-1)y^{\mathrm{\prime }}+3xy=0$$x^3(x-1)y^{\mathrm{\prime }\mathrm{\prime }}-2(x-1)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(x-1)y^{\mathrm{\prime }\mathrm{\prime }}-2(x-1)y^{\mathrm{\prime }}+3xy=0$
   ii) $(3x+1)x{y}^{\mathrm{\prime }\mathrm{\prime }}-(x+1)y^{\mathrm{\prime }}+2y=0$$(3x+1)x{y}^{\mathrm{\prime }\mathrm{\prime }}-(x+1)y^{\mathrm{\prime }}+2y=0$(3x+1)xy^(”)-(x+1)y^(‘)+2y=0(3 x+1) x y^{prime prime}-(x+1) y^{prime}+2 y=0$(3x+1)x{y}^{\mathrm{\prime }\mathrm{\prime }}-(x+1)y^{\mathrm{\prime }}+2y=0$
   b) Show that ${\mathrm{L}}_{\mathrm{n}+1}(\mathrm{x})=(2\mathrm{n}+1-\mathrm{x}){\mathrm{L}}_{\mathrm{n}}(\mathrm{x})-{\mathrm{n}}^2{\mathrm{L}}_{\mathrm{n}-1}(\mathrm{x})$${\mathrm{L}}_{\mathrm{n}+1}(\mathrm{x})=(2\mathrm{n}+1-\mathrm{x}){\mathrm{L}}_{\mathrm{n}}(\mathrm{x})-{\mathrm{n}}^2{\mathrm{L}}_{\mathrm{n}-1}(\mathrm{x})$L_(n+1)(x)=(2n+1-x)L_(n)(x)-n^(2)L_(n-1)(x)mathrm{L}_{mathrm{n}+1}(mathrm{x})=(2 mathrm{n}+1-mathrm{x}) mathrm{L}_{mathrm{n}}(mathrm{x})-mathrm{n}^2 mathrm{~L}_{mathrm{n}-1}(mathrm{x})${\mathrm{L}}_{\mathrm{n}+1}(\mathrm{x})=(2\mathrm{n}+1-\mathrm{x}){\mathrm{L}}_{\mathrm{n}}(\mathrm{x})-{\mathrm{n}}^2{\mathrm{L}}_{\mathrm{n}-1}(\mathrm{x})$.
   c) Show that ${\int }_{-1}^1{x}^2{P}_{n-1}(x)P_{n+1}(x)dx=\frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$${\int }_{-1}^1 x^2{P}_{n-1}(x)P_{n+1}(x)dx=\frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$int_(-1)^(1)x^(2)P_(n-1)(x)P_(n+1)(x)dx=(2n(n+1))/((2n-1)(2n+1)(2n+3))int_{-1}^1 x^2 P_{n-1}(x) P_{n+1}(x) d x=frac{2 n(n+1)}{(2 n-1)(2 n+1)(2 n+3)}${\int }_{-1}^1{x}^2{P}_{n-1}(x)P_{n+1}(x)dx=\frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$.
   d) Construct Green’s function for the differential equation

$$x{y}^{\mathrm{\prime }\mathrm{\prime }}+y^{\mathrm{\prime }}=0,0<x<\ell$$$$x{y}^{\mathrm{\prime }\mathrm{\prime }}+y^{\mathrm{\prime }}=0,0<x<\ell$$xy^(”)+y^(‘)=0,quad0 < x < ℓx y^{prime prime}+y^{prime}=0, quad 0<x<ell$$x{y}^{\mathrm{\prime }\mathrm{\prime }}+y^{\mathrm{\prime }}=0,0<x<\ell$$
under the conditions that $\mathrm{y}(0)$$\mathrm{y}(0)$y(0)mathrm{y}(0)$\mathrm{y}(0)$ is bounded and $\mathrm{y}(\ell )=0$$\mathrm{y}(\ell )=0$y(ℓ)=0mathrm{y}(ell)=0$\mathrm{y}(\ell )=0$.
3. a) Show that between every successive pair of zeros of $J_0(x)$$J_0(x)$J_(0)(x)J_0(x)$J_0(x)$ there exists a zero of $J_1(x)$$J_1(x)$J_(1)(x)J_1(x)$J_1(x)$.
b) Using the transformation $y=x^{1/2}u,2x^{3/2}=3z$$y=x^{1/2}u,2x^{3/2}=3z$y=x^(1//2)u,2x^(3//2)=3zy=x^{1 / 2} u, 2 x^{3 / 2}=3 z$y=x^{1/2}u,2x^{3/2}=3z$ find the solution of the equation $y^{\mathrm{\prime }\mathrm{\prime }}+xy=0$$y^{\mathrm{\prime }\mathrm{\prime }}+xy=0$y^(”)+x quad y=0y^{prime prime}+x quad y=0$y^{\mathrm{\prime }\mathrm{\prime }}+xy=0$ in terms of Bessel’s functions.
c) Show that ${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{a}\mathrm{x}}{\mathrm{J}}_0(\mathrm{b}\mathrm{x})\mathrm{d}\mathrm{x}=\frac{1}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}},\mathrm{a}>0\mathrm{b}>0$${\int }_0^{\mathrm{\infty }} {\mathrm{e}}^{-\mathrm{a}\mathrm{x}}{\mathrm{J}}_0(\mathrm{b}\mathrm{x})\mathrm{d}\mathrm{x}=\frac{1}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}},\mathrm{a}>0\mathrm{b}>0$int_(0)^(oo)e^(-ax)J_(0)(bx)dx=(1)/(sqrt(a^(2)+b^(2))),a > 0b > 0int_0^{infty} mathrm{e}^{-mathrm{ax}} mathrm{J}_0(mathrm{bx}) mathrm{dx}=frac{1}{sqrt{mathrm{a}^2+mathrm{b}^2}}, mathrm{a}>0 mathrm{~b}>0${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{a}\mathrm{x}}{\mathrm{J}}_0(\mathrm{b}\mathrm{x})\mathrm{d}\mathrm{x}=\frac{1}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}},\mathrm{a}>0\mathrm{b}>0$.
4. a) Find the Laplace transform of $\frac{\cos \sqrt{t}}{\sqrt{t}}$$\frac{\cos \sqrt{t}}{\sqrt{t}}$(cos sqrtt)/(sqrtt)frac{cos sqrt{t}}{sqrt{t}}$\frac{\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}}(\mathrm{k}\mathrm{b})=0$${\mathrm{J}}_{\mathrm{p}}(\mathrm{k}\mathrm{b})=0$J_(p)(kb)=0mathrm{J}_{mathrm{p}}(mathrm{kb})=0${\mathrm{J}}_{\mathrm{p}}(\mathrm{k}\mathrm{b})=0$ with $\mathrm{p}\ge 0,\mathrm{b}>0$$\mathrm{p}\ge 0,\mathrm{b}>0$p >= 0,b > 0mathrm{p} geq 0, mathrm{~b}>0$\mathrm{p}\ge 0,\mathrm{b}>0$ then show that
$${\int }_0^{b}x{J}_p\left(k_{m}x\right)J_p\left(k_{n}x\right)dx=\{\begin{array}{lll}0& \text{if}& m\ne n\\ \frac{b^2}{2}\left[J_{p+1}\left(k_{n}b\right)\right]& \text{if}& m=n\end{array}.$$$${\int }_0^b x{J}_p\left(k_{m}x\right)J_p\left(k_{n}x\right)dx=\left\{\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}m\ne n\\ \frac{b^2}{2}\left[J_{p+1}\left(k_{n}b\right)\right]\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}m=n\end{array}.\right.$$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_pleft(k_m xright) J_pleft(k_n xright) d x=left{begin{array}{lll}
0 & text { if } & m neq n \
frac{b^2}{2}left[J_{p+1}left(k_n bright)right] & text { if } & m=n
end{array} .right.$${\int }_0^{b}x{J}_p\left(k_{m}x\right)J_p\left(k_{n}x\right)dx=\{\begin{array}{lll}0& \text{if}& m\ne n\\ \frac{b^2}{2}\left[J_{p+1}\left(k_{n}b\right)\right]& \text{if}& m=n\end{array}.$$

5. a) Solve the following IBVP using the Laplace transform technique:

$$\begin{array}{rl}& {\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}},0<\mathrm{x}<1,\mathrm{t}>0.\\ & \mathrm{u}(0,\mathrm{t})=1,\mathrm{u}(1,\mathrm{t})=1,\mathrm{t}>0\\ & \mathrm{u}(\mathrm{x},0)=1+\sin \pi \mathrm{x},0<\mathrm{x}<1.\end{array}$$$$\begin{array}{r}{\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}},0<\mathrm{x}<1,\mathrm{t}>0.\\ \mathrm{u}(0,\mathrm{t})=1,\mathrm{u}(1,\mathrm{t})=1,\mathrm{t}>0\\ \mathrm{u}(\mathrm{x},0)=1+\sin \pi \mathrm{x},0<\mathrm{x}<1.\end{array}$${:[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}$$\begin{array}{rl}& {\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}},0<\mathrm{x}<1,\mathrm{t}>0.\\ & \mathrm{u}(0,\mathrm{t})=1,\mathrm{u}(1,\mathrm{t})=1,\mathrm{t}>0\\ & \mathrm{u}(\mathrm{x},0)=1+\sin \pi \mathrm{x},0<\mathrm{x}<1.\end{array}$$
b) If the Fourier cosine transform of $f(x)$$f(x)$f(x)f(x)$f(x)$ is ${\alpha }^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{a}\alpha }$${\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
$$f(x)=\frac{2}{\pi }\frac{n!\cos (n+1)\theta }{{(a^2+x^2)}^{\frac{n+1}{2}}}.$$$$f(x)=\frac{2}{\pi }\frac{n!\cos (n+1)\theta }{{\left(a^2+x^2\right)}^{\frac{n+1}{2}}}.$$f(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^2right)^{frac{n+1}{2}}} .$$f(x)=\frac{2}{\pi }\frac{n!\cos (n+1)\theta }{{(a^2+x^2)}^{\frac{n+1}{2}}}.$$

6. a) Find the displacement $u(x,t)$$u(x,t)$u(x,t)u(x, t)$u(x,t)$ of an infinite string using the method of Fourier transform given that the string is initially at rest and that the initial displacement is $\mathrm{f}(\mathrm{x}),-\mathrm{\infty }<\mathrm{x}<\mathrm{\infty }$$\mathrm{f}(\mathrm{x}),-\mathrm{\infty }<\mathrm{x}<\mathrm{\infty }$f(x),-oo < x < oomathrm{f}(mathrm{x}),-infty<mathrm{x}<infty$\mathrm{f}(\mathrm{x}),-\mathrm{\infty }<\mathrm{x}<\mathrm{\infty }$.
   b) Using Fourier integral representation show that

$${\int }_0^{\mathrm{\infty }}\frac{\cos (\alpha \mathrm{x})+\alpha \sin (\alpha \mathrm{x})}{1+{\alpha }^2}\mathrm{d}\alpha =\{\begin{array}{ccc}0& \text{if}& \mathrm{x}<0\\ \pi /2& \text{if}& \mathrm{x}=0.\\ \pi {\mathrm{e}}^{-\mathrm{x}}& \text{if}& \mathrm{x}>0\end{array}$$$${\int }_0^{\mathrm{\infty }} \frac{\cos (\alpha \mathrm{x})+\alpha \sin (\alpha \mathrm{x})}{1+{\alpha }^2}\mathrm{d}\alpha =\left\{\begin{array}{ccc}0& \text{if}& \mathrm{x}<0\\ \pi /2& \text{if}& \mathrm{x}=0.\\ \pi {\mathrm{e}}^{-\mathrm{x}}& \text{if}& \mathrm{x}>0\end{array}\right.$$int_(0)^(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.$${\int }_0^{\mathrm{\infty }}\frac{\cos (\alpha \mathrm{x})+\alpha \sin (\alpha \mathrm{x})}{1+{\alpha }^2}\mathrm{d}\alpha =\{\begin{array}{ccc}0& \text{if}& \mathrm{x}<0\\ \pi /2& \text{if}& \mathrm{x}=0.\\ \pi {\mathrm{e}}^{-\mathrm{x}}& \text{if}& \mathrm{x}>0\end{array}$$

7. a) Using Runge-Kutta second order method with
   (i) $\mathrm{h}=0.1$$\mathrm{h}=0.1$h=0.1mathrm{h}=0.1$\mathrm{h}=0.1$, (ii) $\mathrm{h}=0.2$$\mathrm{h}=0.2$h=0.2mathrm{h}=0.2$\mathrm{h}=0.2$, solve the initial value problem

$${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{y}}^2\sin \mathrm{x},\mathrm{y}(0)=1.$$$${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{y}}^2\sin \mathrm{x},\mathrm{y}(0)=1.$$y^(‘)=y^(2)sin x,quady(0)=1.mathrm{y}^{prime}=mathrm{y}^2 sin mathrm{x}, quad mathrm{y}(0)=1 .$${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{y}}^2\sin \mathrm{x},\mathrm{y}(0)=1.$$
Upto $\mathrm{x}=0.4$$\mathrm{x}=0.4$x=0.4mathrm{x}=0.4$\mathrm{x}=0.4$. If the exact solution is $\mathrm{y}=\sec \mathrm{x}$$\mathrm{y}=\sec \mathrm{x}$y=sec xmathrm{y}=sec mathrm{x}$\mathrm{y}=\sec \mathrm{x}$, obtain the error.
b) Solve the heat conduction equation ${\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}}$${\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}}$u_(t)=u_(xx)mathrm{u}_{mathrm{t}}=mathrm{u}_{mathrm{xx}}${\mathrm{u}}_{\mathrm{t}}={\mathrm{u}}_{\mathrm{x}\mathrm{x}}$ in the region $\mathrm{R}(0\le \mathrm{x}\le 1,\mathrm{t}>0)$$\mathrm{R}(0\le \mathrm{x}\le 1,\mathrm{t}>0)$R(0 <= x <= 1,t > 0)mathrm{R}(0 leq mathrm{x} leq 1, mathrm{t}>0)$\mathrm{R}(0\le \mathrm{x}\le 1,\mathrm{t}>0)$ with the initial and boundary conditions $\mathrm{u}(\mathrm{x},0)=0,\mathrm{u}(0,\mathrm{t})=0,\mathrm{u}(1,\mathrm{t})=\mathrm{t}$$\mathrm{u}(\mathrm{x},0)=0,\mathrm{u}(0,\mathrm{t})=0,\mathrm{u}(1,\mathrm{t})=\mathrm{t}$u(x,0)=0,u(0,t)=0,u(1,t)=tmathrm{u}(mathrm{x}, 0)=0, mathrm{u}(0, mathrm{t})=0, mathrm{u}(1, mathrm{t})=mathrm{t}$\mathrm{u}(\mathrm{x},0)=0,\mathrm{u}(0,\mathrm{t})=0,\mathrm{u}(1,\mathrm{t})=\mathrm{t}$ using Crank-Nicolson method with $\mathrm{h}=0.25$$\mathrm{h}=0.25$h=0.25mathrm{h}=0.25$\mathrm{h}=0.25$ and $\lambda =1$$\lambda =1$lambda=1lambda=1$\lambda =1$ upto two time steps.
8. a) Using second order finite Difference method, solve the boundary value problem
$${\mathrm{y}}^{\mathrm{\prime }\mathrm{\prime }}+5{\mathrm{y}}^{\mathrm{\prime }}+4\mathrm{y}=1,\mathrm{y}(0)=0,\mathrm{y}(1)=0,\mathrm{h}=1/4\text{.}$$$${\mathrm{y}}^{\mathrm{\prime }\mathrm{\prime }}+5{\mathrm{y}}^{\mathrm{\prime }}+4\mathrm{y}=1,\mathrm{y}(0)=0,\mathrm{y}(1)=0,\mathrm{h}=1/4\text{.}$$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 {. }$${\mathrm{y}}^{\mathrm{\prime }\mathrm{\prime }}+5{\mathrm{y}}^{\mathrm{\prime }}+4\mathrm{y}=1,\mathrm{y}(0)=0,\mathrm{y}(1)=0,\mathrm{h}=1/4\text{.}$$
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}(\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{.}$$$$\mathrm{u}(\mathrm{x},0)=0,{\mathrm{u}}_{\mathrm{t}}(\mathrm{x},0)=0,\mathrm{u}(0,\mathrm{t})=0,\mathrm{u}(1,\mathrm{t})=100\sin \pi \mathrm{t}\text{.}$$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 {. }$$\mathrm{u}(\mathrm{x},0)=0,{\mathrm{u}}_{\mathrm{t}}(\mathrm{x},0)=0,\mathrm{u}(0,\mathrm{t})=0,\mathrm{u}(1,\mathrm{t})=100\sin \pi \mathrm{t}\text{.}$$
with $\mathrm{h}=\mathrm{k}=0.25$$\mathrm{h}=\mathrm{k}=0.25$h=k=0.25mathrm{h}=mathrm{k}=0.25$\mathrm{h}=\mathrm{k}=0.25$, using the explicit method upto four time levels.
9. a) Find an approximate value of $y(1.0)$$y(1.0)$y(1.0)y(1.0)$y(1.0)$ for the initial value problem
$${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{x}}^3-{\mathrm{y}}^3,\mathrm{y}(0)=1$$$${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{x}}^3-{\mathrm{y}}^3,\mathrm{y}(0)=1$$y^(‘)=x^(3)-y^(3),y(0)=1mathrm{y}^{prime}=mathrm{x}^3-mathrm{y}^3, mathrm{y}(0)=1$${\mathrm{y}}^{\mathrm{\prime }}={\mathrm{x}}^3-{\mathrm{y}}^3,\mathrm{y}(0)=1$$
using the multiple method
$$y_{n+1}=y_n+\frac{h}{3}[7f_n-2f_{n-1}+f_{n-2}]$$$$y_{n+1}=y_n+\frac{h}{3}\left[7f_n-2f_{n-1}+f_{n-2}\right]$$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]$$y_{n+1}=y_n+\frac{h}{3}[7f_n-2f_{n-1}+f_{n-2}]$$
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}$hmathrm{h}$\mathrm{h}$.
b) Using standard five point formula, solve the Laplace equation ${\mathrm{\nabla }}^{2}u=0$${\mathrm{\nabla }}^{2}u=0$grad^(2)u=0nabla^2 u=0${\mathrm{\nabla }}^{2}u=0$ in $R$$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}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2-{\mathrm{y}}^2$$\mathrm{u}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2-{\mathrm{y}}^2$u(x,y)=x^(2)-y^(2)mathrm{u}(mathrm{x}, mathrm{y})=mathrm{x}^2-mathrm{y}^2$\mathrm{u}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2-{\mathrm{y}}^2$ on $\mathrm{x}=0,\mathrm{y}=0,\mathrm{y}=1$$\mathrm{x}=0,\mathrm{y}=0,\mathrm{y}=1$x=0,y=0,y=1mathrm{x}=0, mathrm{y}=0, mathrm{y}=1$\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$x=1mathrm{x}=1$\mathrm{x}=1$. Assume $\mathrm{h}=\mathrm{k}=1/2$$\mathrm{h}=\mathrm{k}=1/2$h=k=1//2mathrm{h}=mathrm{k}=1 / 2$\mathrm{h}=\mathrm{k}=1/2$.
10. a) Find an approximate value of $y(1.0)$$y(1.0)$y(1.0)y(1.0)$y(1.0)$ for the initial value problem
$${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}-2\mathrm{y},\mathrm{y}(0)=1$$$${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}-2\mathrm{y},\mathrm{y}(0)=1$$y^(‘)=x-2y,quady(0)=1mathrm{y}^{prime}=mathrm{x}-2 mathrm{y}, quad mathrm{y}(0)=1$${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}-2\mathrm{y},\mathrm{y}(0)=1$$
using Milne-Simpson’s method
$$y_{n+1}=y_{n-1}+\frac{h}{3}[f_{n+1}+4f_n+f_{n-1}]$$$$y_{n+1}=y_{n-1}+\frac{h}{3}\left[f_{n+1}+4f_n+f_{n-1}\right]$$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]$$y_{n+1}=y_{n-1}+\frac{h}{3}[f_{n+1}+4f_n+f_{n-1}]$$
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}+\cos \mathrm{y},\mathrm{y}(0)=0$$$${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}+\cos \mathrm{y},\mathrm{y}(0)=0$$y^(‘)=x+cos y,y(0)=0mathrm{y}^{prime}=mathrm{x}+cos mathrm{y}, mathrm{y}(0)=0$${\mathrm{y}}^{\mathrm{\prime }}=\mathrm{x}+\cos \mathrm{y},\mathrm{y}(0)=0$$
upto $\mathrm{x}=1$$\mathrm{x}=1$x=1mathrm{x}=1$\mathrm{x}=1$.


[stray-random categories=trigonometry]

MMT-007 Sample Solution 2024

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

1. a) Show that $f(x,y)=xy$$f(x,y)=xy$f(x,y)=xyf(x, y)=x y$f(x,y)=xy$
   i) satisfies a Lipschitz condition on any rectangle $a\le x\le b$$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$$a\le x\le b$a <= x <= ba leq x leq b$a\le x\le b$ and $-\mathrm{\infty }<y<\mathrm{\infty }$$-\mathrm{\infty }<y<\mathrm{\infty }$-oo < y < oo-infty<y<infty$-\mathrm{\infty }<y<\mathrm{\infty }$;
iii) does not satisfy a Lipschitz condition on the entire plane.
Answer:
To show that the function $f(x,y)=xy$$f(x,y)=xy$f(x,y)=xyf(x, y) = xy$f(x,y)=xy$ satisfies a Lipschitz condition on certain domains, we need to find a constant $L$$L$LL$L$ such that for all points $(x_1,y_1)$$(x_1,y_1)$(x_(1),y_(1))(x_1, y_1)$(x_1,y_1)$ and $(x_2,y_2)$$(x_2,y_2)$(x_(2),y_(2))(x_2, y_2)$(x_2,y_2)$ in the domain,
$$|f(x_1,y_1)-f(x_2,y_2)|\le L\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$$$$|f(x_1,y_1)-f(x_2,y_2)|\le L\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$$|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 Lsqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2}$$|f(x_1,y_1)-f(x_2,y_2)|\le L\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$$
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 $(x_1,y_1)$$(x_1,y_1)$(x_(1),y_(1))(x_1, y_1)$(x_1,y_1)$ and $(x_2,y_2)$$(x_2,y_2)$(x_(2),y_(2))(x_2, y_2)$(x_2,y_2)$ in the rectangle. Then,
$$\begin{array}{rl}|f(x_1,y_1)-f(x_2,y_2)|& =|x_1{y}_1-x_2{y}_2|\\ & =|x_1{y}_1-x_1{y}_2+x_1{y}_2-x_2{y}_2|\\ & \le |x_1||y_1-y_2|+|y_2||x_1-x_2|\\ & \le max\{|x_1|,|y_2|\}(|x_1-x_2|+|y_1-y_2|)\end{array}$$$$\begin{array}{r}|f(x_1,y_1)-f(x_2,y_2)|=|x_1{y}_1-x_2{y}_2|\\ =|x_1{y}_1-x_1{y}_2+x_1{y}_2-x_2{y}_2|\\ \le |x_1||y_1-y_2|+|y_2||x_1-x_2|\\ \le max\{|x_1|,|y_2|\}(|x_1-x_2|+|y_1-y_2|)\end{array}$${:[|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*}$$\begin{array}{rl}|f(x_1,y_1)-f(x_2,y_2)|& =|x_1{y}_1-x_2{y}_2|\\ & =|x_1{y}_1-x_1{y}_2+x_1{y}_2-x_2{y}_2|\\ & \le |x_1||y_1-y_2|+|y_2||x_1-x_2|\\ & \le max\{|x_1|,|y_2|\}(|x_1-x_2|+|y_1-y_2|)\end{array}$$
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\{|a|,|b|,|c|,|d|\}$$L=max\{|a|,|b|,|c|,|d|\}$L=max{|a|,|b|,|c|,|d|}L = max{|a|, |b|, |c|, |d|}$L=max\{|a|,|b|,|c|,|d|\}$. Then,
$$|f(x_1,y_1)-f(x_2,y_2)|\le L(|x_1-x_2|+|y_1-y_2|)\le L\sqrt{2}\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$$$$|f(x_1,y_1)-f(x_2,y_2)|\le L(|x_1-x_2|+|y_1-y_2|)\le L\sqrt{2}\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$$|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 Lsqrt{2}sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2}$$|f(x_1,y_1)-f(x_2,y_2)|\le L(|x_1-x_2|+|y_1-y_2|)\le L\sqrt{2}\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$$
So, $f(x,y)=xy$$f(x,y)=xy$f(x,y)=xyf(x, y) = xy$f(x,y)=xy$ satisfies a Lipschitz condition on any rectangle $a\le x\le b$$a\le x\le b$a <= x <= ba leq x leq b$a\le x\le b$ and $c\le y\le d$$c\le y\le d$c <= y <= dc leq y leq d$c\le y\le d$ with Lipschitz constant $L\sqrt{2}$$L\sqrt{2}$Lsqrt2Lsqrt{2}$L\sqrt{2}$.
ii) On a strip $a\le x\le b$$a\le x\le b$a <= x <= ba leq x leq b$a\le x\le b$ and $-\mathrm{\infty }<y<\mathrm{\infty }$$-\mathrm{\infty }<y<\mathrm{\infty }$-oo < y < oo-infty < y < infty$-\mathrm{\infty }<y<\mathrm{\infty }$:
Using a similar argument as in part (i), we can show that $f(x,y)=xy$$f(x,y)=xy$f(x,y)=xyf(x, y) = xy$f(x,y)=xy$ satisfies a Lipschitz condition on any strip $a\le x\le b$$a\le x\le b$a <= x <= ba leq x leq b$a\le x\le b$ and $-\mathrm{\infty }<y<\mathrm{\infty }$$-\mathrm{\infty }<y<\mathrm{\infty }$-oo < y < oo-infty < y < infty$-\mathrm{\infty }<y<\mathrm{\infty }$ with a Lipschitz constant $L=max\{|a|,|b|\}$$L=max\{|a|,|b|\}$L=max{|a|,|b|}L = max{|a|, |b|}$L=max\{|a|,|b|\}$.
iii) On the entire plane:
To show that $f(x,y)=xy$$f(x,y)=xy$f(x,y)=xyf(x, y) = xy$f(x,y)=xy$ does not satisfy a Lipschitz condition on the entire plane, consider the points $(x_1,y_1)=(1,n)$$(x_1,y_1)=(1,n)$(x_(1),y_(1))=(1,n)(x_1, y_1) = (1, n)$(x_1,y_1)=(1,n)$ and $(x_2,y_2)=(0,0)$$(x_2,y_2)=(0,0)$(x_(2),y_(2))=(0,0)(x_2, y_2) = (0, 0)$(x_2,y_2)=(0,0)$ for some large $n\in \mathbf{N}$$n\in \mathbf{N}$n inNn in mathbf{N}$n\in \mathbf{N}$. Then,
$$|f(x_1,y_1)-f(x_2,y_2)|=|1\cdot n-0\cdot 0|=n$$$$|f(x_1,y_1)-f(x_2,y_2)|=|1\cdot n-0\cdot 0|=n$$|f(x_(1),y_(1))-f(x_(2),y_(2))|=|1*n-0*0|=n|f(x_1, y_1) – f(x_2, y_2)| = |1 cdot n – 0 cdot 0| = n$$|f(x_1,y_1)-f(x_2,y_2)|=|1\cdot n-0\cdot 0|=n$$
However,
$$\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}=\sqrt{1^2+n^2}\approx n$$$$\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}=\sqrt{1^2+n^2}\approx n$$sqrt((x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2))=sqrt(1^(2)+n^(2))~~nsqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2} = sqrt{1^2 + n^2} approx n$$\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}=\sqrt{1^2+n^2}\approx n$$
for large $n$$n$nn$n$. If $f$$f$ff$f$ were Lipschitz on the entire plane, there would exist a constant $L$$L$LL$L$ such that $n\le L\cdot n$$n\le L\cdot n$n <= L*nn leq L cdot n$n\le L\cdot n$ for all $n$$n$nn$n$, which is impossible. Therefore, $f(x,y)=xy$$f(x,y)=xy$f(x,y)=xyf(x, y) = xy$f(x,y)=xy$ 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(1-x)\frac{d^{2}y}{d{x}^2}-(1+3x)\frac{dy}{dx}-y=0.$$$$x(1-x)\frac{d^{2}y}{d{x}^2}-(1+3x)\frac{dy}{dx}-y=0.$$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 .$$x(1-x)\frac{d^{2}y}{d{x}^2}-(1+3x)\frac{dy}{dx}-y=0.$$
Answer:
To solve the differential equation using the Frobenius method, we assume a solution of the form:
$$y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}$$$$y=\sum _{n=0}^{\mathrm{\infty }} a_n{x}^{n+r}$$y=sum_(n=0)^(oo)a_(n)x^(n+r)y = sum_{n=0}^{infty} a_n x^{n+r}$$y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}$$
where $r$$r$rr$r$ is a constant to be determined. We then substitute this series into the differential equation and solve for the coefficients $a_n$$a_n$a_(n)a_n$a_n$.
First, we calculate the derivatives:
$$\frac{dy}{dx}=\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}$$$$\frac{dy}{dx}=\sum _{n=0}^{\mathrm{\infty }} (n+r)a_n{x}^{n+r-1}$$(dy)/(dx)=sum_(n=0)^(oo)(n+r)a_(n)x^(n+r-1)frac{dy}{dx} = sum_{n=0}^{infty} (n+r) a_n x^{n+r-1}$$\frac{dy}{dx}=\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}$$
$$\frac{d^{2}y}{d{x}^2}=\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}$$$$\frac{d^{2}y}{d{x}^2}=\sum _{n=0}^{\mathrm{\infty }} (n+r)(n+r-1)a_n{x}^{n+r-2}$$(d^(2)y)/(dx^(2))=sum_(n=0)^(oo)(n+r)(n+r-1)a_(n)x^(n+r-2)frac{d^2y}{dx^2} = sum_{n=0}^{infty} (n+r)(n+r-1) a_n x^{n+r-2}$$\frac{d^{2}y}{d{x}^2}=\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}$$
Substituting these into the differential equation, we get:
$$x(1-x)\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}-(1+3x)\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}-\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}=0$$$$x(1-x)\sum _{n=0}^{\mathrm{\infty }} (n+r)(n+r-1)a_n{x}^{n+r-2}-(1+3x)\sum _{n=0}^{\mathrm{\infty }} (n+r)a_n{x}^{n+r-1}-\sum _{n=0}^{\mathrm{\infty }} a_n{x}^{n+r}=0$$x(1-x)sum_(n=0)^(oo)(n+r)(n+r-1)a_(n)x^(n+r-2)-(1+3x)sum_(n=0)^(oo)(n+r)a_(n)x^(n+r-1)-sum_(n=0)^(oo)a_(n)x^(n+r)=0x(1-x) sum_{n=0}^{infty} (n+r)(n+r-1) a_n x^{n+r-2} – (1+3x) sum_{n=0}^{infty} (n+r) a_n x^{n+r-1} – sum_{n=0}^{infty} a_n x^{n+r} = 0$$x(1-x)\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}-(1+3x)\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}-\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}=0$$
Expanding and combining like terms, we have:
$$\sum _{n=0}^{\mathrm{\infty }}[(n+r)(n+r-1)-(n+r)+1]a_n{x}^{n+r}-\sum _{n=0}^{\mathrm{\infty }}[3(n+r)+(n+r)(n+r-1)]a_n{x}^{n+r+1}=0$$$$\sum _{n=0}^{\mathrm{\infty }} [(n+r)(n+r-1)-(n+r)+1]a_n{x}^{n+r}-\sum _{n=0}^{\mathrm{\infty }} [3(n+r)+(n+r)(n+r-1)]a_n{x}^{n+r+1}=0$$sum_(n=0)^(oo)[(n+r)(n+r-1)-(n+r)+1]a_(n)x^(n+r)-sum_(n=0)^(oo)[3(n+r)+(n+r)(n+r-1)]a_(n)x^(n+r+1)=0sum_{n=0}^{infty} [(n+r)(n+r-1) – (n+r) + 1] a_n x^{n+r} – sum_{n=0}^{infty} [3(n+r) + (n+r)(n+r-1)] a_n x^{n+r+1} = 0$$\sum _{n=0}^{\mathrm{\infty }}[(n+r)(n+r-1)-(n+r)+1]a_n{x}^{n+r}-\sum _{n=0}^{\mathrm{\infty }}[3(n+r)+(n+r)(n+r-1)]a_n{x}^{n+r+1}=0$$
Equating coefficients of $x^{n+r}$$x^{n+r}$x^(n+r)x^{n+r}$x^{n+r}$ and $x^{n+r+1}$$x^{n+r+1}$x^(n+r+1)x^{n+r+1}$x^{n+r+1}$ to zero, we get the recurrence relations:
$$(n+r)(n+r-2)a_n-(3(n+r)+(n+r)(n+r-1))a_{n+1}=0$$$$(n+r)(n+r-2)a_n-(3(n+r)+(n+r)(n+r-1))a_{n+1}=0$$(n+r)(n+r-2)a_(n)-(3(n+r)+(n+r)(n+r-1))a_(n+1)=0(n+r)(n+r-2) a_n – (3(n+r) + (n+r)(n+r-1)) a_{n+1} = 0$$(n+r)(n+r-2)a_n-(3(n+r)+(n+r)(n+r-1))a_{n+1}=0$$
Simplifying, we find:
$$(n+r)(n+r-2)a_n-(n+r+1)(n+r+3)a_{n+1}=0$$$$(n+r)(n+r-2)a_n-(n+r+1)(n+r+3)a_{n+1}=0$$(n+r)(n+r-2)a_(n)-(n+r+1)(n+r+3)a_(n+1)=0(n+r)(n+r-2) a_n – (n+r+1)(n+r+3) a_{n+1} = 0$$(n+r)(n+r-2)a_n-(n+r+1)(n+r+3)a_{n+1}=0$$
For $n=0$$n=0$n=0n = 0$n=0$, we have the indicial equation:
$$r(r-2)=0$$$$r(r-2)=0$$r(r-2)=0r(r-2) = 0$$r(r-2)=0$$
which gives $r=0$$r=0$r=0r = 0$r=0$ or $r=2$$r=2$r=2r = 2$r=2$.
For $r=0$$r=0$r=0r = 0$r=0$, the recurrence relation becomes:
$$n(n-2)a_n-(n+1)(n+3)a_{n+1}=0$$$$n(n-2)a_n-(n+1)(n+3)a_{n+1}=0$$n(n-2)a_(n)-(n+1)(n+3)a_(n+1)=0n(n-2) a_n – (n+1)(n+3) a_{n+1} = 0$$n(n-2)a_n-(n+1)(n+3)a_{n+1}=0$$
or
$$a_{n+1}=\frac{n(n-2)}{(n+1)(n+3)}{a}_n$$$$a_{n+1}=\frac{n(n-2)}{(n+1)(n+3)}{a}_n$$a_(n+1)=(n(n-2))/((n+1)(n+3))a_(n)a_{n+1} = frac{n(n-2)}{(n+1)(n+3)} a_n$$a_{n+1}=\frac{n(n-2)}{(n+1)(n+3)}{a}_n$$
Since $a_0$$a_0$a_(0)a_0$a_0$ is arbitrary, let’s choose $a_0=1$$a_0=1$a_(0)=1a_0 = 1$a_0=1$ for simplicity. Then, we can find the next few coefficients:
For $n=0$$n=0$n=0n = 0$n=0$:
$$a_1=\frac{0(0-2)}{(0+1)(0+3)}{a}_0=0$$$$a_1=\frac{0(0-2)}{(0+1)(0+3)}{a}_0=0$$a_(1)=(0(0-2))/((0+1)(0+3))a_(0)=0a_1 = frac{0(0-2)}{(0+1)(0+3)} a_0 = 0$$a_1=\frac{0(0-2)}{(0+1)(0+3)}{a}_0=0$$
For $n=1$$n=1$n=1n = 1$n=1$:
$$a_2=\frac{1(1-2)}{(1+1)(1+3)}{a}_1=0$$$$a_2=\frac{1(1-2)}{(1+1)(1+3)}{a}_1=0$$a_(2)=(1(1-2))/((1+1)(1+3))a_(1)=0a_2 = frac{1(1-2)}{(1+1)(1+3)} a_1 = 0$$a_2=\frac{1(1-2)}{(1+1)(1+3)}{a}_1=0$$
For $n=2$$n=2$n=2n = 2$n=2$:
$$a_3=\frac{2(2-2)}{(2+1)(2+3)}{a}_2=0$$$$a_3=\frac{2(2-2)}{(2+1)(2+3)}{a}_2=0$$a_(3)=(2(2-2))/((2+1)(2+3))a_(2)=0a_3 = frac{2(2-2)}{(2+1)(2+3)} a_2 = 0$$a_3=\frac{2(2-2)}{(2+1)(2+3)}{a}_2=0$$
For $n=3$$n=3$n=3n = 3$n=3$:
$$a_4=\frac{3(3-2)}{(3+1)(3+3)}{a}_3=0$$$$a_4=\frac{3(3-2)}{(3+1)(3+3)}{a}_3=0$$a_(4)=(3(3-2))/((3+1)(3+3))a_(3)=0a_4 = frac{3(3-2)}{(3+1)(3+3)} a_3 = 0$$a_4=\frac{3(3-2)}{(3+1)(3+3)}{a}_3=0$$
And so on. It appears that all coefficients after $a_0$$a_0$a_(0)a_0$a_0$ are zero. Therefore, the solution for $r=0$$r=0$r=0r = 0$r=0$ is:
$$y=a_0=1$$$$y=a_0=1$$y=a_(0)=1y = a_0 = 1$$y=a_0=1$$
Now, let’s consider the case $r=2$$r=2$r=2r = 2$r=2$:
For $r=2$$r=2$r=2r = 2$r=2$, the recurrence relation becomes:
$$(n+2)(n)a_n-(n+3)(n+5)a_{n+1}=0$$$$(n+2)(n)a_n-(n+3)(n+5)a_{n+1}=0$$(n+2)(n)a_(n)-(n+3)(n+5)a_(n+1)=0(n+2)(n) a_n – (n+3)(n+5) a_{n+1} = 0$$(n+2)(n)a_n-(n+3)(n+5)a_{n+1}=0$$
or
$$a_{n+1}=\frac{(n+2)(n)}{(n+3)(n+5)}{a}_n$$$$a_{n+1}=\frac{(n+2)(n)}{(n+3)(n+5)}{a}_n$$a_(n+1)=((n+2)(n))/((n+3)(n+5))a_(n)a_{n+1} = frac{(n+2)(n)}{(n+3)(n+5)} a_n$$a_{n+1}=\frac{(n+2)(n)}{(n+3)(n+5)}{a}_n$$
Again, let’s choose $a_0=1$$a_0=1$a_(0)=1a_0 = 1$a_0=1$. Then, we can find the next few coefficients:
For $n=0$$n=0$n=0n = 0$n=0$:
$$a_1=\frac{2(0)}{(0+3)(0+5)}{a}_0=0$$$$a_1=\frac{2(0)}{(0+3)(0+5)}{a}_0=0$$a_(1)=(2(0))/((0+3)(0+5))a_(0)=0a_1 = frac{2(0)}{(0+3)(0+5)} a_0 = 0$$a_1=\frac{2(0)}{(0+3)(0+5)}{a}_0=0$$
For $n=1$$n=1$n=1n = 1$n=1$:
$$a_2=\frac{3(1)}{(1+3)(1+5)}{a}_1=0$$$$a_2=\frac{3(1)}{(1+3)(1+5)}{a}_1=0$$a_(2)=(3(1))/((1+3)(1+5))a_(1)=0a_2 = frac{3(1)}{(1+3)(1+5)} a_1 = 0$$a_2=\frac{3(1)}{(1+3)(1+5)}{a}_1=0$$
And so on. It appears that all coefficients after $a_0$$a_0$a_(0)a_0$a_0$ are also zero in this case. Therefore, the solution for $r=2$$r=2$r=2r = 2$r=2$ is:
$$y=a_0{x}^2=x^2$$$$y=a_0{x}^2=x^2$$y=a_(0)x^(2)=x^(2)y = a_0 x^2 = x^2$$y=a_0{x}^2=x^2$$
In conclusion, the series solutions about $x=0$$x=0$x=0x=0$x=0$ of the equation $x(1-x)\frac{d^{2}y}{d{x}^2}-(1+3x)\frac{dy}{dx}-y=0$$x(1-x)\frac{d^{2}y}{d{x}^2}-(1+3x)\frac{dy}{dx}-y=0$x(1-x)(d^(2)y)/(dx^(2))-(1+3x)(dy)/(dx)-y=0x(1-x) frac{d^2 y}{d x^2}-(1+3 x) frac{d y}{d x}-y=0$x(1-x)\frac{d^{2}y}{d{x}^2}-(1+3x)\frac{dy}{dx}-y=0$ are $y=1$$y=1$y=1y = 1$y=1$ and $y=x^2$$y=x^2$y=x^(2)y = x^2$y=x^2$.


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