BE Computer Engineering (IOE, TU) Engineering Mathematics III (IOE, SH 501) Question Paper 2078 Nepal

(a) Definition, existence and the transform of $e^{-2t}t\cos 3t$

Definition. The Laplace transform of $f(t)$ (defined for $t\ge 0$) is

$$\mathcal{L}\{f(t)\}=F(s)=\int_0^\infty e^{-st}f(t)\,dt,$$

provided the integral converges.

Conditions of existence (sufficient). If $f(t)$ is

1. piecewise continuous on every finite interval $0\le t\le N$, and
2. of exponential order $\alpha$, i.e. there exist $M>0,\;\alpha$ such that $|f(t)|\le M e^{\alpha t}$ for $t\ge N$,

then $F(s)$ exists for all $s>\alpha$.

Transform of $f(t)=e^{-2t}\,t\cos 3t$.

Start from $\mathcal{L}\{\cos 3t\}=\dfrac{s}{s^2+9}.$

Multiplication by $t$ (differentiation theorem $\mathcal{L}\{t\,g(t)\}=-G'(s)$):

$$\mathcal{L}\{t\cos 3t\}=-\frac{d}{ds}\!\left(\frac{s}{s^2+9}\right)=-\frac{(s^2+9)-s(2s)}{(s^2+9)^2}=\frac{s^2-9}{(s^2+9)^2}.$$

First shifting theorem ($\mathcal{L}\{e^{-2t}g(t)\}=G(s+2)$): replace $s\to s+2$:

$$\boxed{\;\mathcal{L}\{e^{-2t}t\cos 3t\}=\frac{(s+2)^2-9}{\big((s+2)^2+9\big)^2}=\frac{s^2+4s-5}{(s^2+4s+13)^2}.\;}$$

(b) IVP $y''+4y'+4y=e^{-2t},\;y(0)=0,\;y'(0)=1$

Taking Laplace transforms with $Y=\mathcal{L}\{y\}$:

$$\big(s^2Y - s\,y(0)-y'(0)\big)+4\big(sY-y(0)\big)+4Y=\frac{1}{s+2}.$$

Using $y(0)=0,\;y'(0)=1$:

$$(s^2+4s+4)Y-1=\frac{1}{s+2}\quad\Rightarrow\quad (s+2)^2 Y = 1+\frac{1}{s+2}.$$

Hence

$$Y=\frac{1}{(s+2)^2}+\frac{1}{(s+2)^3}.$$

Inverting term by term using $\mathcal{L}^{-1}\{(s+a)^{-n}\}=\dfrac{t^{n-1}}{(n-1)!}e^{-at}$:

$$\boxed{\;y(t)=t\,e^{-2t}+\tfrac12 t^2 e^{-2t}=e^{-2t}\left(t+\tfrac{t^2}{2}\right).\;}$$

Check: $y(0)=0$ and $y'(t)=e^{-2t}(1+t-t^2)\Rightarrow y'(0)=1.$ ✓

(a) Fourier series of $f(x)=|x|$ on $(-\pi,\pi)$

Here $f(x)=-x$ for $-\pi<x<0$ and $f(x)=x$ for $0<x<\pi$, i.e. $f(x)=|x|$, which is an even function. Hence all $b_n=0$ and only cosine terms appear.

Constant term.

$$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|\,dx=\frac{2}{\pi}\int_0^\pi x\,dx=\frac{2}{\pi}\cdot\frac{\pi^2}{2}=\pi.$$

Cosine coefficients.

$$a_n=\frac{2}{\pi}\int_0^\pi x\cos nx\,dx=\frac{2}{\pi}\left[\frac{x\sin nx}{n}+\frac{\cos nx}{n^2}\right]_0^\pi=\frac{2}{\pi}\cdot\frac{\cos n\pi-1}{n^2}=\frac{2}{\pi n^2}\big((-1)^n-1\big).$$

Thus $a_n=0$ for even $n$ and $a_n=-\dfrac{4}{\pi n^2}$ for odd $n$. Writing $n=2k-1$:

$$\boxed{\;f(x)=\frac{\pi}{2}-\frac{4}{\pi}\sum_{k=1}^{\infty}\frac{\cos(2k-1)x}{(2k-1)^2}.\;}$$

Deduction. Put $x=0$; $f(0)=0$ and $\cos 0=1$:

$$0=\frac{\pi}{2}-\frac{4}{\pi}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}\;\Rightarrow\;\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}=\frac{\pi}{2}\cdot\frac{\pi}{4}=\boxed{\frac{\pi^2}{8}.}$$

(b) Half-range cosine series of $f(x)=x$ on $(0,2)$

Here $L=2$. For a half-range cosine series $b_n=0$ and

$$a_0=\frac{2}{L}\int_0^L x\,dx=\frac{2}{2}\int_0^2 x\,dx=2,$$ $$a_n=\frac{2}{L}\int_0^L x\cos\frac{n\pi x}{L}\,dx=\int_0^2 x\cos\frac{n\pi x}{2}\,dx.$$

Integrating by parts:

$$a_n=\left[\frac{2x}{n\pi}\sin\frac{n\pi x}{2}+\frac{4}{n^2\pi^2}\cos\frac{n\pi x}{2}\right]_0^2=\frac{4}{n^2\pi^2}\big(\cos n\pi-1\big)=\frac{4}{n^2\pi^2}\big((-1)^n-1\big).$$

So $a_n=0$ (even $n$), $a_n=-\dfrac{8}{n^2\pi^2}$ (odd $n$). With $a_0/2=1$:

$$\boxed{\;x=1-\frac{8}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}\cos\frac{(2k-1)\pi x}{2},\qquad 0<x<2.\;}$$

(a) Separation of variables for the heat equation

Solve $u_t=c^2u_{xx}$, $0<x<L$, with $u(0,t)=u(L,t)=0$ and $u(x,0)=f(x)$.

Separation. Assume $u(x,t)=X(x)T(t)$. Substituting,

$$X T' = c^2 X'' T\;\Rightarrow\;\frac{T'}{c^2T}=\frac{X''}{X}=-\lambda\ (\text{constant}).$$

This gives two ODEs:

$$X''+\lambda X=0,\qquad T'+c^2\lambda T=0.$$

Spatial problem (eigenvalues). Boundary conditions force $X(0)=X(L)=0$. Only $\lambda>0$ gives non-trivial solutions. Writing $\lambda=k^2$, $X=A\cos kx+B\sin kx$; $X(0)=0\Rightarrow A=0$; $X(L)=0\Rightarrow \sin kL=0\Rightarrow k=\dfrac{n\pi}{L}$. Hence

$$\lambda_n=\left(\frac{n\pi}{L}\right)^2,\qquad X_n(x)=\sin\frac{n\pi x}{L},\quad n=1,2,3,\dots$$

Time part. $T'+c^2\lambda_n T=0\Rightarrow T_n(t)=e^{-c^2(n\pi/L)^2 t}.$

Superposition.

$$u(x,t)=\sum_{n=1}^{\infty} b_n\sin\frac{n\pi x}{L}\,e^{-c^2(n\pi/L)^2 t}.$$

Initial condition. At $t=0$, $f(x)=\sum b_n\sin\dfrac{n\pi x}{L}$, so the $b_n$ are Fourier sine coefficients:

$$\boxed{\;b_n=\frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx.\;}$$

(b) Particular case $f(x)=3\sin\frac{\pi x}{L}-5\sin\frac{4\pi x}{L}$

The initial profile is already a sine sum, so by inspection $b_1=3,\;b_4=-5$ and all other $b_n=0$. Therefore

$$\boxed{\;u(x,t)=3\sin\frac{\pi x}{L}\,e^{-c^2(\pi/L)^2 t}-5\sin\frac{4\pi x}{L}\,e^{-16c^2(\pi/L)^2 t}.\;}$$

(a) Frobenius solution of $2x y''+(1-x)y'-y=0$

Write in standard form $y''+\dfrac{1-x}{2x}y'-\dfrac{1}{2x}y=0$. At $x=0$, $xp(x)=\tfrac12(1-x)$ and $x^2q(x)=-\tfrac12 x$ are analytic, so $x=0$ is a regular singular point.

Assume $y=\sum_{k=0}^{\infty} a_k x^{k+r}$, $a_0\ne 0$. Then

$$y'=\sum (k+r)a_k x^{k+r-1},\qquad y''=\sum (k+r)(k+r-1)a_k x^{k+r-2}.$$

Substitute:

$$\sum 2(k+r)(k+r-1)a_k x^{k+r-1}+\sum (k+r)a_k x^{k+r-1}-\sum (k+r)a_k x^{k+r}-\sum a_k x^{k+r}=0.$$

Indicial equation (lowest power $x^{r-1}$, $k=0$):

$$2r(r-1)a_0+r a_0=0\;\Rightarrow\; r(2r-1)=0\;\Rightarrow\; \boxed{r=0\ \text{or}\ r=\tfrac12.}$$

Recurrence relation. Collecting the coefficient of $x^{k+r}$:

$$\big[2(k+1+r)(k+r)+(k+1+r)\big]a_{k+1}-\big[(k+r)+1\big]a_k=0,$$ $$\boxed{\;a_{k+1}=\frac{(k+r+1)}{(k+r+1)\big(2(k+r)+1\big)}a_k=\frac{a_k}{2(k+r)+1}.\;}$$

For $r=0$: $a_{k+1}=\dfrac{a_k}{2k+1}$, giving $a_1=a_0,\;a_2=\tfrac{a_0}{3},\;a_3=\tfrac{a_0}{15},\dots$ so

$$y_1=a_0\left(1+x+\frac{x^2}{3}+\frac{x^3}{15}+\cdots\right).$$

For $r=\tfrac12$: $a_{k+1}=\dfrac{a_k}{2k+2}=\dfrac{a_k}{2(k+1)}$, giving $a_k=\dfrac{a_0}{2^k k!}$, so

$$y_2=a_0\,x^{1/2}\sum_{k=0}^{\infty}\frac{x^k}{2^k k!}=a_0\,x^{1/2}e^{x/2}.$$

The general solution is $y=c_1 y_1+c_2 y_2$.

(b) Bessel equation for $n=\tfrac12$

For $x^2y''+xy'+(x^2-n^2)y=0$ with $n=\tfrac12$, substitute $y=\dfrac{u(x)}{\sqrt{x}}$. Then

$$y'=\frac{u'}{\sqrt{x}}-\frac{u}{2x^{3/2}},\qquad y''=\frac{u''}{\sqrt{x}}-\frac{u'}{x^{3/2}}+\frac{3u}{4x^{5/2}}.$$

Substituting and simplifying, the equation reduces to the elementary harmonic form

$$u''+u=0,$$

whose general solution is $u=A\cos x+B\sin x$. Hence

$$y=\frac{A\cos x+B\sin x}{\sqrt{x}}.$$

The solution bounded/standard at the origin (Bessel function $J_{1/2}$) is

$$\boxed{\;J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\,\sin x\ \propto\ \frac{\sin x}{\sqrt{x}}.\;}$$

(a) Convolution theorem

Statement. If $\mathcal{L}\{f\}=F(s)$ and $\mathcal{L}\{g\}=G(s)$, then with the convolution $(f*g)(t)=\displaystyle\int_0^t f(\tau)g(t-\tau)\,d\tau$,

$$\mathcal{L}\{(f*g)(t)\}=F(s)\,G(s).$$

Proof.

$$F(s)G(s)=\int_0^\infty e^{-s\tau}f(\tau)\,d\tau\int_0^\infty e^{-s\sigma}g(\sigma)\,d\sigma=\int_0^\infty\!\!\int_0^\infty e^{-s(\tau+\sigma)}f(\tau)g(\sigma)\,d\sigma\,d\tau.$$

Substitute $t=\tau+\sigma$ (so $\sigma=t-\tau$, $t$ from $\tau$ to $\infty$) and interchange the order of integration:

$$F(s)G(s)=\int_0^\infty e^{-st}\left(\int_0^t f(\tau)g(t-\tau)\,d\tau\right)dt=\mathcal{L}\{(f*g)(t)\}.\qquad\blacksquare$$

Application. Write $\dfrac{1}{(s^2+a^2)^2}=\dfrac{1}{s^2+a^2}\cdot\dfrac{1}{s^2+a^2}$. Since $\mathcal{L}^{-1}\{\tfrac{1}{s^2+a^2}\}=\tfrac{1}{a}\sin at$, by convolution

$$\mathcal{L}^{-1}\!\left\{\frac{1}{(s^2+a^2)^2}\right\}=\frac{1}{a^2}\int_0^t \sin a\tau\,\sin a(t-\tau)\,d\tau.$$

Using $\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]$:

$$=\frac{1}{2a^2}\int_0^t\big[\cos a(2\tau-t)-\cos at\big]d\tau=\frac{1}{2a^2}\left[\frac{\sin at}{a}-t\cos at\right].$$ $$\boxed{\;\mathcal{L}^{-1}\!\left\{\frac{1}{(s^2+a^2)^2}\right\}=\frac{1}{2a^3}\big(\sin at-at\cos at\big).\;}$$

(b) Laplace transform of $f(t)=|\sin t|$ (period $\pi$)

For a periodic function of period $T$, $\mathcal{L}\{f\}=\dfrac{1}{1-e^{-sT}}\displaystyle\int_0^T e^{-st}f(t)\,dt$. Here $T=\pi$ and $f(t)=\sin t$ on $(0,\pi)$:

$$\int_0^\pi e^{-st}\sin t\,dt=\left[\frac{e^{-st}(-s\sin t-\cos t)}{s^2+1}\right]_0^\pi=\frac{1+e^{-\pi s}}{s^2+1}.$$

Hence

$$\mathcal{L}\{|\sin t|\}=\frac{1}{1-e^{-\pi s}}\cdot\frac{1+e^{-\pi s}}{s^2+1}=\boxed{\frac{1}{s^2+1}\coth\!\frac{\pi s}{2}.}$$

(a) Line integral by Green's theorem

Green's theorem: $\displaystyle\oint_C(P\,dx+Q\,dy)=\iint_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$

Here $P=3x^2-8y^2,\;Q=4y-6xy$, so

$$\frac{\partial Q}{\partial x}=-6y,\qquad \frac{\partial P}{\partial y}=-16y,\qquad \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=10y.$$

Region. $y=\sqrt{x}$ and $y=x^2$ meet at $(0,0)$ and $(1,1)$; for $0\le x\le 1$, $x^2\le y\le\sqrt{x}$. Thus

$$\oint_C=\int_0^1\!\!\int_{x^2}^{\sqrt{x}} 10y\,dy\,dx=\int_0^1 5\big[y^2\big]_{x^2}^{\sqrt{x}}dx=\int_0^1 5\big(x-x^4\big)dx.$$ $$=5\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1=5\left(\frac12-\frac15\right)=5\cdot\frac{3}{10}=\boxed{\frac{3}{2}.}$$

(b) Conservative field and potential

$\vec F=(2xy+z^3)\hat i+x^2\hat j+3xz^2\hat k$. Compute the curl:

$$\nabla\times\vec F=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)\hat i+\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right)\hat j+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\hat k.$$

• $\partial_y F_3-\partial_z F_2=0-0=0$
• $\partial_z F_1-\partial_x F_3=3z^2-3z^2=0$
• $\partial_x F_2-\partial_y F_1=2x-2x=0$

Since $\nabla\times\vec F=\vec 0$, the field is conservative. Find $\phi$ with $\nabla\phi=\vec F$:

$$\phi=\int(2xy+z^3)\,dx=x^2y+xz^3+g(y,z).$$

$\partial_y\phi=x^2+g_y=x^2\Rightarrow g_y=0$; $\partial_z\phi=3xz^2+g_z=3xz^2\Rightarrow g_z=0$. So $g=$ const.

$$\boxed{\;\phi(x,y,z)=x^2y+xz^3+C.\;}$$

(a) Verify the Divergence theorem for $\vec F=x^2\hat i+y^2\hat j+z^2\hat k$ over the unit cube

Gauss theorem: $\displaystyle\iint_S \vec F\cdot\hat n\,dS=\iiint_V\nabla\cdot\vec F\,dV.$

Volume integral. $\nabla\cdot\vec F=2x+2y+2z$.

$$\iiint_V(2x+2y+2z)\,dV=2\int_0^1\!\!\int_0^1\!\!\int_0^1(x+y+z)\,dx\,dy\,dz.$$

By symmetry each of $\int\!\!\int\!\!\int x\,dV=\tfrac12$, so the sum $=3\cdot\tfrac12=\tfrac32$ and the integral $=2\cdot\tfrac32=\boxed{3}.$

Surface integral. Evaluate $\iint \vec F\cdot\hat n\,dS$ over the six faces.

• Face $x=1$ ($\hat n=\hat i$): $\iint x^2\,dS=\iint 1\,dy\,dz=1$. Face $x=0$: $\hat n=-\hat i$, $x^2=0\Rightarrow 0$.
• Face $y=1$: $1$; face $y=0$: $0$.
• Face $z=1$: $1$; face $z=0$: $0$.

Total $=1+1+1=3$. Since both sides equal $3$, the theorem is verified. ✓

(b) Stokes' theorem and physical significance

Statement. For a smooth oriented surface $S$ bounded by a closed curve $C$ (oriented consistently with $\hat n$ by the right-hand rule),

$$\oint_C \vec F\cdot d\vec r=\iint_S(\nabla\times\vec F)\cdot\hat n\,dS.$$

Physical significance. The line integral $\oint_C\vec F\cdot d\vec r$ is the circulation of the field around the boundary $C$. Stokes' theorem says this circulation equals the total flux of the curl (local rotation / vortex density) of the field through any surface spanning $C$. Thus the curl measures the microscopic rotation of the field, and summing it over the surface reproduces the net macroscopic circulation around the edge — the basis, for example, of Ampere's and Faraday's laws in electromagnetism.

(a) Rodrigues' formula and $P_2,P_3$

Rodrigues' formula.

$$P_n(x)=\frac{1}{2^n\,n!}\frac{d^n}{dx^n}\big[(x^2-1)^n\big].$$

$P_2(x)$ ($n=2$): $(x^2-1)^2=x^4-2x^2+1$, second derivative $=12x^2-4$.

$$P_2(x)=\frac{1}{2^2\cdot 2!}(12x^2-4)=\frac{12x^2-4}{8}=\boxed{\tfrac12(3x^2-1).}$$

$P_3(x)$ ($n=3$): $(x^2-1)^3=x^6-3x^4+3x^2-1$; third derivative $=120x^3-72x$.

$$P_3(x)=\frac{1}{2^3\cdot 3!}(120x^3-72x)=\frac{120x^3-72x}{48}=\boxed{\tfrac12(5x^3-3x).}$$

(b) Orthogonality of Legendre polynomials

Each $P_n$ satisfies Legendre's equation, written in Sturm–Liouville form:

$$\frac{d}{dx}\!\left[(1-x^2)P_n'\right]+n(n+1)P_n=0.$$

Similarly $\dfrac{d}{dx}\!\left[(1-x^2)P_m'\right]+m(m+1)P_m=0.$

Multiply the first by $P_m$ and the second by $P_n$, subtract, and integrate over $[-1,1]$:

$$\int_{-1}^{1}\!\Big(P_m\big[(1-x^2)P_n'\big]'-P_n\big[(1-x^2)P_m'\big]'\Big)dx+\big[n(n+1)-m(m+1)\big]\!\int_{-1}^{1}P_mP_n\,dx=0.$$

The first integral, integrated by parts, equals

$$\Big[(1-x^2)\big(P_m P_n'-P_n P_m'\big)\Big]_{-1}^{1}=0,$$

because the factor $(1-x^2)$ vanishes at $x=\pm1$. Therefore

$$\big[n(n+1)-m(m+1)\big]\int_{-1}^{1}P_m(x)P_n(x)\,dx=0.$$

For $m\neq n$ the bracket $n(n+1)-m(m+1)\neq 0$, hence

$$\boxed{\;\int_{-1}^{1}P_m(x)P_n(x)\,dx=0,\qquad m\neq n.\;}\qquad\blacksquare$$

(a) Cauchy–Riemann equations and analyticity of $f(z)=z^2$

Let $f(z)=u(x,y)+iv(x,y)$, $z=x+iy$. If $f$ is differentiable at $z$, the derivative must be the same along the real and imaginary directions:

$$f'(z)=\lim_{\Delta x\to0}\frac{\Delta f}{\Delta x}=u_x+iv_x,\qquad f'(z)=\lim_{\Delta y\to0}\frac{\Delta f}{i\Delta y}=\frac{1}{i}(u_y+iv_y)=v_y-iu_y.$$

Equating real and imaginary parts gives the Cauchy–Riemann (CR) equations:

$$\boxed{\;u_x=v_y,\qquad u_y=-v_x.\;}$$

Apply to $f(z)=z^2$. $z^2=(x+iy)^2=(x^2-y^2)+i(2xy)$, so $u=x^2-y^2,\;v=2xy$.

$$u_x=2x=v_y,\qquad u_y=-2y=-v_x.$$

The CR equations hold for all $x,y$ and the partials are continuous, so $f(z)=z^2$ is analytic everywhere (entire). Its derivative:

$$f'(z)=u_x+iv_x=2x+i(2y)=2(x+iy)=\boxed{2z.}$$

(b) Conjugate of $u=x^3-3xy^2$

First check $u$ is harmonic: $u_{xx}=6x,\;u_{yy}=-6x\Rightarrow u_{xx}+u_{yy}=0$. ✓

Using CR: $v_y=u_x=3x^2-3y^2$. Integrate w.r.t. $y$:

$$v=3x^2y-y^3+h(x).$$

Also $v_x=-u_y$. Now $u_y=-6xy$, so $v_x=6xy$. From the expression, $v_x=6xy+h'(x)\Rightarrow h'(x)=0\Rightarrow h=$ const $=C$. Thus

$$\boxed{\;v(x,y)=3x^2y-y^3+C.\;}$$

Analytic function.

$$f(z)=u+iv=(x^3-3xy^2)+i(3x^2y-y^3)+iC=(x+iy)^3+iC=\boxed{z^3+iC.}$$

(a) Form the PDE by eliminating arbitrary functions from $z=f(x+ct)+g(x-ct)$

Let $\xi=x+ct,\;\eta=x-ct$. Differentiate:

$$z_x=f'+g',\qquad z_{xx}=f''+g'',$$ $$z_t=c f'-c g',\qquad z_{tt}=c^2 f''+c^2 g''=c^2(f''+g'').$$

Comparing, $z_{tt}=c^2 z_{xx}$, i.e.

$$\boxed{\;\frac{\partial^2 z}{\partial t^2}=c^2\frac{\partial^2 z}{\partial x^2}\;}$$

which is the one-dimensional wave equation (the arbitrary functions $f,g$ are eliminated).

(b) d'Alembert's solution

The wave equation $y_{tt}=c^2 y_{xx}$ has general solution (from part a) $y(x,t)=\phi_1(x+ct)+\phi_2(x-ct)$ for arbitrary twice-differentiable $\phi_1,\phi_2$.

Apply initial conditions $y(x,0)=\phi(x)$, $y_t(x,0)=0$.

$$y(x,0)=\phi_1(x)+\phi_2(x)=\phi(x),$$ $$y_t(x,0)=c\phi_1'(x)-c\phi_2'(x)=0\;\Rightarrow\;\phi_1'(x)=\phi_2'(x)\;\Rightarrow\;\phi_1(x)-\phi_2(x)=\text{const}.$$

Hence $\phi_1(x)=\phi_2(x)=\tfrac12\phi(x)$ (the constant is absorbed). Therefore

$$\boxed{\;y(x,t)=\tfrac12\big[\phi(x+ct)+\phi(x-ct)\big].\;}$$

The initial displacement splits into two equal waves travelling with speed $c$ in the $+x$ and $-x$ directions.