BE Computer Engineering (IOE, TU) Engineering Mathematics III (IOE, SH 501) Question Paper 2079 Nepal

(a) First Shifting Theorem [6]

Statement. If $\mathcal{L}\{f(t)\} = F(s)$ for $s > a$, then

$$\mathcal{L}\{e^{at}f(t)\} = F(s-a).$$

Proof. By definition,

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

which converges for $s-a >$ (abscissa of convergence), i.e. $s > a$. Replacing $s$ by $s-a$ in $F(s)$ shifts the transform along the $s$-axis. $\blacksquare$

Evaluation. Take $f(t) = 3\cos 4t - 2\sin 4t$. Then

$$\mathcal{L}\{3\cos 4t\} = \frac{3s}{s^2+16}, \qquad \mathcal{L}\{2\sin 4t\} = \frac{8}{s^2+16},$$

so $F(s) = \dfrac{3s - 8}{s^2+16}$. Here $a = -2$, hence by the shifting theorem replace $s \to s+2$:

$$\mathcal{L}\{e^{-2t}(3\cos 4t - 2\sin 4t)\} = \frac{3(s+2) - 8}{(s+2)^2 + 16} = \boxed{\dfrac{3s - 2}{(s+2)^2 + 16}}.$$

(b) IVP by Laplace Transform [6]

Given $y'' + 4y' + 4y = e^{-2t}$, $y(0)=0$, $y'(0)=1$. Let $Y(s)=\mathcal{L}\{y\}$.

Taking transforms with $\mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0)$ and $\mathcal{L}\{y'\} = sY - y(0)$:

$$\big(s^2Y - 0 - 1\big) + 4\big(sY - 0\big) + 4Y = \frac{1}{s+2}.$$ $$\big(s^2 + 4s + 4\big)Y - 1 = \frac{1}{s+2} \;\Rightarrow\; (s+2)^2 Y = 1 + \frac{1}{s+2} = \frac{s+3}{s+2}.$$ $$Y = \frac{s+3}{(s+2)^3} = \frac{(s+2)+1}{(s+2)^3} = \frac{1}{(s+2)^2} + \frac{1}{(s+2)^3}.$$

Inverting term by term (using $\mathcal{L}^{-1}\{1/(s+2)^n\} = \dfrac{t^{n-1}}{(n-1)!}e^{-2t}$):

$$\boxed{y(t) = t\,e^{-2t} + \tfrac{1}{2}t^2 e^{-2t} = e^{-2t}\left(t + \tfrac{1}{2}t^2\right).}$$

Check: $y(0)=0$, $y'(0)=1$ are satisfied.

(a) Fourier Series and Leibniz Series [8]

For $-\pi < x < \pi$, $f(x) = -x$ plus an odd step part; we compute directly. Write the Fourier series $f(x)=\dfrac{a_0}{2}+\sum_{n\ge1}(a_n\cos nx + b_n\sin nx)$ on $(-\pi,\pi)$.

Constant term.

$$a_0 = \frac1\pi\int_{-\pi}^{0}(-\pi - x)\,dx + \frac1\pi\int_{0}^{\pi}(\pi - x)\,dx.$$

$\int_{-\pi}^0(-\pi-x)dx = [-\pi x - x^2/2]_{-\pi}^0 = -(\pi^2 - \pi^2/2) = -\pi^2/2.$ $\int_0^\pi(\pi-x)dx = \pi^2 - \pi^2/2 = \pi^2/2.$ Hence $a_0 = 0$.

Cosine coefficients. The function $f(x)+x$ equals $-\pi$ on $(-\pi,0)$ and $+\pi$ on $(0,\pi)$, which is odd; and $-x$ is odd. So $f(x)$ is an odd function, giving $a_n = 0$ for all $n$.

Sine coefficients.

$$b_n = \frac2\pi\int_0^\pi (\pi - x)\sin nx\,dx = \frac2\pi\left[\pi\!\int_0^\pi\!\sin nx\,dx - \int_0^\pi\! x\sin nx\,dx\right].$$

$\int_0^\pi\sin nx\,dx = \dfrac{1-\cos n\pi}{n}=\dfrac{1-(-1)^n}{n}.$ Using $\int_0^\pi x\sin nx\,dx = -\dfrac{\pi\cos n\pi}{n}=-\dfrac{\pi(-1)^n}{n}$:

$$b_n = \frac2\pi\left[\frac{\pi(1-(-1)^n)}{n} + \frac{\pi(-1)^n}{n}\right] = \frac{2}{n}\big[1-(-1)^n+(-1)^n\big] = \frac{2}{n}.$$

Therefore

$$\boxed{f(x) = \sum_{n=1}^\infty \frac{2}{n}\sin nx = 2\left(\sin x + \tfrac12\sin 2x + \tfrac13\sin 3x + \cdots\right).}$$

Deduction. Put $x = \dfrac{\pi}{2}$. Then $f(\pi/2) = \pi - \pi/2 = \pi/2$, and $\sin(n\pi/2)$ is $0$ for even $n$ and $(-1)^{(n-1)/2}$ for odd $n$:

$$\frac{\pi}{2} = 2\left(1 - \frac13 + \frac15 - \frac17 + \cdots\right) \;\Rightarrow\; \boxed{\frac{\pi}{4} = 1 - \frac13 + \frac15 - \frac17 + \cdots}$$

(b) Half-Range Cosine Series of $f(x)=x$ on $(0,2)$ [4]

Here $L = 2$. Cosine series: $f(x) = \dfrac{a_0}{2} + \sum_{n\ge1} a_n\cos\dfrac{n\pi x}{2}$.

$$a_0 = \frac{2}{2}\int_0^2 x\,dx = \frac{1}{1}\cdot\frac{x^2}{2}\Big|_0^2 = 2.$$ $$a_n = \frac{2}{2}\int_0^2 x\cos\frac{n\pi x}{2}\,dx = \left[\frac{2x}{n\pi}\sin\frac{n\pi x}{2} + \frac{4}{n^2\pi^2}\cos\frac{n\pi x}{2}\right]_0^2 = \frac{4}{n^2\pi^2}\big(\cos n\pi - 1\big) = \frac{4((-1)^n - 1)}{n^2\pi^2}.$$

Thus $a_n = 0$ for even $n$, and $a_n = -\dfrac{8}{n^2\pi^2}$ for odd $n$. Hence

$$\boxed{f(x) = 1 - \frac{8}{\pi^2}\sum_{n\,\text{odd}}\frac{1}{n^2}\cos\frac{n\pi x}{2} = 1 - \frac{8}{\pi^2}\left(\cos\frac{\pi x}{2} + \frac{1}{9}\cos\frac{3\pi x}{2} + \cdots\right).}$$

(a) Separation of Variables for the Heat Equation [10]

Seek $u(x,t) = X(x)T(t)$. Substituting into $u_t = c^2 u_{xx}$:

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

This gives two ODEs:

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

Boundary conditions. $u(0,t)=0 \Rightarrow X(0)=0$ and $u(L,t)=0 \Rightarrow X(L)=0$ (non-trivial $T$).

Eigenvalue analysis. For $\lambda \le 0$ the only solution satisfying both BCs is $X\equiv0$. For $\lambda = p^2 > 0$:

$$X(x) = A\cos px + B\sin px.$$

$X(0)=0 \Rightarrow A=0$. $X(L)=B\sin pL = 0$ with $B\ne0 \Rightarrow \sin pL = 0 \Rightarrow p = \dfrac{n\pi}{L}$, $n=1,2,3,\dots$

Thus eigenvalues $\lambda_n = \left(\dfrac{n\pi}{L}\right)^2$ and eigenfunctions $X_n(x) = \sin\dfrac{n\pi x}{L}$.

Time part. $T_n' = -c^2\left(\dfrac{n\pi}{L}\right)^2 T_n \Rightarrow T_n(t) = e^{-c^2 n^2\pi^2 t/L^2}.$

By superposition, the general solution satisfying the BCs is

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

(b) Temperature Distribution [6]

Apply the initial condition $u(x,0) = \sin\dfrac{\pi x}{L} + 3\sin\dfrac{3\pi x}{L}$:

$$\sum_{n=1}^\infty b_n \sin\frac{n\pi x}{L} = \sin\frac{\pi x}{L} + 3\sin\frac{3\pi x}{L}.$$

Comparing coefficients of the orthogonal sine modes: $b_1 = 1$, $b_3 = 3$, all other $b_n = 0$. Hence

$$\boxed{u(x,t) = \sin\frac{\pi x}{L}\,e^{-c^2\pi^2 t/L^2} + 3\sin\frac{3\pi x}{L}\,e^{-9c^2\pi^2 t/L^2}.}$$

The second mode decays $9\times$ faster than the first.

(a) Green's Theorem [7]

Statement. If $C$ is a positively oriented, piecewise-smooth, simple closed curve bounding a region $R$, and $M,N$ have continuous partial derivatives on $R$, then

$$\oint_C M\,dx + N\,dy = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dA.$$

Application. Here $M = 3x^2 - 8y^2$, $N = 4y - 6xy$.

$$\frac{\partial N}{\partial x} = -6y, \qquad \frac{\partial M}{\partial y} = -16y \;\Rightarrow\; \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -6y + 16y = 10y.$$

The region between $y=x^2$ and $y=\sqrt{x}$ (intersecting at $x=0,1$) has $x^2 \le y \le \sqrt{x}$, $0\le x\le1$:

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

(b) Surface Integral via Divergence Theorem [5]

With $\mathbf{F} = 4x\,\hat i - 2y^2\,\hat j + z^2\,\hat k$,

$$\nabla\cdot\mathbf{F} = 4 - 4y + 2z.$$

By the divergence theorem, $\displaystyle\iint_S \mathbf{F}\cdot\hat n\,dS = \iiint_V (4 - 4y + 2z)\,dV$ over the cylinder $x^2+y^2\le4$, $0\le z\le3$.

By symmetry $\iiint_V (-4y)\,dV = 0$ (odd in $y$). Volume $V = \pi(2)^2(3) = 12\pi$.

$$\iiint_V 4\,dV = 4(12\pi) = 48\pi.$$ $$\iiint_V 2z\,dV = 2\cdot(\pi\cdot4)\int_0^3 z\,dz = 8\pi\cdot\frac{9}{2} = 36\pi.$$

Total:

$$\boxed{\iint_S \mathbf{F}\cdot\hat n\,dS = 48\pi + 36\pi = 84\pi.}$$

(a) Laplace Transform of a Square Wave [4]

For a periodic function of period $p$, $\mathcal{L}\{f\} = \dfrac{1}{1-e^{-ps}}\displaystyle\int_0^p e^{-st}f(t)\,dt$. Here $p = 2a$.

$$\int_0^{2a} e^{-st}f(t)\,dt = k\int_0^a e^{-st}dt - k\int_a^{2a} e^{-st}dt = \frac{k}{s}(1 - e^{-as}) - \frac{k}{s}(e^{-as} - e^{-2as}).$$ $$= \frac{k}{s}\big(1 - 2e^{-as} + e^{-2as}\big) = \frac{k}{s}(1 - e^{-as})^2.$$

Hence

$$\mathcal{L}\{f\} = \frac{k(1-e^{-as})^2}{s(1-e^{-2as})} = \frac{k(1-e^{-as})^2}{s(1-e^{-as})(1+e^{-as})} = \frac{k}{s}\cdot\frac{1-e^{-as}}{1+e^{-as}}.$$ $$\boxed{\mathcal{L}\{f\} = \frac{k}{s}\tanh\frac{as}{2}.}$$

(b) Inverse Transform by Convolution [4]

Write $\dfrac{1}{s^2(s^2+1)} = \dfrac{1}{s^2}\cdot\dfrac{1}{s^2+1}$ with $\mathcal{L}^{-1}\{1/s^2\}=t$ and $\mathcal{L}^{-1}\{1/(s^2+1)\}=\sin t$.

By the convolution theorem $\mathcal{L}^{-1}\{F(s)G(s)\} = \int_0^t f(\tau)g(t-\tau)\,d\tau$:

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

Integrating by parts (or directly): $\int_0^t \tau\sin(t-\tau)d\tau = \big[\tau\cos(t-\tau)\big]_0^t - \int_0^t \cos(t-\tau)d\tau = t - \big[\sin(t-\tau)(-1)\big]...$ evaluating gives

$$= t - \sin t.$$ $$\boxed{\mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+1)}\right\} = t - \sin t.}$$

(Check by partial fractions: $\frac{1}{s^2(s^2+1)} = \frac{1}{s^2} - \frac{1}{s^2+1}$, inverting gives $t - \sin t$.)

Frobenius Series Solution of $2xy'' + y' + y = 0$ [8]

The point $x=0$ is a regular singular point. Assume $y = \sum_{n=0}^\infty a_n x^{n+r}$, $a_0\ne0$. Then

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

Substitute:

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

Combine the first two: $\sum (n+r)\big[2(n+r-1)+1\big]a_n x^{n+r-1} = \sum (n+r)(2n+2r-1)a_n x^{n+r-1}$.

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

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

Recurrence relation. Matching $x^{n+r}$ (i.e. shifting index $n\to n-1$ in the first sum):

$$(n+r)(2n+2r-1)a_n + a_{n-1} = 0 \;\Rightarrow\; a_n = \frac{-a_{n-1}}{(n+r)(2n+2r-1)}.$$

Solution for $r = \tfrac12$. Then $(n+\tfrac12)(2n) = n(2n+1)$, so $a_n = \dfrac{-a_{n-1}}{n(2n+1)}$.

$$a_1 = \frac{-a_0}{1\cdot3} = -\frac{a_0}{3!},\quad a_2 = \frac{-a_1}{2\cdot5} = \frac{a_0}{5!},\quad a_3 = \frac{-a_2}{3\cdot7}=-\frac{a_0}{7!},\dots$$

In general $a_n = \dfrac{(-1)^n a_0}{(2n+1)!}$. With $a_0=1$:

$$y_1 = x^{1/2}\sum_{n=0}^\infty \frac{(-1)^n x^n}{(2n+1)!} = \frac{1}{\sqrt{x}}\sum_{n=0}^\infty \frac{(-1)^n (\sqrt{x})^{2n+1}}{(2n+1)!} = \boxed{\frac{\sin\sqrt{x}}{\sqrt{x}}.}$$

Solution for $r = 0$. Here $a_n = \dfrac{-a_{n-1}}{n(2n-1)}$, giving $a_n = \dfrac{(-1)^n a_0}{(2n)!}$, so

$$y_2 = \sum_{n=0}^\infty \frac{(-1)^n x^n}{(2n)!} = \cos\sqrt{x}.$$

The two linearly independent solutions are $\sin\sqrt x$ and $\cos\sqrt x$ (up to the $1/\sqrt x$ factor noted above); general solution $y = c_1\cos\sqrt x + c_2\sin\sqrt x$.

(a) Bessel Recurrence Relation [4]

Use the series $J_n(x) = \sum_{k=0}^\infty \dfrac{(-1)^k}{k!\,\Gamma(n+k+1)}\left(\dfrac{x}{2}\right)^{2k+n}$.

Then

$$x^n J_n(x) = \sum_{k=0}^\infty \frac{(-1)^k}{k!\,\Gamma(n+k+1)}\frac{x^{2k+2n}}{2^{2k+n}}.$$

Differentiate term by term:

$$\frac{d}{dx}\big[x^n J_n(x)\big] = \sum_{k=0}^\infty \frac{(-1)^k (2k+2n)}{k!\,\Gamma(n+k+1)}\frac{x^{2k+2n-1}}{2^{2k+n}}.$$

Since $2k+2n = 2(n+k)$ and $\Gamma(n+k+1) = (n+k)\Gamma(n+k)$:

$$= \sum_{k=0}^\infty \frac{(-1)^k}{k!\,\Gamma(n+k)}\frac{x^{2k+2n-1}}{2^{2k+n-1}} = x^n\sum_{k=0}^\infty \frac{(-1)^k}{k!\,\Gamma((n-1)+k+1)}\frac{x^{2k+n-1}}{2^{2k+n-1}} = x^n J_{n-1}(x).$$ $$\boxed{\frac{d}{dx}\big[x^n J_n(x)\big] = x^n J_{n-1}(x).} \quad\blacksquare$$

(b) Legendre Polynomials via Rodrigues' Formula [4]

Rodrigues' formula: $P_n(x) = \dfrac{1}{2^n n!}\dfrac{d^n}{dx^n}\big(x^2-1\big)^n$.

$P_2$: $(x^2-1)^2 = x^4 - 2x^2 + 1$. Second derivative: $12x^2 - 4$. So

$$P_2(x) = \frac{1}{2^2\cdot2!}(12x^2-4) = \frac{12x^2-4}{8} = \boxed{\tfrac12(3x^2-1)}.$$

$P_3$: $(x^2-1)^3 = x^6 - 3x^4 + 3x^2 - 1$. Third derivative: $120x^3 - 72x$. So

$$P_3(x) = \frac{1}{2^3\cdot3!}(120x^3 - 72x) = \frac{120x^3-72x}{48} = \boxed{\tfrac12(5x^3-3x)}.$$

Orthogonality check. The product $P_2 P_3 = \tfrac14(3x^2-1)(5x^3-3x)$ is an odd function of $x$. Integrating an odd function over the symmetric interval $[-1,1]$ gives

$$\int_{-1}^1 P_2(x)P_3(x)\,dx = 0,$$

verifying orthogonality. (Directly: $\tfrac14\int_{-1}^1(15x^5 - 9x^3 - 5x^3 + 3x)dx = 0$ since each integrand term is odd.)

(a) Cauchy-Riemann Equations; $f(z)=\bar z$ [3]

For $f(z) = u(x,y) + i\,v(x,y)$ to be analytic, the Cauchy-Riemann (CR) equations must hold and the partials be continuous:

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$$

For $f(z) = \bar z = x - iy$: $u = x$, $v = -y$. Then

$$u_x = 1, \quad v_y = -1 \;\Rightarrow\; u_x \ne v_y.$$

The first CR equation fails at every point, so $f(z)=\bar z$ is nowhere analytic. $\blacksquare$

(b) Harmonic Function and Conjugate [5]

Given $u = x^3 - 3xy^2 + 3x^2 - 3y^2 + 1$.

$$u_x = 3x^2 - 3y^2 + 6x, \quad u_{xx} = 6x + 6; \qquad u_y = -6xy - 6y, \quad u_{yy} = -6x - 6.$$ $$u_{xx} + u_{yy} = (6x+6) + (-6x-6) = 0,$$

so $u$ is harmonic.

Find $v$ via CR equations. $v_y = u_x = 3x^2 - 3y^2 + 6x$. Integrate w.r.t. $y$:

$$v = 3x^2 y - y^3 + 6xy + g(x).$$

Differentiate w.r.t. $x$: $v_x = 6xy + 6y + g'(x)$. CR requires $v_x = -u_y = 6xy + 6y$. Hence $g'(x) = 0 \Rightarrow g(x) = C$.

$$\boxed{v(x,y) = 3x^2 y - y^3 + 6xy + C.}$$

Analytic function. $f(z) = u + iv = z^3 + 3z^2 + 1 + iC$ (one checks $z^3 = (x^3-3xy^2) + i(3x^2y - y^3)$ and $3z^2 = (3x^2-3y^2)+i(6xy)$).

(a) Cauchy's Residue Theorem [2]

If $f(z)$ is analytic inside and on a simple closed contour $C$ except at a finite number of isolated singular points $z_1,z_2,\dots,z_k$ inside $C$, then

$$\oint_C f(z)\,dz = 2\pi i \sum_{j=1}^k \operatorname{Res}_{z=z_j} f(z),$$

where $C$ is traversed in the positive (counter-clockwise) sense.

(b) Evaluation [6]

$$f(z) = \frac{z+1}{z^2 - 2z} = \frac{z+1}{z(z-2)}.$$

Simple poles at $z=0$ and $z=2$, both inside $|z|=3$.

Residue at $z=0$:

$$\operatorname{Res}_{z=0} = \lim_{z\to0} z\cdot\frac{z+1}{z(z-2)} = \frac{0+1}{0-2} = -\frac12.$$

Residue at $z=2$:

$$\operatorname{Res}_{z=2} = \lim_{z\to2}(z-2)\cdot\frac{z+1}{z(z-2)} = \frac{2+1}{2} = \frac32.$$

By the residue theorem,

$$\oint_C f(z)\,dz = 2\pi i\left(-\frac12 + \frac32\right) = 2\pi i \cdot 1 = \boxed{2\pi i}.$$

(a) Form the PDE from $z = f(x^2+y^2+z^2)$ [3]

Let $w = x^2+y^2+z^2$, so $z = f(w)$. Differentiate:

$$p = z_x = f'(w)(2x + 2z\,p) \;\Rightarrow\; p = 2f'(w)(x + zp),$$ $$q = z_y = f'(w)(2y + 2z\,q) \;\Rightarrow\; q = 2f'(w)(y + zq).$$

Dividing to eliminate $f'(w)$:

$$\frac{p}{q} = \frac{x + zp}{y + zq} \;\Rightarrow\; p(y + zq) = q(x + zp) \;\Rightarrow\; py + pzq = qx + qzp.$$

The $pzq$ terms cancel:

$$\boxed{py - qx = 0 \quad\text{i.e.}\quad y\,p - x\,q = 0.}$$

(b) Lagrange's Method [5]

The equation $x(y-z)p + y(z-x)q = z(x-y)$ has auxiliary (Lagrange) equations

$$\frac{dx}{x(y-z)} = \frac{dy}{y(z-x)} = \frac{dz}{z(x-y)}.$$

First multipliers $(1,1,1)$: numerators sum $dx+dy+dz$, denominators sum

$$x(y-z)+y(z-x)+z(x-y) = xy-xz+yz-xy+xz-yz = 0.$$

So $dx+dy+dz = 0 \Rightarrow x+y+z = c_1$.

Second multipliers $(1/x, 1/y, 1/z)$: numerator $\dfrac{dx}{x}+\dfrac{dy}{y}+\dfrac{dz}{z}$, denominators sum

$$(y-z)+(z-x)+(x-y) = 0.$$

So $\dfrac{dx}{x}+\dfrac{dy}{y}+\dfrac{dz}{z} = 0 \Rightarrow \ln(xyz) = \text{const} \Rightarrow xyz = c_2$.

General solution:

$$\boxed{\Phi(x+y+z,\; xyz) = 0,}$$

or equivalently $x+y+z = F(xyz)$ for an arbitrary function $F$.