BT-202 Mathematics - II - Nov 2022 Question Paper

Under each question, How to answer lists the points an answer should cover, outlined by CampusPrep from the syllabus. Check the details against your textbook.

1. 7 Marks
   aSolve $\cos x\,dy = y(\sin x - y)\,dx$ using Bernoulli's.
   How to answer

   • Rearrange into standard Bernoulli form: $\frac{dy}{dx} - (\tan x)y = -(\sec x)y^2$
   • Divide by $y^2$ and substitute $v = y^{-1}$ so that $-\frac{dv}{dx} = y^{-2}\frac{dy}{dx}$
   • Transform into linear ODE in $v$: $\frac{dv}{dx} + (\tan x)v = \sec x$
   • Find integrating factor: $I.F. = e^{\int \tan x\,dx} = \sec x$
   • Integrate to obtain $v \cdot \sec x = \int \sec^2 x\,dx + C = \tan x + C$
   • Substitute back $v = y^{-1}$ to write final solution: $\frac{\sec x}{y} = \tan x + C$
2. 7 Marks
   bSolve the Linear differential equation $$\sin 2x\frac{dy}{dx} - y = \tan x$$
   How to answer

   • Convert to standard linear form: $\frac{dy}{dx} - (\csc 2x)y = \frac{\tan x}{\sin 2x} = \frac{1}{2}\sec^2 x$
   • Find integrating factor: $I.F. = e^{-\int \csc 2x\,dx} = e^{-\frac{1}{2}\ln|\tan x|} = \frac{1}{\sqrt{\tan x}}$
   • Multiply and integrate: $y \cdot \frac{1}{\sqrt{\tan x}} = \int \frac{1}{2}\sec^2 x (\tan x)^{-1/2}\,dx + C$
   • Evaluate right-hand integral: $\int (\tan x)^{-1/2} d(\tan x) = 2\sqrt{\tan x}$
   • Write final general solution: $y = \tan x + C\sqrt{\tan x}$
3. 7 Marks
   aSolve $(r + \sin\theta - \cos\theta)dr + r(\sin\theta + \cos\theta)d\theta = 0$.
   How to answer

   • Identify $M = r + \sin\theta - \cos\theta$ and $N = r(\sin\theta + \cos\theta)$
   • Check condition of exactness or rearrange into linear ODE: $\frac{d\theta}{dr} + \frac{\sin\theta + \cos\theta}{r(\sin\theta + \cos\theta)}\dots$ or test $\frac{\partial M}{\partial \theta} = \cos\theta + \sin\theta$ and $\frac{\partial N}{\partial r} = \sin\theta + \cos\theta$
   • Since $\frac{\partial M}{\partial \theta} = \frac{\partial N}{\partial r}$, the differential equation is exact
   • Integrate $M$ with respect to $r$ treating $\theta$ as constant: $\int (r + \sin\theta - \cos\theta)dr = \frac{r^2}{2} + r(\sin\theta - \cos\theta)$
   • Terms in $N$ not containing $r$: none (0)
   • Write final general solution: $\frac{r^2}{2} + r(\sin\theta - \cos\theta) = C$
4. 7 Marks
   bSolve the differential equation. $$(D^3 - 7D^2 + 14D - 8)y = e^x \cos 2x$$
   How to answer

   • Auxiliary equation: $m^3 - 7m^2 + 14m - 8 = 0 \implies (m-1)(m-2)(m-4) = 0$
   • Find Complementary Function: $y_c = C_1 e^x + C_2 e^{2x} + C_3 e^{4x}$
   • Apply exponential shift for Particular Integral: $y_p = \frac{1}{D^3 - 7D^2 + 14D - 8} (e^x \cos 2x) = e^x \frac{1}{(D+1)^3 - 7(D+1)^2 + 14(D+1) - 8} \cos 2x$
   • Simplify the polynomial in $D$ and replace $D^2 \to -2^2 = -4$
   • Rationalize the operator with respect to $D$ to obtain $y_p$
   • Total solution: $y = y_c + y_p$
5. 14 Marks
   Solve $(D^2 + 4)y = \tan 2x$ by using method of variation of parameters.
   How to answer

   • Find $y_c$: Auxiliary equation $m^2 + 4 = 0 \implies m = \pm 2i$, so $y_1 = \cos 2x, y_2 = \sin 2x$
   • Compute Wronskian: $W = y_1 y_2' - y_1' y_2 = \cos 2x(2\cos 2x) - (-2\sin 2x)(\sin 2x) = 2$
   • Compute $u(x) = -\int \frac{y_2 X}{W}\,dx = -\int \frac{\sin 2x \tan 2x}{2}\,dx = -\frac{1}{2}\int (\sec 2x - \cos 2x)\,dx$
   • Compute $v(x) = \int \frac{y_1 X}{W}\,dx = \int \frac{\cos 2x \tan 2x}{2}\,dx = \frac{1}{2}\int \sin 2x\,dx = -\frac{1}{4}\cos 2x$
   • Write $y_p = u(x)y_1 + v(x)y_2$ and combine with $y_c$ for complete solution $y = y_c + y_p$
6. 7 Marks
   aShow that $\frac{\vec{r}}{r^3}$ is solenoidal.
   How to answer

   • Define solenoidal field: $\nabla \cdot \vec{F} = 0$
   • Use identity: $\nabla \cdot (\phi \vec{A}) = (\nabla \phi) \cdot \vec{A} + \phi (\nabla \cdot \vec{A})$ with $\phi = r^{-3}$ and $\vec{A} = \vec{r}$
   • Calculate $\nabla \cdot \vec{r} = 3$
   • Calculate $\nabla(r^{-3}) = -3r^{-4} \nabla r = -3r^{-4}\left(\frac{\vec{r}}{r}\right) = -3\frac{\vec{r}}{r^5}$
   • Evaluate divergence: $\nabla \cdot (r^{-3}\vec{r}) = \left(-3\frac{\vec{r}}{r^5}\right) \cdot \vec{r} + r^{-3}(3) = -3\frac{r^2}{r^5} + 3r^{-3} = 0$
   • Conclude that $\frac{\vec{r}}{r^3}$ is solenoidal
7. 7 Marks
   bShow that the vector $$(x^2 - yz)\hat{i} + (y^2 - zx)\hat{j} + (z^2 - xy)\hat{k}$$ is Irrotational. Find it's scalar potential.
   How to answer

   • Compute $\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2-yz & y^2-zx & z^2-xy \end{vmatrix}$
   • Show each component vanishes: $(-x - (-x))\hat{i} - (-y - (-y))\hat{j} + (-z - (-z))\hat{k} = \vec{0}$, proving $\vec{F}$ is irrotational
   • Set $\vec{F} = \nabla \phi \implies \frac{\partial \phi}{\partial x} = x^2 - yz$, $\frac{\partial \phi}{\partial y} = y^2 - zx$, $\frac{\partial \phi}{\partial z} = z^2 - xy$
   • Integrate to find the scalar potential: $\phi(x,y,z) = \frac{x^3}{3} + \frac{y^3}{3} + \frac{z^3}{3} - xyz + C$
8. 14 Marks
   Verify Green's theorem for $\oint_C [(3x^2 - 8y^2)\,dx + (4y - 6xy)\,dy]$, Where $C$ is the region bounded by $x = 0, y = 0$ and $x + y = 1$.
   How to answer

   • State Green's theorem: $\oint_C (M\,dx + N\,dy) = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dx\,dy$
   • Identify $M = 3x^2 - 8y^2$ and $N = 4y - 6xy$
   • Compute $\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -6y - (-16y) = 10y$
   • Evaluate double integral: $\int_{x=0}^1 \int_{y=0}^{1-x} 10y\,dy\,dx = \int_0^1 5(1-x)^2\,dx = \frac{5}{3}$
   • Evaluate line integral along three segments: $C_1 (y=0, x:0\to 1)$, $C_2 (x+y=1, x:1\to 0)$, $C_3 (x=0, y:1\to 0)$
   • Sum line integrals to obtain $\frac{5}{3}$, confirming both sides are equal
9. 7 Marks
   aShow that $f(Z) = z\bar{z}$ is differentiable but not analytic at origin.
   How to answer

   • Write $f(z) = z\bar{z} = x^2 + y^2$, so $u(x,y) = x^2 + y^2$ and $v(x,y) = 0$
   • Compute derivative at origin using definition: $f'(0) = \lim_{z\to 0}\frac{z\bar{z} - 0}{z} = \lim_{z\to 0}\bar{z} = 0$, so $f(z)$ is differentiable at $z=0$
   • Check CR equations: $u_x = 2x, u_y = 2y, v_x = 0, v_y = 0$
   • CR equations $u_x = v_y$ and $u_y = -v_x$ hold only at $(0,0)$ and fail in any neighborhood around the origin
   • Conclude that since differentiability does not hold in any open neighborhood of the origin, $f(z)$ is not analytic at the origin
10. 7 Marks
    bShow that $u(x,y) = e^{-2x}\sin 2y$ is harmonic and determine it's Harmonic conjugate.
    How to answer

    • Find first partial derivatives: $u_x = -2e^{-2x}\sin 2y$, $u_y = 2e^{-2x}\cos 2y$
    • Find second partial derivatives: $u_{xx} = 4e^{-2x}\sin 2y$, $u_{yy} = -4e^{-2x}\sin 2y$
    • Verify Laplace equation: $u_{xx} + u_{yy} = 0$, hence $u$ is harmonic
    • Use CR equations $v_x = -u_y = -2e^{-2x}\cos 2y$ and $v_y = u_x = -2e^{-2x}\sin 2y$
    • Integrate to find $v(x,y) = e^{-2x}\cos 2y + C$, which is the harmonic conjugate
11. 7 Marks
    aBy Residue theorem, Evaluate $\oint_C \frac{\tan z}{z^2 - 1}\,dz$, where $C:|Z|=2$.
    How to answer

    • Identify poles of $f(z) = \frac{\sin z}{\cos z (z-1)(z+1)}$
    • Poles of denominator $z^2-1=0$ are $z = 1, -1$; poles of $\tan z$ occur at $z = \pm \frac{\pi}{2} \approx \pm 1.571$
    • Determine which poles lie inside $C: |z| = 2$: $z = 1, -1, \frac{\pi}{2}, -\frac{\pi}{2}$
    • Calculate residues at $z = 1$: $\text{Res}(1) = \frac{\tan 1}{2}$ and at $z = -1$: $\text{Res}(-1) = \frac{\tan(-1)}{-2} = \frac{\tan 1}{2}$
    • Calculate residues at $z = \pm \frac{\pi}{2}$: $\text{Res}(\pm\pi/2) = \frac{-\sin(\pm\pi/2)}{\sin(\pm\pi/2)((\pm\pi/2)^2 - 1)} = -\frac{1}{\pi^2/4 - 1}$
    • Sum all enclosed residues and multiply by $2\pi i$ to get the final result
12. 7 Marks
    bUsing Cauchy integral theorem, to evaluate the integral $$\int_C \frac{e^{2z}}{(z-1)^2(z-3)}\,dz$$, where $C$ is the circle $|Z|=2$.
    How to answer

    • Identify singularities of integrand: $z = 1$ (pole of order 2) and $z = 3$ (simple pole)
    • Check location with respect to circle $|z| = 2$: $z = 1$ is inside, $z = 3$ is outside
    • Rewrite integral in Cauchy Integral Formula form: $\oint_C \frac{g(z)}{(z-1)^2}\,dz$ where $g(z) = \frac{e^{2z}}{z-3}$, which is analytic inside and on $C$
    • By Cauchy's integral formula for derivatives: $\oint_C \frac{g(z)}{(z-1)^2}\,dz = 2\pi i \cdot g'(1)$
    • Compute $g'(z) = \frac{2e^{2z}(z-3) - e^{2z}}{(z-3)^2} = \frac{e^{2z}(2z-7)}{(z-3)^2}$
    • Evaluate at $z = 1$: $g'(1) = \frac{e^2(2 - 7)}{(1 - 3)^2} = -\frac{5e^2}{4}$
    • Final value: $2\pi i \left(-\frac{5e^2}{4}\right) = -\frac{5\pi i e^2}{2}$
13. 7 Marks
    aSolve $x^2 p^2 + y^2 q^2 = z^2$.
    How to answer

    • Rewrite the PDE as $\left(\frac{x}{z}p\right)^2 + \left(\frac{y}{z}q\right)^2 = 1$
    • Let $X = \ln x, Y = \ln y, Z = \ln z$, so $\frac{\partial Z}{\partial X} = \frac{x}{z}p = P$ and $\frac{\partial Z}{\partial Y} = \frac{y}{z}q = Q$
    • Equation reduces to standard form: $P^2 + Q^2 = 1$
    • Complete solution for constant derivatives: $Z = aX + bY + c$ where $a^2 + b^2 = 1$, so $b = \sqrt{1 - a^2}$
    • Substitute back $X, Y, Z$: $\ln z = a\ln x + \sqrt{1 - a^2}\ln y + c$
    • Express in exponentiated form: $z = c' x^a y^{\sqrt{1-a^2}}$
14. 7 Marks
    bSolve $(D^2 - 4DD' + 4{D'}^2)Z = \cos(x - 2y)$
    How to answer

    • Factor operator: $(D - 2D')^2 Z = \cos(x - 2y)$
    • Find Complementary Function: roots of $m - 2 = 0$ are repeated ($m = 2, 2$), so $Z_c = \phi_1(y + 2x) + x\phi_2(y + 2x)$
    • Find Particular Integral: $Z_p = \frac{1}{(D - 2D')^2} \cos(x - 2y)$
    • Substitute $a = 1, b = -2$; note that $D - 2D' \to 1 - 2(-2) = 5 \neq 0$, so $(D - 2D')^2 \to 5^2 = 25$
    • Compute $Z_p = \frac{\cos(x - 2y)}{25}$
    • Complete solution: $Z = \phi_1(y + 2x) + x\phi_2(y + 2x) + \frac{1}{25}\cos(x - 2y)$