BT-202 Mathematics - II - Dec 2023 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 $(1 + y^2)dx = (\tan^{-1}y - x)dy$ using Leibnitz linear method.
   How to answer

   • Rearrange equation to standard linear form $\frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{\tan^{-1}y}{1+y^2}$
   • Compute integrating factor: $I.F. = e^{\int \frac{1}{1+y^2}dy} = e^{\tan^{-1}y}$
   • Set up general solution: $x \cdot e^{\tan^{-1}y} = \int \frac{\tan^{-1}y}{1+y^2} e^{\tan^{-1}y} dy + C$
   • Evaluate integral using substitution $t = \tan^{-1}y$, yielding $x e^{\tan^{-1}y} = (\tan^{-1}y - 1)e^{\tan^{-1}y} + C$
2. 7 Marks
   bSolve $(e^y + 1)\cos x dx + e^y \sin x dy = 0$.
   How to answer

   • Separate the variables: $\frac{\cos x}{\sin x} dx + \frac{e^y}{e^y + 1} dy = 0$
   • Integrate both terms: $\int \cot x dx + \int \frac{e^y}{e^y + 1} dy = C_1$
   • Obtain $\ln|\sin x| + \ln(e^y + 1) = \ln C$
   • Simplify to $(e^y + 1)\sin x = C$
3. 7 Marks
   aSolve $(D^2 - 4D + 3)y = \cos 2x$.
   How to answer

   • Write auxiliary equation: $m^2 - 4m + 3 = 0 \implies m = 1, 3$
   • Complementary function: $y_c = c_1 e^x + c_2 e^{3x}$
   • Find particular integral: $y_p = \frac{1}{D^2 - 4D + 3}\cos 2x = \frac{1}{-4 - 4D + 3}\cos 2x = \frac{1}{-(4D+1)}\cos 2x$
   • Multiply numerator and denominator by $(4D-1)$ and replace $D^2$ with $-4$: $y_p = -\frac{4D-1}{16D^2-1}\cos 2x = \frac{-8\sin 2x - \cos 2x}{65}$
   • Write complete solution: $y = y_c + y_p$
4. 7 Marks
   bShow that $J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}} \sin x$.
   How to answer

   • State definition: $J_n(x) = \sum_{r=0}^{\infty} \frac{(-1)^r}{r! \Gamma(n+r+1)} \left(\frac{x}{2}\right)^{n+2r}$
   • Substitute $n = 1/2$: $J_{1/2}(x) = \sqrt{\frac{x}{2}} \sum_{r=0}^{\infty} \frac{(-1)^r}{r! \Gamma(r + 3/2)} \left(\frac{x}{2}\right)^{2r}$
   • Expand terms for $r = 0, 1, 2, \dots$ using $\Gamma(1/2) = \sqrt{\pi}$, $\Gamma(3/2) = \frac{1}{2}\sqrt{\pi}$, etc.
   • Factor out $\sqrt{\frac{2}{\pi x}}$ to identify the Taylor series of $\sin x$
   • Conclude $J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \sin x$
5. 14 Marks
   Solve $(D^2 + 9)y = \tan 3x$ by using method of variation of parameters.
   How to answer

   • Find complementary function: $m^2 + 9 = 0 \implies m = \pm 3i$, so $y_c = c_1 \cos 3x + c_2 \sin 3x$
   • Identify $y_1 = \cos 3x$, $y_2 = \sin 3x$, and $R(x) = \tan 3x$
   • Calculate Wronskian: $W = \cos 3x (3\cos 3x) - (-3\sin 3x)(\sin 3x) = 3$
   • Compute $u = -\int \frac{\sin 3x \tan 3x}{3} dx$ and $v = \int \frac{\cos 3x \tan 3x}{3} dx$
   • Integrate to obtain $u$ and $v$, then write total solution $y = y_c + u y_1 + v y_2$
6. 7 Marks
   aSolve the partial differential equation $(x - y)p + (x + y)q = 2xz$.
   How to answer

   • Form Lagrange's auxiliary equations: $\frac{dx}{x-y} = \frac{dy}{x+y} = \frac{dz}{2xz}$
   • Choose multipliers or combine terms to find first independent integral $u(x, y, z) = c_1$
   • Find second independent integral $v(x, y, z) = c_2$
   • State final solution in the form $f(u, v) = 0$ or $u = \phi(v)$
7. 7 Marks
   bSolve $(p^2 + q^2)y = qz$ by using Charpit's method.
   How to answer

   • Write $f(x,y,z,p,q) = (p^2+q^2)y - qz = 0$
   • Compute partial derivatives: $f_x=0$, $f_y=p^2+q^2$, $f_z=-q$, $f_p=2py$, $f_q=2qy-z$
   • Set up Charpit's auxiliary equations: $\frac{dp}{-pq} = \frac{dq}{(p^2+q^2) - q^2} = \frac{dq}{p^2}$
   • Obtain relation between $p$ and $q$, e.g., $p dp + q dq = 0 \implies p^2 + q^2 = a^2$
   • Substitute into given PDE to express $p$ and $q$ in terms of coordinates, and integrate $dz = p dx + q dy$ to obtain complete integral
8. 7 Marks
   aSolve $(D^2 + 4DD' - 5D'^2)Z = \sin(2x + 3y)$.
   How to answer

   • Form auxiliary equation: $m^2 + 4m - 5 = 0 \implies (m+5)(m-1) = 0 \implies m = 1, -5$
   • Write complementary function: $Z_c = \phi_1(y + x) + \phi_2(y - 5x)$
   • Find particular integral: replace $D^2 \to -2^2 = -4$, $DD' \to -(2)(3) = -6$, $D'^2 \to -3^2 = -9$
   • Denominator value: $\phi(-4, -6, -9) = -4 + 4(-6) - 5(-9) = -4 - 24 + 45 = 17$
   • Obtain $Z_p = \frac{1}{17} \sin(2x + 3y)$
   • Combine to give $Z = Z_c + Z_p$
9. 7 Marks
   bDetermine $p$ so that the function $f(z) = \frac{1}{2}\log(x^2 + y^2) + i\tan^{-1}\left(\frac{px}{y}\right)$ is analytic function.
   How to answer

   • Identify $u = \frac{1}{2}\ln(x^2 + y^2)$ and $v = \tan^{-1}\left(\frac{px}{y}\right)$
   • Compute partial derivatives of $u$: $\frac{\partial u}{\partial x} = \frac{x}{x^2+y^2}$, $\frac{\partial u}{\partial y} = \frac{y}{x^2+y^2}$
   • Compute partial derivatives of $v$: $\frac{\partial v}{\partial x} = \frac{py}{y^2 + p^2 x^2}$, $\frac{\partial v}{\partial y} = \frac{-px}{y^2 + p^2 x^2}$
   • Apply Cauchy-Riemann equation $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$: $\frac{x}{x^2+y^2} = \frac{-px}{y^2 + p^2 x^2}$
   • Solve for $p$, finding $p = -1$ (or using $\tan^{-1}(y/x)$ relationship to yield consistency)
10. 7 Marks
    aShow that the function $u(x, y) = e^x \cos y$ is Harmonic. Determine it's Harmonic conjugate.
    How to answer

    • Find second partial derivatives: $u_{xx} = e^x \cos y$, $u_{yy} = -e^x \cos y$
    • Verify Laplace's equation: $u_{xx} + u_{yy} = 0$, hence $u$ is harmonic
    • Use CR equations $dv = \frac{\partial v}{\partial x}dx + \frac{\partial v}{\partial y}dy = -\frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy$
    • Substitute: $dv = e^x \sin y dx + e^x \cos y dy$
    • Integrate to find harmonic conjugate: $v(x, y) = e^x \sin y + C$
11. 7 Marks
    bFind the residue of $\frac{Z e^z}{(Z-1)^3}$ at it's pole.
    How to answer

    • Identify pole at $z = 1$ of order $m = 3$
    • Apply formula: $\text{Res}(f, 1) = \frac{1}{2!} \lim_{z \to 1} \frac{d^2}{dz^2} [z e^z]$
    • First derivative: $\frac{d}{dz}[z e^z] = e^z + z e^z = (1+z)e^z$
    • Second derivative: $\frac{d^2}{dz^2}[z e^z] = e^z + (1+z)e^z = (2+z)e^z$
    • Evaluate limit at $z = 1$: $\text{Res} = \frac{1}{2}(2+1)e^1 = \frac{3e}{2}$
12. 14 Marks
    Verify Gauss divergence theorem for $\overline{F} = x^3 \overline{i} + y^3 \overline{j} + z^3 \overline{k}$ taken over the cube bounded by $x = 0$, $x = a$, $y = 0$, $y = a$, $z = 0$, $z = a$.
    How to answer

    • Compute divergence: $\nabla \cdot \overline{F} = 3(x^2 + y^2 + z^2)$
    • Evaluate volume integral: $\iiint_V 3(x^2 + y^2 + z^2) dx dy dz = 3 \cdot 3 \left(\frac{a^3}{3} \cdot a \cdot a\right) = 3a^5$
    • Evaluate surface integrals over the 6 faces of the cube
    • Sum integrals for $x=0,a$, $y=0,a$, $z=0,a$ to show total surface flux equals $a^5 + a^5 + a^5 = 3a^5$
    • Conclude that LHS = RHS, verifying the theorem
13. 7 Marks
    aProve that $\text{curl}(r^n \overline{r}) = \overline{0}$
    How to answer

    • Use vector identity: $\text{curl}(\phi \overline{r}) = (\nabla \phi) \times \overline{r} + \phi (\text{curl } \overline{r})$
    • Set $\phi = r^n$: $\nabla(r^n) = n r^{n-1} \nabla r = n r^{n-2} \overline{r}$
    • Show $\text{curl } \overline{r} = \overline{0}$
    • Calculate $(n r^{n-2} \overline{r}) \times \overline{r} = \overline{0}$ since cross product of parallel vectors is zero
    • Conclude $\text{curl}(r^n \overline{r}) = \overline{0}$
14. 7 Marks
    bWrite short note on: i) Cauchy Riemann equations ii) Stokes theorem
    How to answer

    • Cauchy-Riemann equations: statement of conditions $u_x = v_y$ and $u_y = -v_x$ in Cartesian and polar forms, necessary condition for analyticity
    • Stokes' theorem: statement $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\text{curl } \vec{F}) \cdot \hat{n} dS$, physical significance relating circulation and flux of curl