BT-202 Mathematics - II - Jun 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 $\left(1+e^{\frac{x}{y}}\right)dx + e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)dy = 0$.
   How to answer

   • Identify form $M dx + N dy = 0$ or substitute $x = vy$
   • Check for exactness: compute $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$
   • Show that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, hence equation is exact
   • Integrate $M$ with respect to $x$ (treating $y$ as constant)
   • Integrate terms in $N$ not containing $x$ with respect to $y$
   • Obtain general solution $x + y e^{x/y} = C$
2. 7 Marks
   bSolve the Linear differential equation $$\left(1+y^{2}\right) + \left(x - e^{-\tan^{-1}y}\right)\frac{dy}{dx} = 0$$
   How to answer

   • Rewrite into standard linear form in $x$: $\frac{dx}{dy} + \frac{1}{1+y^2} x = \frac{e^{-\tan^{-1}y}}{1+y^2}$
   • Identify $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{e^{-\tan^{-1}y}}{1+y^2}$
   • Compute integrating factor $I.F. = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1}y}$
   • Multiply and integrate: $x e^{\tan^{-1}y} = \int \frac{1}{1+y^2} dy + C$
   • Final solution: $x e^{\tan^{-1}y} = \tan^{-1}y + C$
3. 7 Marks
   aSolve $\frac{dy}{dx} + y \tan x = y^{2}\sec x$ using Bernoulli's.
   How to answer

   • Divide by $y^2$: $y^{-2} \frac{dy}{dx} + y^{-1} \tan x = \sec x$
   • Substitute $v = y^{-1}$, so $\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$
   • Obtain linear equation: $\frac{dv}{dx} - v\tan x = -\sec x$
   • Calculate $I.F. = e^{-\int \tan x dx} = \cos x$
   • Solve $v \cos x = \int -\sec x \cos x dx = -x + C$
   • Substitute back $v = \frac{1}{y}$ to get $\frac{1}{y}\cos x = C - x$
4. 7 Marks
   bSolve the differential equation $$(D^{2} - 2D - 3)y = x^{3}e^{-3x}.$$
   How to answer

   • Write auxiliary equation: $m^2 - 2m - 3 = 0 \implies (m-3)(m+1) = 0$
   • Determine Complementary Function: $y_c = C_1 e^{3x} + C_2 e^{-x}$
   • Set up Particular Integral: $y_p = \frac{1}{D^2 - 2D - 3} (e^{-3x} x^3)$
   • Apply shift rule: $y_p = e^{-3x} \frac{1}{(D-3)^2 - 2(D-3) - 3} x^3 = e^{-3x} \frac{1}{D^2 - 8D + 12} x^3$
   • Expand $\frac{1}{12}\left(1 - \frac{8D - D^2}{12}\right)^{-1}$ in powers of $D$ and apply to $x^3$
   • Sum $y = y_c + y_p$
5. 14 Marks
   Solve $(D^{2} + a^{2})y = \tan ax$ by using method of variation of parameters.
   How to answer

   • Find complementary function: $y_c = C_1 \cos ax + C_2 \sin ax$, so $y_1 = \cos ax, y_2 = \sin ax$
   • Calculate Wronskian $W = \cos ax (a \cos ax) - (-a \sin ax)\sin ax = a$
   • Use variation of parameters formulas: $u(x) = -\int \frac{y_2 X}{W} dx = -\int \frac{\sin ax \tan ax}{a} dx$
   • Integrate $u(x)$ and $v(x) = \int \frac{\cos ax \tan ax}{a} dx = \int \frac{\sin ax}{a} dx$
   • Write particular integral $y_p = u y_1 + v y_2$
   • State general solution $y = y_c + y_p$
6. 7 Marks
   aFind the angle between the surfaces $x^{2} + y^{2} + z^{2} = 9$ and $Z = x^{2} + y^{2} - 3$ at the point $(2, -1, 2)$.
   How to answer

   • Find gradient $\nabla \phi_1 = 2x\hat{i} + 2y\hat{j} + 2z\hat{k}$ at $(2, -1, 2) \implies 4\hat{i} - 2\hat{j} + 4\hat{k}$
   • Find gradient $\nabla \phi_2 = 2x\hat{i} + 2y\hat{j} - \hat{k}$ at $(2, -1, 2) \implies 4\hat{i} - 2\hat{j} - \hat{k}$
   • Calculate magnitude $|\nabla \phi_1| = \sqrt{16+4+16} = 6$ and $|\nabla \phi_2| = \sqrt{16+4+1} = \sqrt{21}$
   • Compute dot product $\nabla \phi_1 \cdot \nabla \phi_2 = (4)(4) + (-2)(-2) + (4)(-1) = 16 + 4 - 4 = 16$
   • Determine $\cos \theta = \frac{16}{6\sqrt{21}} = \frac{8}{3\sqrt{21}}$, $\theta = \cos^{-1}\left(\frac{8}{3\sqrt{21}}\right)$
7. 7 Marks
   bProve that $r^{n}\overline{r}$ is Solenoidal if $n = -3$.
   How to answer

   • Define condition for solenoidal vector: $\nabla \cdot \vec{F} = 0$
   • Apply divergence formula: $\nabla \cdot (r^n \vec{r}) = r^n (\nabla \cdot \vec{r}) + \vec{r} \cdot \nabla(r^n)$
   • Recall $\nabla \cdot \vec{r} = 3$ and $\nabla(r^n) = n r^{n-1} \frac{\vec{r}}{r} = n r^{n-2}\vec{r}$
   • Simplify: $\nabla \cdot (r^n \vec{r}) = 3 r^n + n r^{n-2}(\vec{r} \cdot \vec{r}) = 3 r^n + n r^n = (n+3)r^n$
   • Set $(n+3)r^n = 0 \implies n = -3$
8. 14 Marks
   Verify Gauss divergence theorem for $\overline{F} = x^{2}\overline{i} + y^{2}\overline{j} + z^{2}\overline{k}$ over the cube formed by the planes $x = 0, x = a, y = 0, y = b, z = 0, z = c$.
   How to answer

   • State Gauss Divergence Theorem: $\iint_S \vec{F} \cdot \hat{n}\, dS = \iiint_V (\nabla \cdot \vec{F})\, dV$
   • Evaluate volume integral: $\nabla \cdot \vec{F} = 2(x + y + z)$
   • Calculate $\int_0^a \int_0^b \int_0^c 2(x+y+z) dz dy dx = abc(a+b+c)$
   • Evaluate surface integral over all six faces ($x=0, x=a, y=0, y=b, z=0, z=c$)
   • Sum the six surface flux contributions to obtain $a^2bc + ab^2c + abc^2 = abc(a+b+c)$
   • Conclude that L.H.S. = R.H.S., verifying the theorem
9. 7 Marks
   aProve that $\left[\frac{d^{2}}{dx^{2}} + \frac{d^{2}}{dy^{2}}\right] |f(z)|^{2} = 4|f'(z)|^{2}$.
   How to answer

   • Let $f(z) = u + iv$, so $|f(z)|^2 = u^2 + v^2$
   • Differentiate with respect to $x$: $\frac{\partial}{\partial x}(u^2+v^2) = 2uu_x + 2vv_x$
   • Compute $\frac{\partial^2}{\partial x^2}(u^2+v^2) = 2(u_x^2 + v_x^2) + 2(u u_{xx} + v v_{xx})$
   • Similarly compute $\frac{\partial^2}{\partial y^2}(u^2+v^2) = 2(u_y^2 + v_y^2) + 2(u u_{yy} + v v_{yy})$
   • Add both equations and use harmonic properties ($u_{xx}+u_{yy}=0, v_{xx}+v_{yy}=0$) and Cauchy-Riemann equations
   • Arrive at $4(u_x^2 + v_x^2) = 4|f'(z)|^2$
10. 7 Marks
    bConstruct the analytic function $f(z)$, whose real part is $e^{x}\cos y$.
    How to answer

    • Given $u = e^x \cos y$
    • Find $\phi_1(x, y) = \frac{\partial u}{\partial x} = e^x \cos y$
    • Find $\phi_2(x, y) = \frac{\partial u}{\partial y} = -e^x \sin y$
    • Apply Milne-Thomson method: set $x = z, y = 0$
    • $\phi_1(z, 0) = e^z$, $\phi_2(z, 0) = 0$
    • Integrate $f'(z) = \phi_1(z, 0) - i\phi_2(z, 0) = e^z \implies f(z) = e^z + C$
11. 7 Marks
    aUsing Cauchy's integral formula, find $\int_{C} \frac{e^{2z}}{(z+1)^{3}} dz$ where $C$ is the curve $|z| = 2$.
    How to answer

    • Identify singularity: $z = -1$, which lies inside the circle $|z| = 2$
    • Let $f(z) = e^{2z}$, with pole of order $n+1 = 3 \implies n = 2$
    • Use Cauchy's Integral Formula for derivatives: $\oint_C \frac{f(z)}{(z-z_0)^3} dz = \frac{2\pi i}{2!} f''(z_0)$
    • Differentiate: $f'(z) = 2e^{2z}$, $f''(z) = 4e^{2z}$
    • Evaluate at $z_0 = -1$: $f''(-1) = 4e^{-2}$
    • Compute integral: $\pi i \cdot 4e^{-2} = 4\pi i e^{-2}$
12. 7 Marks
    bEvaluate $\int_{C} \frac{1}{(z+4)z^{8}} dz$ where $C$ is the circle $|z| = 2$.
    How to answer

    • Identify poles of $f(z) = \frac{1}{(z+4)z^8}$: $z = 0$ (order 8) and $z = -4$ (simple pole)
    • Note that $z = 0$ is inside $|z| = 2$, while $z = -4$ lies outside $|z| = 2$
    • Expand $\frac{1}{z+4}$ around $z = 0$: $\frac{1}{4}\sum_{k=0}^{\infty} (-1)^k \left(\frac{z}{4}\right)^k$
    • Find residue at $z = 0$ as coefficient of $z^7$ in the expansion of $\frac{1}{z+4}$, which is $\frac{1}{4} \cdot \frac{(-1)^7}{4^7} = -\frac{1}{4^8}$
    • By Cauchy's residue theorem: $\oint_C f(z) dz = 2\pi i \left(-\frac{1}{4^8}\right) = -\frac{2\pi i}{4^8} = -\frac{\pi i}{2^{15}}$
13. 7 Marks
    aForm the partial differential equation by eliminating arbitrary function from $Z = f(y/x)$.
    How to answer

    • Given $Z = f(u)$ where $u = \frac{y}{x}$
    • Differentiate partially with respect to $x$: $p = \frac{\partial Z}{\partial x} = f'(y/x) \left(-\frac{y}{x^2}\right)$
    • Differentiate partially with respect to $y$: $q = \frac{\partial Z}{\partial y} = f'(y/x) \left(\frac{1}{x}\right)$
    • Eliminate $f'(y/x)$ by taking ratio or substituting: $p = -\frac{y}{x} q$
    • Obtain PDE: $x p + y q = 0$
14. 7 Marks
    bSolve the $(D^{3} - 3D^{2}D' + 4{D'}^{3})z = e^{x+2y}$.
    How to answer

    • Form auxiliary equation: put $D = m, D' = 1 \implies m^3 - 3m^2 + 4 = 0$
    • Factorize: $(m+1)(m-2)^2 = 0 \implies m = -1, 2, 2$
    • Write Complementary Function: $C.F. = \phi_1(y - x) + \phi_2(y + 2x) + x\phi_3(y + 2x)$
    • Find Particular Integral: $P.I. = \frac{1}{D^3 - 3D^2 D' + 4{D'}^3} e^{x+2y}$
    • Substitute $D = 1, D' = 2$: $F(1, 2) = 1^3 - 3(1)^2(2) + 4(2)^3 = 1 - 6 + 32 = 27 \neq 0$
    • So $P.I. = \frac{1}{27} e^{x+2y}$
    • General solution: $z = C.F. + P.I.$