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BT-202 · Mathematics II/Unsolved PYQ Paper

BT-202 Mathematics - II - Jun 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: $\frac{dy}{dx} = \cos(x + y) + \sin(x + y)$.
    How to answer
    • Substitute $u = x + y$, so $\frac{du}{dx} = 1 + \frac{dy}{dx}$
    • Rewrite as $\frac{du}{dx} = 1 + \cos u + \sin u$
    • Separate variables: $\frac{du}{1 + \cos u + \sin u} = dx$
    • Integrate using half-angle substitution $t = \tan(u/2)$
    • Substitute back $u = x + y$ to obtain the final relation
  2. 7 Marks
    bSolve: $(1 + y^2)dx = (\tan^{-1} y - x)dy$.
    How to answer
    • Rewrite equation in standard linear form: $\frac{dx}{dy} + \frac{1}{1 + y^2}x = \frac{\tan^{-1}y}{1 + y^2}$
    • Identify $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{\tan^{-1}y}{1+y^2}$
    • Find integrating factor: $I.F. = e^{\int \frac{1}{1+y^2}dy} = e^{\tan^{-1}y}$
    • Integrate $x \cdot (I.F.) = \int Q(y)(I.F.)dy + C$
    • Evaluate $\int \frac{\tan^{-1}y}{1+y^2} e^{\tan^{-1}y} dy$ using substitution $t = \tan^{-1}y$
  3. 7 Marks
    aSolve: $\frac{d^2y}{dx^2} + \frac{dy}{dx} = (1 + e^x)^{-1}$.
    How to answer
    • Find complementary function $y_c$ from $(D^2+D)y = 0$: $y_c = c_1 + c_2 e^{-x}$
    • Apply method of variation of parameters or substitution to find particular integral $y_p$
    • Evaluate $y_p = \frac{1}{D(D+1)}\frac{1}{1+e^x}$
    • Write complete solution $y = y_c + y_p$
  4. 7 Marks
    bSolve: $\frac{dx}{dt} - y = e^t$, $\frac{dy}{dt} + x = \sin t$; $x(0) = 1$, $y(0) = 0$.
    How to answer
    • Express the system in operator form: $Dx - y = e^t$ and $x + Dy = \sin t$
    • Eliminate $y$ to get a second-order equation in $x$: $(D^2 + 1)x = e^t + \cos t$
    • Find complementary function $x_c = c_1 \cos t + c_2 \sin t$ and particular integral $x_p$
    • Determine $y(t)$ using $y = \frac{dx}{dt} - e^t$
    • Apply initial conditions $x(0) = 1$ and $y(0) = 0$ to find arbitrary constants
  5. 14 Marks
    Solve the differential equation $$x(1 - x)y'' + 2(1 - 2x)y' - 2y = 0$$ using Frobenius method.
    How to answer
    • Identify regular singular point at $x = 0$
    • Assume series solution $y = \sum_{m=0}^\infty a_m x^{m+r}$ and find $y', y''$
    • Substitute into equation and equate the lowest power of $x$ to find the indicial equation for $r$
    • Find the roots of the indicial equation
    • Determine recurrence relation for coefficients $a_m$
    • Write the two linearly independent series solutions
  6. 7 Marks
    aProve that $J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}} \sin x$
    How to answer
    • Write definition of $J_n(x)$ in series form
    • Substitute $n = 1/2$
    • Express $\Gamma(r + 3/2)$ using properties of the Gamma function and $\Gamma(1/2) = \sqrt{\pi}$
    • Expand the terms to match the power series of $\sin x$
    • Factor out $\sqrt{\frac{2}{\pi x}}$ to conclude the proof
  7. 7 Marks
    bSolve by Charpit's method, the P.D.E $(p^2 + q^2)y = qz$.
    How to answer
    • Define $f(x, y, z, p, q) = (p^2 + q^2)y - qz = 0$
    • Set up Charpit's auxiliary equations
    • Find suitable integrable combination to obtain a relation between $p$ and $q$
    • Solve for $p$ and $q$ in terms of $x, y, z$
    • Substitute into $dz = p dx + q dy$ and integrate to obtain complete integral
  8. 7 Marks
    aSolve: $(D^2 - 6DD' + 9D'^2)z = 12x^2 + 36xy$.
    How to answer
    • Form auxiliary equation: $m^2 - 6m + 9 = 0 \implies (m-3)^2 = 0$
    • Roots are $m_1 = m_2 = 3$, so $C.F. = \phi_1(y + 3x) + x \phi_2(y + 3x)$
    • Calculate $P.I. = \frac{1}{(D - 3D')^2}(12x^2 + 36xy)$ using binomial expansion or integration
    • Combine to obtain general solution $z = C.F. + P.I.$
  9. 7 Marks
    bProve that an analytic function with constant modulus is constant.
    How to answer
    • Let $f(z) = u + iv$ be analytic, so it satisfies Cauchy-Riemann equations: $u_x = v_y$, $u_y = -v_x$
    • Given $|f(z)| = c$, then $u^2 + v^2 = c^2$
    • Differentiate partially with respect to $x$ and $y$: $u u_x + v v_x = 0$ and $u u_y + v v_y = 0$
    • Use CR equations to express derivatives in terms of $u_x$ and $v_x$, and show $u_x = 0, v_x = 0, u_y = 0, v_y = 0$
    • Conclude that $u$ and $v$ are constant, hence $f(z)$ is constant
  10. 7 Marks
    aUse Cauchy Integral formula to solve $$\oint_C \frac{\sin \pi z^2 + \cos \pi z^2}{(z - 1)(z - 2)} dz$$ where $C$ is the circle $|z| = 3$.
    How to answer
    • Identify poles of the integrand at $z = 1$ and $z = 2$, both lying inside $|z| = 3$
    • Decompose integrand into partial fractions: $\frac{1}{(z-1)(z-2)} = \frac{1}{z-2} - \frac{1}{z-1}$
    • Split the integral into two integrals around $z = 2$ and $z = 1$
    • Apply Cauchy's Integral Formula to each part with $f(z) = \sin \pi z^2 + \cos \pi z^2$
    • Evaluate $f(2) = \sin 4\pi + \cos 4\pi = 1$ and $f(1) = \sin \pi + \cos \pi = -1$
    • Calculate final result: $2\pi i [f(2) - f(1)] = 2\pi i [1 - (-1)] = 4\pi i$
  11. 7 Marks
    bUsing complex integration method, solve: $\int_0^{2\pi} \frac{\cos 4\theta}{5 + 4\cos\theta} d\theta$
    How to answer
    • Substitute $z = e^{i\theta}$, $\cos\theta = \frac{z+z^{-1}}{2}$, $\cos 4\theta = \text{Re}(z^4)$, and $d\theta = \frac{dz}{iz}$ over unit circle $|z|=1$
    • Express integrand in terms of $z$
    • Find the poles of the integrand inside the unit circle $|z| < 1$
    • Calculate residues at these enclosed poles
    • Apply Cauchy Residue Theorem: $\oint = 2\pi i \sum \text{Res}$ and find the real part
  12. 7 Marks
    aSolve: $\int_0^{1+i} (x - y + ix^2) dz$ along the real axis from $z = 0$ to $z = 1$ and then along a line parallel to imaginary axis from $z = 1$ to $z = 1 + i$.
    How to answer
    • Break integral into two paths: $C_1$ (along real axis) and $C_2$ (parallel to imaginary axis)
    • Along $C_1$: $y = 0 \implies dy = 0, dz = dx$, $x$ varies from $0$ to $1$: $\int_0^1 (x + ix^2)dx$
    • Evaluate $\int_0^1 (x + ix^2)dx = \frac{1}{2} + \frac{i}{3}$
    • Along $C_2$: $x = 1 \implies dx = 0, dz = i dy$, $y$ varies from $0$ to $1$: $\int_0^1 (1 - y + i)(i dy)$
    • Evaluate $i \int_0^1 (1 - y + i) dy = i(1 - 1/2) - 1 = -1 + \frac{i}{2}$
    • Sum both results: $(\frac{1}{2} + \frac{i}{3}) + (-1 + \frac{i}{2}) = -\frac{1}{2} + \frac{5i}{6}$
  13. 7 Marks
    bProve that: $\nabla^2 f(r) = f''(r) + \frac{2}{r} f'(r)$
    How to answer
    • Note $r^2 = x^2 + y^2 + z^2$, giving $\frac{\partial r}{\partial x} = \frac{x}{r}$
    • Compute first derivative: $\frac{\partial}{\partial x} f(r) = f'(r) \frac{x}{r}$
    • Compute second derivative: $\frac{\partial^2}{\partial x^2} f(r) = f''(r)\frac{x^2}{r^2} + f'(r)\left(\frac{1}{r} - \frac{x^2}{r^3}\right)$
    • Sum symmetrically for $x, y, z$ to form $\nabla^2 f(r) = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}$
    • Simplify using $x^2 + y^2 + z^2 = r^2$ to obtain $f''(r) + \frac{2}{r}f'(r)$
  14. 7 Marks
    aFind the directional derivative of $f(x, y, z) = e^{2x}\cos yz$ at $(0, 0, 0)$ in the direction of the tangent to the curve $x = a\sin t$, $y = a\cos t$, $z = at$ at $t = \frac{\pi}{4}$.
    How to answer
    • Calculate $\nabla f = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k}$
    • Evaluate $\nabla f$ at $(0, 0, 0)$: $\nabla f(0,0,0) = 2\hat{i}$
    • Find the tangent vector $\vec{v} = \frac{d\vec{r}}{dt} = a\cos t\hat{i} - a\sin t\hat{j} + a\hat{k}$ at $t = \pi/4$
    • Compute unit vector $\hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j} + \hat{k}}{\sqrt{1/2 + 1/2 + 1}} = \frac{1}{2}(\hat{i} - \hat{j} + \sqrt{2}\hat{k})$
    • Compute directional derivative as $\nabla f \cdot \hat{u} = 2 \times \frac{1}{2} = 1$ (or $\frac{2}{\sqrt{3}}$ depending on curve normalization convention)
  15. 7 Marks
    bUsing Green's theorem, find the area of the region in the first quadrant bounded by the curve $y = x$, $y = \frac{1}{x}$, $y = \frac{x}{4}$.
    How to answer
    • State Green's theorem formula for area: $\text{Area} = \frac{1}{2} \oint_C (x dy - y dx)$
    • Determine the points of intersection of the curves $y = x$, $y = 1/x$, and $y = x/4$
    • Trace the boundary curve oriented counterclockwise
    • Set up the line integrals along each boundary segment
    • Sum the line integrals and multiply by $1/2$ to compute the area
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