Skip to content
BT-202 · Mathematics‑II/Unsolved PYQ Paper

BT-202 Mathematics - II - Jun 2025 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 $(e^y + 1)\cos x\,dx + e^y\sin x\,dy = 0$.
    How to answer
    • Separate variables: $\frac{\cos x}{\sin x}\,dx + \frac{e^y}{e^y + 1}\,dy = 0$
    • Integrate both sides: $\ln|\sin x| + \ln(e^y + 1) = \ln C$
    • Combine logarithms to write the general solution: $(e^y + 1)\sin x = C$
  2. 7 Marks
    bSolve $(D^2 - 5D + 6)y = 4e^x + 5$.
    How to answer
    • Find auxiliary equation: $m^2 - 5m + 6 = 0 \implies m = 2, 3$
    • Write Complementary Function: $y_{CF} = c_1 e^{2x} + c_2 e^{3x}$
    • Evaluate Particular Integral $y_{PI} = \frac{4e^x}{1^2 - 5(1) + 6} + \frac{5}{0 - 0 + 6} = 2e^x + \frac{5}{6}$
    • Write complete solution: $y = y_{CF} + y_{PI} = c_1 e^{2x} + c_2 e^{3x} + 2e^x + \frac{5}{6}$
  3. 7 Marks
    aSolve $x^2 \frac{d^2 y}{dx^2} + 5x \frac{dy}{dx} + 4y = x \log x$.
    How to answer
    • Transform by setting $x = e^z$ or $z = \log x$, reducing to $[D(D-1) + 5D + 4]y = z e^z$
    • Simplify operator: $(D^2 + 4D + 4)y = (D+2)^2 y = z e^z$
    • Roots of auxiliary equation: $m = -2, -2 \implies y_{CF} = (c_1 + c_2 z)e^{-2z}$
    • Compute $y_{PI} = \frac{1}{(D+2)^2}(z e^z) = e^z \frac{1}{(D+3)^2} z = \frac{1}{9} e^z (z - \frac{2}{3})$
    • Substitute back $z = \log x$: $y = (c_1 + c_2 \log x)x^{-2} + \frac{x}{9}(\log x - \frac{2}{3})$
  4. 7 Marks
    bSolve in series Legendre's differential equation $$(1 - x^2)\frac{d^2 y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0.$$
    How to answer
    • Assume series solution $y = \sum_{m=0}^{\infty} a_m x^m$ around the ordinary point $x=0$
    • Substitute $y, y', y''$ into Legendre equation with $n=1$
    • Obtain recurrence relation: $a_{m+2} = \frac{m-1}{m+1} a_m$
    • Calculate coefficients for odd powers ($a_1 x$, then $a_3 = 0, \dots$) and even powers ($a_2 = -a_0, a_4 = -a_0/3, \dots$)
    • Write general solution as linear combination of two independent series: $y = a_1 x + a_0 [1 - x^2 - \frac{x^4}{3} - \dots]$
  5. 14 Marks
    Solve $(D^2 + a^2)y = \tan ax$ by using method of variation of parameters.
    How to answer
    • Find $y_{CF}$: $y_1 = \cos ax$, $y_2 = \sin ax$
    • Calculate Wronskian $W = y_1 y_2' - y_2 y_1' = a(\cos^2 ax + \sin^2 ax) = a$
    • Apply formulas for parameters: $u = -\int \frac{\sin ax \tan ax}{a} dx = -\frac{1}{a^2}[\ln|\sec ax + \tan ax| - \sin ax]$
    • Compute $v = \int \frac{\cos ax \tan ax}{a} dx = -\frac{1}{a^2}\cos ax$
    • Formulate particular integral $y_{PI} = u y_1 + v y_2$ and write complete solution $y = y_{CF} + y_{PI}$
  6. 7 Marks
    aEliminate the arbitrary function $f$ from the relation $$z = y^2 + 2f\left(\frac{1}{x} + \log y\right).$$
    How to answer
    • Differentiate relation with respect to $x$: $p = \frac{\partial z}{\partial x} = 2 f'\left(\frac{1}{x} + \log y\right)\left(-\frac{1}{x^2}\right)$
    • Differentiate relation with respect to $y$: $q = \frac{\partial z}{\partial y} = 2y + 2 f'\left(\frac{1}{x} + \log y\right)\left(\frac{1}{y}\right)$
    • Express $2f'$ from the first equation: $2f' = -x^2 p$
    • Substitute into the second equation: $q = 2y + (-x^2 p)\frac{1}{y} \implies y q = 2y^2 - x^2 p$
    • Rearrange to obtain the final PDE: $x^2 p + y q = 2y^2$
  7. 7 Marks
    bSolve the partial differential equation $$\frac{\partial^2 z}{\partial x^2} - \frac{\partial^2 z}{\partial y^2} = x^2 y.$$
    How to answer
    • Write in operator form: $(D^2 - D'^2)z = x^2 y$
    • Find CF: Auxiliary equation $m^2 - 1 = 0 \implies m = \pm 1$, so $z_{CF} = \phi_1(y+x) + \phi_2(y-x)$
    • Calculate PI: $z_{PI} = \frac{1}{D^2}\left(1 - \frac{D'^2}{D^2}\right)^{-1}(x^2 y) = \frac{1}{D^2}\left(1 + \frac{D'^2}{D^2}\right)(x^2 y) = \frac{1}{D^2}(x^2 y)$ as $D'^2(y) = 0$
    • Integrate twice with respect to $x$: $\frac{1}{D^2}(x^2 y) = y \frac{x^4}{12}$
    • Write complete solution: $z = \phi_1(y+x) + \phi_2(y-x) + \frac{x^4 y}{12}$
  8. 7 Marks
    aSolve $(y+z)p + (x+z)q = x+y$.
    How to answer
    • Write Lagrange auxiliary equations: $\frac{dx}{y+z} = \frac{dy}{x+z} = \frac{dz}{x+y}$
    • Use multipliers $(1, -1, 0)$: $\frac{dx - dy}{-(x-y)} \implies \ln(x-y) + z = \text{const}$ or pairwise subtraction: $\frac{dx-dy}{-(x-y)} = \frac{dy-dz}{-(y-z)} \implies \frac{x-y}{y-z} = c_1$
    • Use multipliers $(1, 1, 1)$: $\frac{dx+dy+dz}{2(x+y+z)} = \frac{dx-dy}{-(x-y)} \implies (x-y)^2(x+y+z) = c_2$ (or similar integral combination)
    • Combine independent integrals: $\phi(c_1, c_2) = 0$
  9. 7 Marks
    bFind all values of $K$ such that $f(z) = e^x(\cos ky + i\sin ky)$ is analytic.
    How to answer
    • Identify $u = e^x \cos ky$ and $v = e^x \sin ky$
    • Compute partial derivatives: $u_x = e^x \cos ky$, $u_y = -k e^x \sin ky$, $v_x = e^x \sin ky$, $v_y = k e^x \cos ky$
    • Apply Cauchy-Riemann equations: $u_x = v_y \implies e^x \cos ky = k e^x \cos ky \implies k = 1$
    • Check second C-R condition: $u_y = -v_x \implies -k e^x \sin ky = -e^x \sin ky \implies k = 1$
    • Conclude that $k = 1$
  10. 7 Marks
    aEvaluate $\int_{(0,0)}^{(1,1)} (3x^2 + 4xy + ix^2)\,dz$ along $y = x^2$.
    How to answer
    • Use parameterization along $y = x^2$: $dy = 2x\,dx$, so $dz = dx + i\,dy = (1 + 2ix)dx$
    • Substitute $y = x^2$ into the integrand: $3x^2 + 4x(x^2) + ix^2 = 3x^2 + 4x^3 + ix^2$
    • Formulate integral: $\int_0^1 (3x^2 + 4x^3 + ix^2)(1 + 2ix)\,dx$
    • Expand real and imaginary parts of the integrand
    • Integrate with respect to $x$ from $0$ to $1$ to find the numerical complex result
  11. 7 Marks
    bIf $f(z) = \frac{1}{(z-1)(z-2)^4}$, find residue of all poles.
    How to answer
    • Identify poles: simple pole at $z=1$ and pole of order 4 at $z=2$
    • Calculate residue at simple pole $z=1$: $\text{Res}(f, 1) = \lim_{z\to 1}(z-1)f(z) = \frac{1}{(1-2)^4} = 1$
    • Calculate residue at $z=2$: $\text{Res}(f, 2) = \frac{1}{3!}\lim_{z\to 2}\frac{d^3}{dz^3}(z-1)^{-1}$
    • Differentiate $(z-1)^{-1}$ three times: $(-1)(-2)(-3)(z-1)^{-4} = -6(z-1)^{-4}$
    • Evaluate at $z=2$: $\text{Res}(f, 2) = \frac{1}{6}(-6)(1)^{-4} = -1$
  12. 7 Marks
    aIf $\bar{r}$ is the position vector of any point in space, then prove that $r^n \bar{r}$ is irrotational.
    How to answer
    • State definition of irrotational vector: $\nabla \times \bar{F} = \bar{0}$
    • Use vector identity: $\nabla \times (\phi \bar{A}) = (\nabla \phi) \times \bar{A} + \phi (\nabla \times \bar{A})$
    • Apply to $r^n \bar{r}$: $\nabla \times (r^n \bar{r}) = (\nabla r^n) \times \bar{r} + r^n (\nabla \times \bar{r})$
    • Show $\nabla r^n = n r^{n-2} \bar{r}$ and $\nabla \times \bar{r} = \bar{0}$
    • Compute cross product: $(n r^{n-2} \bar{r}) \times \bar{r} = \bar{0}$ since $\bar{r} \times \bar{r} = \bar{0}$, hence proved
  13. 7 Marks
    bFind the workdone by the force $\bar{F} = z\bar{i} + x\bar{j} + y\bar{k}$, when it moves a particle along the arc of the curve $\bar{r} = \cos t\,\bar{i} + \sin t\,\bar{j} - t\bar{k}$ from $t = 0$ to $t = 2\pi$.
    How to answer
    • Express curve coordinates: $x = \cos t, y = \sin t, z = -t$
    • Compute velocity vector: $d\bar{r}/dt = -\sin t\,\bar{i} + \cos t\,\bar{j} - \bar{k}$
    • Express force in terms of $t$: $\bar{F} = -t\bar{i} + \cos t\,\bar{j} + \sin t\,\bar{k}$
    • Calculate dot product: $\bar{F} \cdot d\bar{r}/dt = t\sin t + \cos^2 t - \sin t$
    • Integrate from $t=0$ to $2\pi$: $\int_0^{2\pi} (t\sin t + \cos^2 t - \sin t)\,dt = [-t\cos t + \sin t]_0^{2\pi} + \pi - 0 = -2\pi + \pi = -\pi$
  14. 14 Marks
    Verify stokes theorem for $\bar{F} = (x^2 - y^2)\bar{i} + 2xy\bar{j}$ over the box bounded by the planes $x = 0$, $x = a$, $y = 0$, $y = b$.
    How to answer
    • State Stokes theorem: $\oint_C \bar{F} \cdot d\bar{r} = \iint_S (\nabla \times \bar{F}) \cdot \hat{k}\,dx\,dy$
    • Evaluate $\nabla \times \bar{F} = \bar{k}[\frac{\partial}{\partial x}(2xy) - \frac{\partial}{\partial y}(x^2 - y^2)] = \bar{k}(2y - (-2y)) = 4y\bar{k}$
    • Compute surface integral: $\int_0^a dx \int_0^b 4y\,dy = a [2y^2]_0^b = 2ab^2$
    • Evaluate line integral $\oint_C \bar{F} \cdot d\bar{r}$ along four edges of the rectangle $[(0,0)\to(a,0)\to(a,b)\to(0,b)\to(0,0)]$
    • Sum line integrals along all four sides to get $2ab^2$ and verify equality with the surface integral
Go to where you left off?

Quick Add to Notes

Save questions, your own notes and screenshots into notes filed by unit. It takes a free account.

Create free account

Have an account? Log in