KIX1001 2017/2018 Semester 1

Original Question Paper (PDF)

Paper details

Academic session 2017/2018, Semester 1 · 4 questions · 7 printed pages · 2 hours (Jan 2018). Answer all questions. Questions are reproduced in English from the bilingual original. Solutions are worked in full; numbers are given to a sensible precision.

Reference only — these are not official answers

These solutions are revision aids, not examiner answers. They were produced with the help of a large language model and then checked, but an LLM can still be wrong: expect the occasional arithmetic slip, a different method than your lecturer's, or a misreading of the printed question. The question wording is reproduced by hand from a scan of the original paper, so it too may differ in small ways from what was printed.

Use this page as a worked suggestion to check, not as ground truth. Always compare against the original paper (linked above) and your own lecture notes, and if something here disagrees with your lecturer, trust your lecturer. Marks shown are the paper's allocations, not a marking scheme.

Question 1

Question 1(a)(i) (2 marks)

Question

Given function of $f(x) = \dfrac{2x+1}{x-2}$.

(i) Find $f'(x)$.

Solution

Differentiate using the quotient rule, with $u = 2x+1$ and $v = x-2$, so that $u' = 2$ and $v' = 1$:

$$f'(x) = \frac{u'v - uv'}{v^{2}} = \frac{2(x-2) - (2x+1)(1)}{(x-2)^{2}}$$

Expand the numerator:

$$f'(x) = \frac{2x - 4 - 2x - 1}{(x-2)^{2}} = \frac{-5}{(x-2)^{2}}$$

Answer: $f'(x) = -\dfrac{5}{(x-2)^{2}}$, valid for $x \ne 2$.

---

Question 1(a)(ii) (1 mark)

Question

Calculate $\displaystyle \lim_{x \to 3} f(x)$.

Solution

The function is a rational function whose denominator is non-zero at $x = 3$, so it is continuous there and the limit is found by direct substitution:

$$\lim_{x \to 3} \frac{2x+1}{x-2} = \frac{2(3)+1}{3-2} = \frac{7}{1} = 7$$

(No indeterminate form arises, so no factoring or L'Hôpital's rule is needed.)

Answer: $\displaystyle \lim_{x \to 3} f(x) = 7$.

---

Question 1(b)(i) (3 marks)

Question

A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec as shown in Figure Q1. Its height is given by $s = -16t^{2} + 160t$. Note: The graphs of $s$ and $v$ as functions of time; $s$ is largest when $v = \dfrac{ds}{dt} = 0$.

(i) How high does the rock go?

Solution

The height is greatest at the instant the velocity vanishes. Differentiate the height function with respect to time:

$$v = \frac{ds}{dt} = \frac{d}{dt}\left(-16t^{2} + 160t\right) = -32t + 160$$

Set $v = 0$ and solve for the time of maximum height:

$$-32t + 160 = 0 \quad \Longrightarrow \quad t = 5 \ \text{s}$$

Substitute $t = 5$ back into $s$:

$$s_{\max} = -16(5)^{2} + 160(5) = -400 + 800 = 400 \ \text{ft}$$

This agrees with the peak of the parabola drawn in Figure Q1 (the vertex at $t = 5$ gives $s = 400$ ft), and with $v = 0$ at that instant as marked on the figure.

Answer: The rock rises to a maximum height of $s_{\max} = 400$ ft (reached at $t = 5$ s).

---

Question 1(b)(ii) (3 marks)

Question

What are the velocities when the rock is 256 ft. above the ground on the way up and on the way down?

Solution

Set $s = 256$ and solve the quadratic for the two times:

$$-16t^{2} + 160t = 256$$$$16t^{2} - 160t + 256 = 0 \quad \Longrightarrow \quad t^{2} - 10t + 16 = 0$$$$(t-2)(t-8) = 0 \quad \Longrightarrow \quad t = 2 \ \text{s} \ \text{(up)}, \qquad t = 8 \ \text{s} \ \text{(down)}$$

Now use $v = -32t + 160$ at each time:

$$v(2) = -32(2) + 160 = +96 \ \text{ft/s}$$$$v(8) = -32(8) + 160 = -96 \ \text{ft/s}$$

The positive sign means the rock is still rising, the negative sign means it is falling.

Answer: On the way up, $v = +96$ ft/s (96 ft/s upward) at $t = 2$ s; on the way down, $v = -96$ ft/s (96 ft/s downward) at $t = 8$ s. The speeds are equal: 96 ft/s.

---

Question 1(b)(iii) (3 marks)

Question

What is the acceleration of the rock at 256 ft?

Solution

Acceleration is the derivative of velocity (the second derivative of displacement):

$$a = \frac{dv}{dt} = \frac{d}{dt}\left(-32t + 160\right) = -32 \ \text{ft/s}^{2}$$

This is a constant — it does not depend on $t$ — so it has the same value at every instant, in particular at $t = 2$ s and at $t = 8$ s when the rock is 256 ft above the ground.

Physically, the value $-32 \ \text{ft/s}^{2}$ is the (constant) acceleration due to gravity in imperial units, directed downwards.

Answer: $a = -32 \ \text{ft/s}^{2}$, i.e. a constant downward acceleration of $32 \ \text{ft/s}^{2}$ — the same at 256 ft as at any other height.

---

Question 1(b)(iv) (3 marks)

Question

When does the rock hit the ground? At what velocity?

Solution

The rock hits the ground when $s = 0$:

$$-16t^{2} + 160t = 0 \quad \Longrightarrow \quad -16t\,(t - 10) = 0$$$$t = 0 \ \text{s} \quad \text{(launch)} \qquad \text{or} \qquad t = 10 \ \text{s} \ \text{(impact)}$$

The physically relevant root for the impact is $t = 10$ s. Its velocity is

$$v(10) = -32(10) + 160 = -160 \ \text{ft/s}$$

The negative sign confirms downward motion, and the speed of impact equals the launch speed of 160 ft/s, as expected for motion under a constant acceleration with $s(0) = s(10) = 0$.

note

The velocity label printed in Figure Q1 reads $v = \dfrac{dx}{dt} = 160 - 32t$. The derivative must be taken with respect to the height variable $s$, i.e. $v = \dfrac{ds}{dt} = 160 - 32t$, which is the form used above and the form stated in the question text.

Answer: The rock hits the ground at $t = 10$ s with velocity $v = -160$ ft/s, i.e. a speed of 160 ft/s directed downwards.

---

Question 2

Question 2(a) (10 marks)

Question

Determine the first and second order partial derivatives of the function $2x^{4}\sin 10y$.

Solution

Write the function as $f(x,y) = 2x^{4}\sin 10y$. Treat the other variable as a constant each time and use the product/chain rules.

First order partial derivatives. Differentiating with respect to $x$ (treat $\sin 10y$ as a constant):

$$f_x = \frac{\partial f}{\partial x} = 2(4x^{3})\sin 10y = 8x^{3}\sin 10y$$

Differentiating with respect to $y$ (treat $2x^{4}$ as a constant, and $\frac{d}{dy}\sin 10y = 10\cos 10y$):

$$f_y = \frac{\partial f}{\partial y} = 2x^{4}\,(10\cos 10y) = 20x^{4}\cos 10y$$

Second order partial derivatives. Differentiate each first derivative once more:

$$f_{xx} = \frac{\partial}{\partial x}\left(8x^{3}\sin 10y\right) = 24x^{2}\sin 10y$$$$f_{yy} = \frac{\partial}{\partial y}\left(20x^{4}\cos 10y\right) = 20x^{4}\left(-10\sin 10y\right) = -200x^{4}\sin 10y$$$$f_{xy} = \frac{\partial}{\partial y}\left(8x^{3}\sin 10y\right) = 8x^{3}\left(10\cos 10y\right) = 80x^{3}\cos 10y$$$$f_{yx} = \frac{\partial}{\partial x}\left(20x^{4}\cos 10y\right) = 80x^{3}\cos 10y$$

The mixed partial derivatives agree, $f_{xy} = f_{yx} = 80x^{3}\cos 10y$, as guaranteed by Clairaut's theorem since both are continuous.

Derivative	Result
$f_x$	$8x^{3}\sin 10y$
$f_y$	$20x^{4}\cos 10y$
$f_{xx}$	$24x^{2}\sin 10y$
$f_{yy}$	$-200x^{4}\sin 10y$
$f_{xy} = f_{yx}$	$80x^{3}\cos 10y$

Answer: $f_x = 8x^{3}\sin 10y$, $f_y = 20x^{4}\cos 10y$; $f_{xx} = 24x^{2}\sin 10y$, $f_{yy} = -200x^{4}\sin 10y$, $f_{xy} = f_{yx} = 80x^{3}\cos 10y$.

---

Question 2(b) (5 marks)

Question

Find the tangent plane and normal line to the sphere $x^{2} + y^{2} + z^{2} = 36$ at point $(2,-2, 4)$.

Solution

Write the surface as a level surface of

$$F(x,y,z) = x^{2} + y^{2} + z^{2} - 36 = 0$$

The gradient $\nabla F$ is normal to the surface, and therefore normal to the tangent plane:

$$\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right) = (2x,\ 2y,\ 2z)$$

Evaluate at the point $(2,-2,4)$:

$$\nabla F(2,-2,4) = (4,\ -4,\ 8)$$

Tangent plane. A plane through $(x_0,y_0,z_0)$ with normal $(A,B,C)$ has equation $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$:

$$4(x-2) - 4(y+2) + 8(z-4) = 0$$$$4x - 8 - 4y - 8 + 8z - 32 = 0 \quad \Longrightarrow \quad 4x - 4y + 8z = 48$$

Dividing through by 4:

$$x - y + 2z = 12$$

Check: $2 - (-2) + 2(4) = 2 + 2 + 8 = 12$, so the printed point does lie on the plane found — as it must, since the plane was constructed through it. (The point does not, however, lie on the printed sphere; see the note after the answer.)

Normal line. The normal line passes through $(2,-2,4)$ with direction vector $(4,-4,8)$, equivalently $(1,-1,2)$. In symmetric form:

$$\frac{x-2}{1} = \frac{y+2}{-1} = \frac{z-4}{2}$$

or in parametric form:

$$(x,y,z) = (2 + t,\ -2 - t,\ 4 + 2t), \qquad t \in \mathbb{R}$$

Answer: Tangent plane: $x - y + 2z = 12$. Normal line: $\dfrac{x-2}{1} = \dfrac{y+2}{-1} = \dfrac{z-4}{2}$, i.e. $(x,y,z) = (2+t,\ -2-t,\ 4+2t)$.

A misprint in the printed question

The point $(2,-2,4)$ is not on the sphere $x^{2}+y^{2}+z^{2}=36$: substituting gives $2^{2} + (-2)^{2} + 4^{2} = 4 + 4 + 16 = 24 \neq 36$. It lies on the concentric sphere $x^{2}+y^{2}+z^{2}=24$, of radius $\sqrt{24} = 2\sqrt{6} \approx 4.899$.

Strictly, a sphere has no tangent plane at a point that is not on it, so the printed data are mutually inconsistent. The solution above is the standard reading: it applies the tangent-plane formula with $\nabla F(2,-2,4)$ exactly as the question instructs, which gives the answer the paper expects, $x - y + 2z = 12$ (this is the tangent plane of the level surface $x^{2}+y^{2}+z^{2}=24$ through the printed point). Had the point been taken on the printed sphere along the same radial direction, $\left(\sqrt{6},\,-\sqrt{6},\,2\sqrt{6}\right)$, the tangent plane would instead be $x - y + 2z = 6\sqrt{6} \approx 14.70$. The normal line is unaffected by the misprint: it always has direction $\nabla F$ at the printed point.

---

Question 3

Question 3(a)(i) (5 marks)

Question

Movement path of an autopilot ship is governed by the following equation: $\dfrac{dy}{dx} + 6y = 5x$, where the $y$ and $x$ variables are the distances travelled in $y$-axis and $x$-axis respectively.

(i) Given the initial position of the ship is at $(x, y)$ coordinate = $(0,3)$, determine the particular solution of the problem, $y(x)$.

Solution

This is a first-order linear differential equation $\dfrac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = 6$ and $Q(x) = 5x$. Solve it with an integrating factor.

$$I(x) = e^{\int P(x)\,dx} = e^{\int 6\,dx} = e^{6x}$$

Multiply the whole equation by $e^{6x}$; the left side becomes an exact derivative:

$$e^{6x}\frac{dy}{dx} + 6e^{6x}y = 5x e^{6x} \quad \Longrightarrow \quad \frac{d}{dx}\left(y e^{6x}\right) = 5x e^{6x}$$

Integrate both sides. Use integration by parts on the right with $u = x$, $dv = e^{6x}dx$, so $du = dx$ and $v = \frac{1}{6}e^{6x}$:

$$\int 5x e^{6x}\,dx = 5\left[\frac{x e^{6x}}{6} - \int \frac{e^{6x}}{6}\,dx\right] = 5\left[\frac{x e^{6x}}{6} - \frac{e^{6x}}{36}\right] + C$$

Hence

$$y e^{6x} = \left(\frac{5x}{6} - \frac{5}{36}\right)e^{6x} + C$$$$y = \frac{5x}{6} - \frac{5}{36} + C e^{-6x}$$

Apply the initial condition $y(0) = 3$:

$$3 = 0 - \frac{5}{36} + C \quad \Longrightarrow \quad C = 3 + \frac{5}{36} = \frac{108 + 5}{36} = \frac{113}{36}$$

Therefore the particular solution is

$$y(x) = \frac{5x}{6} - \frac{5}{36} + \frac{113}{36}e^{-6x}$$

Check: $y(0) = -\frac{5}{36} + \frac{113}{36} = \frac{108}{36} = 3$. ✓ And $y' + 6y = \frac{5}{6} - \frac{113}{6}e^{-6x} + 5x - \frac{5}{6} + \frac{113}{6}e^{-6x} = 5x$. ✓

Answer: $y(x) = \dfrac{5x}{6} - \dfrac{5}{36} + \dfrac{113}{36}e^{-6x}$, i.e. $y(x) = \dfrac{5}{36}\left(6x - 1\right) + \dfrac{113}{36}e^{-6x}$.

---

Question 3(a)(ii) (2 marks)

Question

Predict if the ship will hit with the rock located at $(x, y)$ coordinate = $(5,5)$ when it continue to travel from $x=2$ to $x=5$ as shown in the Figure Q3(a).

Solution

Evaluate the particular solution at the rock's $x$-coordinate, $x = 5$:

$$y(5) = \frac{5(5)}{6} - \frac{5}{36} + \frac{113}{36}e^{-6(5)} = \frac{25}{6} - \frac{5}{36} + \frac{113}{36}e^{-30}$$

The exponential term is negligible: $e^{-30} \approx 9.36 \times 10^{-14}$, so $\frac{113}{36}e^{-30} \approx 2.9 \times 10^{-13} \approx 0$. Thus

$$y(5) \approx 4.1667 - 0.1389 = 4.0278 \ \text{units}$$

The rock sits at $y = 5$, but the ship is at $y \approx 4.03$ when $x = 5$ — it passes about $0.97$ units below the rock. (For comparison, $y(2) \approx 1.528$ units, matching the initial path plotted in Figure Q3(a).)

So the ship does not hit the rock.

Answer: $y(5) \approx 4.03 \ne 5$; the ship passes about $0.97$ units below the rock, so it does not collide with the rock at $(5,5)$.

---

Question 3(b) (8 marks)

Question

A truck is carrying goods with mass $m = 25$ kg as shown in Figure Q3(b). A vibration suppression system with stiffness $k = 40000$ N/m and damping coefficient $c = 2000$ Ns/m has been installed to minimize the damage on the goods due to vibration. Given the governing equation: $m\ddot{x} + c\dot{x} + kx = F(t)$ where $F(t)$ is the forcing function and $x$, $\dot{x}$ & $\ddot{x}$ are the displacement, velocity and acceleration of the mass respectively. Determine the general solution (i.e. total complementary and particular solutions) of this equation: $25\ddot{x} + 2000\dot{x} + 40000x = 5\sin(10t)$. Hint: Euler's Formula: $e^{\pm ix} = \cos x \pm i(\sin x)$.

Solution

Step 1 — normalise the equation. Divide through by $m = 25$:

$$\ddot{x} + 80\dot{x} + 1600x = 0.2\sin(10t)$$

Step 2 — complementary solution $x_c$. Solve the homogeneous equation $\ddot{x} + 80\dot{x} + 1600x = 0$ with the trial solution $x = e^{rt}$; the auxiliary equation is

$$r^{2} + 80r + 1600 = 0 \quad \Longrightarrow \quad (r + 40)^{2} = 0 \quad \Longrightarrow \quad r = -40 \ \text{(repeated)}$$

A repeated real root gives

$$x_c(t) = (A + Bt)e^{-40t}$$

Step 3 — particular solution $x_p$. Try $x_p = C\sin 10t + D\cos 10t$ (the forcing has frequency $\omega = 10$ rad/s, which is not a root of the auxiliary equation). Then

$$\dot{x}_p = 10C\cos 10t - 10D\sin 10t, \qquad \ddot{x}_p = -100C\sin 10t - 100D\cos 10t$$

Substitute into the normalised equation:

$$\left(-100C - 800D + 1600C\right)\sin 10t + \left(-100D + 800C + 1600D\right)\cos 10t = 0.2\sin 10t$$

Comparing coefficients of $\sin 10t$ and $\cos 10t$:

$$\begin{aligned} 1500C - 800D &= 0.2\\ 800C + 1500D &= 0 \end{aligned}$$

From the second equation $D = -\frac{800}{1500}C = -\frac{8}{15}C$. Substituting into the first:

$$1500C - 800\left(-\frac{8}{15}C\right) = 0.2 \quad \Longrightarrow \quad 1500C + \frac{6400}{15}C = 0.2 \quad \Longrightarrow \quad \frac{28900}{15}C = 0.2$$$$C = \frac{3}{28900} \approx 1.0381 \times 10^{-4}, \qquad D = -\frac{8}{15}\cdot\frac{3}{28900} = -\frac{2}{36125} \approx -5.5363 \times 10^{-5}$$

So

$$x_p(t) = \frac{3}{28900}\sin 10t - \frac{2}{36125}\cos 10t = \frac{15\sin 10t - 8\cos 10t}{144500}$$

Step 4 — general solution. Add the complementary and particular solutions:

$$x(t) = x_c(t) + x_p(t) = (A + Bt)e^{-40t} + \frac{3}{28900}\sin 10t - \frac{2}{36125}\cos 10t$$

with $A$ and $B$ arbitrary constants fixed by the initial displacement and velocity.

Check and remarks. Substituting $x_p$ back into $25\ddot{x} + 2000\dot{x} + 40000x$ returns exactly $5\sin(10t)$. Since $\zeta = \dfrac{c}{2\sqrt{km}} = \dfrac{2000}{2\sqrt{40000 \cdot 25}} = 1$, the system is critically damped, which is why $x_c$ has the repeated-root form $(A+Bt)e^{-40t}$ and dies away as $t \to \infty$, leaving only the steady-state oscillation $x_p$. Written as a single sinusoid, $x_p \approx 1.18 \times 10^{-4}\sin(10t - 0.490)$ m.

Answer: $x(t) = (A + Bt)e^{-40t} + \dfrac{3}{28900}\sin 10t - \dfrac{2}{36125}\cos 10t$, equivalently $x(t) = (A+Bt)e^{-40t} + \dfrac{15\sin 10t - 8\cos 10t}{144500}$.

---

Question 4

Question 4(a) (10 marks)

Question

Solve the following differential equation using Power Series:

$$y'' - 3y' + 2y = 0$$

Solution

Assume a power-series solution about the ordinary point $x = 0$:

$$y = \sum_{n=0}^{\infty} a_n x^{n}, \qquad y' = \sum_{n=1}^{\infty} n a_n x^{n-1}, \qquad y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$$

Substituting into the differential equation:

$$\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} - 3\sum_{n=1}^{\infty} n a_n x^{n-1} + 2\sum_{n=0}^{\infty} a_n x^{n} = 0$$

Re-index every sum to the common power $x^{m}$ (first sum: $m = n-2$; second sum: $m = n-1$; third sum already in $x^{m}$):

$$\sum_{m=0}^{\infty}\Big[(m+2)(m+1)a_{m+2} - 3(m+1)a_{m+1} + 2a_m\Big]x^{m} = 0$$

Since the coefficient of each power of $x$ must vanish, we obtain the recurrence relation

$$(m+2)(m+1)a_{m+2} - 3(m+1)a_{m+1} + 2a_m = 0$$$$a_{m+2} = \frac{3(m+1)a_{m+1} - 2a_m}{(m+2)(m+1)}, \qquad m = 0, 1, 2, \dots$$

with $a_0$ and $a_1$ arbitrary. Evaluating the first few coefficients:

$$\begin{aligned} a_2 &= \frac{3a_1 - 2a_0}{2} = \frac{3}{2}a_1 - a_0\\ a_3 &= \frac{6a_2 - 2a_1}{6} = \frac{7}{6}a_1 - a_0\\ a_4 &= \frac{9a_3 - 2a_2}{12} = \frac{5}{8}a_1 - \frac{7}{12}a_0\\ a_5 &= \frac{12a_4 - 2a_3}{20} = \frac{31}{120}a_1 - \frac{1}{4}a_0 \end{aligned}$$

Hence

$$y = a_0\left(1 - x^{2} - x^{3} - \frac{7}{12}x^{4} - \frac{1}{4}x^{5} - \cdots\right) + a_1\left(x + \frac{3}{2}x^{2} + \frac{7}{6}x^{3} + \frac{5}{8}x^{4} + \frac{31}{120}x^{5} + \cdots\right)$$

Now recognise the two series by expanding the elementary functions:

$$2e^{x} - e^{2x} = 1 - x^{2} - x^{3} - \frac{7}{12}x^{4} - \frac{1}{4}x^{5} - \cdots$$$$e^{2x} - e^{x} = x + \frac{3}{2}x^{2} + \frac{7}{6}x^{3} + \frac{5}{8}x^{4} + \frac{31}{120}x^{5} + \cdots$$

so the series solution closes up as

$$y = a_0\left(2e^{x} - e^{2x}\right) + a_1\left(e^{2x} - e^{x}\right) = \left(2a_0 - a_1\right)e^{x} + \left(a_1 - a_0\right)e^{2x}$$

Writing $c_1 = 2a_0 - a_1$ and $c_2 = a_1 - a_0$ gives the general solution $y = c_1 e^{x} + c_2 e^{2x}$.

Check: the series solution is entire (the recurrence has no singularities for $m \ge 0$, so the radius of convergence is infinite), and direct substitution of $y = e^{rx}$ into $y'' - 3y' + 2y = 0$ gives $r^{2} - 3r + 2 = (r-1)(r-2) = 0$, i.e. $r = 1$ and $r = 2$ — exactly the two independent solutions found above.

Answer: $y = c_1 e^{x} + c_2 e^{2x}$, where the power series gives the two independent solutions $y_1 = 2e^{x} - e^{2x}$ (from $a_0 = 1$, $a_1 = 0$) and $y_2 = e^{2x} - e^{x}$ (from $a_0 = 0$, $a_1 = 1$).

---

Question 4(b) (5 marks)

Question

Determine the singular point(s) of the given differential equation and classify it/them as regular or irregular. Subsequently, find the indicial roots for the differential equation.

$$xy'' + 2y' + 4xy = 0$$

Solution

Step 1 — write the equation in standard form. Divide by $x$:

$$y'' + \frac{2}{x}y' + 4y = 0 \quad \Longrightarrow \quad P(x) = \frac{2}{x}, \qquad Q(x) = 4$$

Step 2 — locate the singular points. $P(x) = \frac{2}{x}$ is undefined (not analytic) at $x = 0$, while $Q(x) = 4$ is a polynomial and analytic everywhere. Hence $x = 0$ is the only singular point of the equation.

Step 3 — classify it. Test the two conditions for a regular singular point:

$$xP(x) = x \cdot \frac{2}{x} = 2 \quad \text{(analytic at } x = 0\text{)}$$$$x^{2}Q(x) = x^{2} \cdot 4 = 4x^{2} \quad \text{(analytic at } x = 0\text{)}$$

Both are finite and analytic at $x = 0$, so $x = 0$ is a regular singular point (the equation is of Frobenius type and a series solution of the form $y = x^{r}\sum_{n=0}^{\infty} a_n x^{n}$ exists).

Step 4 — indicial equation. With

$$p_0 = \lim_{x \to 0} xP(x) = 2, \qquad q_0 = \lim_{x \to 0} x^{2}Q(x) = 0$$

the indicial equation is

$$r(r-1) + p_0 r + q_0 = 0$$$$r(r-1) + 2r + 0 = 0 \quad \Longrightarrow \quad r^{2} + r = 0 \quad \Longrightarrow \quad r(r+1) = 0$$$$r_1 = 0, \qquad r_2 = -1$$

The two roots differ by an integer ($r_1 - r_2 = 1$). This is the borderline case in which a logarithmic term may appear — but here it does not. The recurrence for the smaller root $r_2 = -1$ never degenerates, so two independent Frobenius series exist and no logarithm is needed. In fact, since $(xy)'' = xy'' + 2y'$, the equation says exactly $\dfrac{d^{2}}{dx^{2}}(xy) + 4(xy) = 0$; writing $u = xy$ gives $u'' + 4u = 0$, so

$$y_1 = \frac{\sin 2x}{x} \quad (r = 0), \qquad y_2 = \frac{\cos 2x}{x} \quad (r = -1)$$

and the general solution is $y = \dfrac{c_1 \sin 2x + c_2 \cos 2x}{x}$ — with no logarithmic term at all.

Answer: The only singular point is $x = 0$, and it is a regular singular point ($xP(x) = 2$ and $x^{2}Q(x) = 4x^{2}$ are both analytic at $x = 0$). The indicial roots are $r_1 = 0$ and $r_2 = -1$ (from $r^{2} + r = 0$).