KIX1001 2016/2017 Semester 1

Original Question Paper (PDF)

Paper details

Academic session 2016/2017, Semester 1 · December 2016 / January 2017 · 4 questions · 5 printed pages · 2 hours. 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) (3 marks)

Question

An airplane takes off from the runway and its altitude (in feet) at time $t$ (in minutes) is given by $h = 3000 \ln(t+3)$. Find the rate of climb at time $t = 2$ min.

Solution

The rate of climb is the time derivative of the altitude, $\dfrac{dh}{dt}$.

Differentiate $h = 3000\ln(t+3)$ with respect to $t$, using $\dfrac{d}{dt}\ln(t+3) = \dfrac{1}{t+3}$:

$$\frac{dh}{dt} = 3000 \cdot \frac{1}{t+3} = \frac{3000}{t+3} \ \text{ft/min}.$$

Evaluate at $t = 2$ min:

$$\left.\frac{dh}{dt}\right|_{t=2} = \frac{3000}{2+3} = \frac{3000}{5} = 600 \ \text{ft/min}.$$

So at $t = 2$ minutes the aircraft is climbing at 600 feet per minute.

Answer: $600$ ft/min (i.e. $\left.\dfrac{dh}{dt}\right|_{t=2} = 600$ ft/min)

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Question 1(b) (12 marks)

Question

A 2 meter material is used to build two window frames. The material is divided into 2 pieces where one piece is bent into a square while the other piece is bent into an equilateral triangle. What must the dimensions (length of each side of the square and triangle frames), in meter, of the windows be in order to let in least light coming in through both windows? Provide justifications for the resulted dimensions.

Solution

Interpretation. "Least light coming in" means the smallest total glazed area of the two frames. The 2 m of material is the total perimeter, which is fixed; we must choose how to split it.

Step 1 — Set up the variables and the constraint.

Let $x$ = side of the square frame (m) and $y$ = side of the equilateral triangle frame (m).

The square uses $4x$ metres of material and the triangle uses $3y$ metres, and together they use all 2 m:

$$4x + 3y = 2 \quad\Longrightarrow\quad y = \frac{2-4x}{3}, \qquad 0 \le x \le \frac{1}{2}.$$

Step 2 — Write the total area as a function of one variable.

The area of a square of side $x$ is $x^2$, and the area of an equilateral triangle of side $y$ is $\dfrac{\sqrt3}{4}y^2$. Hence

$$A(x) = x^2 + \frac{\sqrt3}{4}y^2 = x^2 + \frac{\sqrt3}{4}\left(\frac{2-4x}{3}\right)^{\!2} = x^2 + \frac{\sqrt3}{36}(2-4x)^2 .$$

Step 3 — Differentiate and find the stationary point.

$$\frac{dA}{dx} = 2x + \frac{\sqrt3}{36}\cdot 2(2-4x)(-4) = 2x - \frac{2\sqrt3}{9}(2-4x).$$

Setting $\dfrac{dA}{dx} = 0$:

$$2x = \frac{2\sqrt3}{9}(2-4x) \;\Longrightarrow\; 9x = \sqrt3\,(2-4x) \;\Longrightarrow\; 9x + 4\sqrt3\,x = 2\sqrt3$$$$x = \frac{2\sqrt3}{9+4\sqrt3} = \frac{2\sqrt3\,(9-4\sqrt3)}{(9+4\sqrt3)(9-4\sqrt3)} = \frac{18\sqrt3 - 24}{33} = \frac{6\sqrt3 - 8}{11} \approx 0.2175 \ \text{m}.$$

Substituting back into the constraint:

$$y = \frac{2-4x}{3} = \frac{18 - 8\sqrt3}{11} \approx 0.3767 \ \text{m}.$$

Step 4 — Justify that this is a minimum.

$$\frac{d^2A}{dx^2} = 2 + \frac{8\sqrt3}{9} \approx 3.54 > 0,$$

so the stationary point is a local minimum. Since $A(x)$ is a convex quadratic, it is the global minimum on the interval. Checking the endpoints confirms it: as $x \to 0^{+}$ (all material to the triangle) $A \to \dfrac{\sqrt3}{9} \approx 0.1925$ m², and at $x = \tfrac12$ (all material to the square) $A = 0.25$ m². Both exceed the stationary value

$$A_{\min} \approx 0.1087 \ \text{m}^2 .$$

Step 5 — Independent check by Lagrange multipliers.

Minimise $A = x^2 + \dfrac{\sqrt3}{4}y^2$ subject to $g(x,y) = 4x + 3y - 2 = 0$. With $\nabla A = \mu\nabla g$:

$$2x = 4\mu, \qquad \frac{\sqrt3}{2}y = 3\mu \quad\Longrightarrow\quad \mu = \frac{x}{2} = \frac{\sqrt3}{6}y \quad\Longrightarrow\quad y = \sqrt3\,x .$$

Substituting $y = \sqrt3\,x$ into the constraint, $4x + 3\sqrt3\,x = 2$, gives $x = \dfrac{2}{4+3\sqrt3} = \dfrac{6\sqrt3-8}{11}$ and $y = \dfrac{2\sqrt3}{4+3\sqrt3} = \dfrac{18-8\sqrt3}{11}$ — exactly the values found above.

Conclusion. The optimal split satisfies $y = \sqrt3\,x$: the triangle's side is $\sqrt3$ times the square's side. This is the classic result that, for a fixed total perimeter, the shape with fewer sides should be given relatively more wire when the total area is to be minimised.

Answer: Square side $x = \dfrac{6\sqrt3-8}{11} \approx 0.2175$ m; equilateral-triangle side $y = \dfrac{18-8\sqrt3}{11} \approx 0.3767$ m (so $y = \sqrt3\,x$). The minimum total window area is $\approx 0.1087$ m², confirmed as a minimum by $\dfrac{d^2A}{dx^2} = 2 + \dfrac{8\sqrt3}{9} > 0$.

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Question 2

Question 2(a) (9 marks)

Question

By using Jacobian, find the values of the constant parameter $\lambda$ for which the functions:

$$u = \cos x \cos y - \lambda \sin x \sin y, \qquad v = \sin x \cos y + \lambda \cos x \sin y$$

are functionally dependent.

Solution

Method. Two functions $u(x,y)$ and $v(x,y)$ are functionally dependent if and only if their Jacobian determinant vanishes identically:

$$J = \frac{\partial(u,v)}{\partial(x,y)} = \det \begin{pmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\[6pt] \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{pmatrix} = 0 .$$

Step 1 — First partial derivatives.

$$\begin{aligned} \frac{\partial u}{\partial x} &= -\sin x \cos y - \lambda \cos x \sin y, &\qquad \frac{\partial u}{\partial y} &= -\cos x \sin y - \lambda \sin x \cos y, \\[4pt] \frac{\partial v}{\partial x} &= \cos x \cos y - \lambda \sin x \sin y, &\qquad \frac{\partial v}{\partial y} &= -\sin x \sin y + \lambda \cos x \cos y . \end{aligned}$$

Step 2 — Expand the determinant.

To keep the algebra readable write $s = \sin x$, $c = \cos x$, $S = \sin y$, $C = \cos y$. Then

$$J = (-sC - \lambda cS)(-sS + \lambda cC) - (-cS - \lambda sC)(cC - \lambda sS).$$

Expand the first product:

$$(-sC - \lambda cS)(-sS + \lambda cC) = s^2CS - \lambda sc\,C^2 + \lambda sc\,S^2 - \lambda^2 c^2 SC .$$

Expand the second product:

$$(-cS - \lambda sC)(cC - \lambda sS) = -c^2SC + \lambda cs\,S^2 - \lambda cs\,C^2 + \lambda^2 s^2 SC .$$

Subtracting the second from the first, the mixed terms cancel in pairs:

$$\begin{aligned} J &= \left(s^2CS - \lambda sc\,C^2 + \lambda sc\,S^2 - \lambda^2 c^2SC\right) - \left(-c^2SC + \lambda cs\,S^2 - \lambda cs\,C^2 + \lambda^2 s^2SC\right) \\[4pt] &= \left(s^2 + c^2\right)SC - \lambda^2\left(s^2 + c^2\right)SC \\[4pt] &= SC\left(1 - \lambda^2\right) \qquad \text{(since } s^2 + c^2 = 1\text{)} . \end{aligned}$$

Therefore

$$J = \left(1-\lambda^2\right)\sin y \cos y = \frac{1}{2}\left(1-\lambda^2\right)\sin 2y .$$

Step 3 — Impose functional dependence.

For $u$ and $v$ to be functionally dependent we need $J \equiv 0$ for all $(x,y)$. Since $\sin y\cos y$ is not identically zero, this forces

$$1 - \lambda^2 = 0 \quad\Longrightarrow\quad \lambda = \pm 1 .$$

Step 4 — Verify the two values.

• $\lambda = 1$: $u = \cos x\cos y - \sin x\sin y = \cos(x+y)$ and $v = \sin x\cos y + \cos x\sin y = \sin(x+y)$, so $u^2 + v^2 = 1$ — dependent.
• $\lambda = -1$: $u = \cos x\cos y + \sin x\sin y = \cos(x-y)$ and $v = \sin x\cos y - \cos x\sin y = \sin(x-y)$, so again $u^2 + v^2 = 1$ — dependent.

Answer: $\lambda = \pm 1$. (Then $J = \dfrac{1}{2}(1-\lambda^2)\sin 2y \equiv 0$, and indeed $u^2+v^2 = 1$ in both cases.)

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Question 2(b) (6 marks)

Question

Find the equation of the tangent plane and normal line to $xyz^2 = 6$ at $(3,2,1)$.

Solution

Method. Write the surface as a level surface $F(x,y,z) = xyz^2 - 6 = 0$. Its normal vector at a point is $\nabla F$ evaluated there; the tangent plane is then $\nabla F \cdot \left(\mathbf{r} - \mathbf{r}_0\right) = 0$ and the normal line runs along $\nabla F$.

Step 1 — Gradient of $F$.

$$\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right) = \left(yz^2,\; xz^2,\; 2xyz\right).$$

Step 2 — Evaluate at $P(3,2,1)$.

$$\left.\nabla F\right|_{(3,2,1)} = \left(2\cdot 1^2,\; 3\cdot 1^2,\; 2\cdot 3\cdot 2\cdot 1\right) = (2,\;3,\;12).$$

Step 3 — Tangent plane. With normal $(2,3,12)$ through $(3,2,1)$:

$$2(x-3) + 3(y-2) + 12(z-1) = 0$$$$2x + 3y + 12z = 6 + 6 + 12 = 24 .$$

Check: $2(3) + 3(2) + 12(1) = 6 + 6 + 12 = 24$, so the given point does lie on the plane.

Step 4 — Normal line. The normal line passes through $(3,2,1)$ with direction vector $(2,3,12)$:

$$\frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{12},$$

or in parametric form

$$x = 3 + 2t, \qquad y = 2 + 3t, \qquad z = 1 + 12t, \qquad t \in \mathbb{R}.$$

Answer: Tangent plane: $2x + 3y + 12z = 24$. Normal line: $\dfrac{x-3}{2} = \dfrac{y-2}{3} = \dfrac{z-1}{12}$ (i.e. $\mathbf{r} = (3,2,1) + t(2,3,12)$).

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Question 3

Question 3(a) (2 marks)

Question

Identify the dependent and independent variables for each equation respectively. Classify each equation according to its order, linearity or non-linearity, and homogeneity or non-homogeneity.

(i) $2\dfrac{d^2y}{dt^2} - 4t^2\dfrac{dy}{dt} - \tan 2t = 0$

(ii) $2z\dfrac{dx}{dz} - 3x = 0$

Solution

(i) $2\dfrac{d^2y}{dt^2} - 4t^2\dfrac{dy}{dt} - \tan 2t = 0$, i.e. $2y'' - 4t^2y' = \tan 2t$.

• Variables: the derivative is taken with respect to $t$, so $t$ is the independent variable and $y$ is the dependent variable.
• Order: the highest derivative present is $\dfrac{d^2y}{dt^2}$, so the equation is second order.
• Linearity: $y$, $\dfrac{dy}{dt}$ and $\dfrac{d^2y}{dt^2}$ each occur to the first power only, are not multiplied together, and are not arguments of any nonlinear function; the coefficients $2$ and $-4t^2$ depend on $t$ alone. Hence the equation is linear.
• Homogeneity: the term $\tan 2t$ depends on $t$ alone and is not identically zero, so the equation is non-homogeneous.

(ii) $2z\dfrac{dx}{dz} - 3x = 0$.

• Variables: $z$ is the independent variable and $x$ is the dependent variable.
• Order: the highest derivative is $\dfrac{dx}{dz}$, so the equation is first order.
• Linearity: $x$ and $\dfrac{dx}{dz}$ appear to the first power with coefficients $-3$ and $2z$ that involve only $z$, so the equation is linear.
• Homogeneity: every term contains $x$ or one of its derivatives (there is no term free of $x$), so the equation is homogeneous. (Equivalently, $x \equiv 0$ is a solution.)

Answer: (i) dependent $y$, independent $t$; second order, linear, non-homogeneous. (ii) dependent $x$, independent $z$; first order, linear, homogeneous.

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Question 3(b) (5 marks)

Question

A detective discovers a murder victim in a hotel room at 10.00 am one morning. The temperature of the body is $80^\circ F$. One hour later, at 11.00 am, the body has cooled to $75^\circ F$. Assume that the victim had a normal temperature of $98.6^\circ F$ at the time of death. Given the governing equation: $\dfrac{du}{dt} = k(u-70)$ where $u = u(t)$ is the temperature of the body after $t$ hour, $k$ is a constant and the room is kept at environmental temperature of $70^\circ F$. Determine the time when the murder took place.

Solution

Method. This is Newton's law of cooling. The equation is separable and linear; we solve it, use the two temperature readings to find $k$, then solve for the time at which $u = 98.6^\circ F$.

Step 1 — Solve the differential equation.

$$\frac{du}{dt} = k(u - 70).$$

Let $\theta = u - 70$, so $\dfrac{d\theta}{dt} = \dfrac{du}{dt}$ and the equation becomes $\dfrac{d\theta}{dt} = k\theta$. Separating and integrating,

$$\int \frac{d\theta}{\theta} = \int k\,dt \quad\Longrightarrow\quad \ln\lvert \theta \rvert = kt + C \quad\Longrightarrow\quad \theta = \theta_0 e^{kt},$$

that is

$$u(t) = 70 + \left(u_0 - 70\right)e^{kt}.$$

Step 2 — Fit the data. Take $t = 0$ at 10.00 am, so $u(0) = 80^\circ F$ and $\theta_0 = 80 - 70 = 10$:

$$u(t) = 70 + 10e^{kt}.$$

At 11.00 am, $t = 1$ and $u(1) = 75^\circ F$:

$$75 = 70 + 10e^{k} \quad\Longrightarrow\quad 10e^{k} = 5 \quad\Longrightarrow\quad e^{k} = \frac{1}{2} \quad\Longrightarrow\quad k = -\ln 2 \approx -0.6931 \ \text{h}^{-1}.$$

So

$$u(t) = 70 + 10\left(\frac{1}{2}\right)^{t}.$$

Step 3 — Find the time of death. At the moment of death the body temperature was $98.6^\circ F$:

$$98.6 = 70 + 10\left(\frac{1}{2}\right)^{t} \quad\Longrightarrow\quad 10\left(\frac{1}{2}\right)^{t} = 28.6 \quad\Longrightarrow\quad \left(\frac{1}{2}\right)^{t} = 2.86 .$$

Taking logarithms,

$$t = \frac{\ln 2.86}{\ln(1/2)} = -\frac{\ln 2.86}{\ln 2} = -\frac{1.05082}{0.69315} \approx -1.516 \ \text{hours}.$$

The negative sign means the death occurred before 10.00 am, by

$$1.516 \ \text{h} = 1 \ \text{hour} + 0.516\times 60 \ \text{min} \approx 1 \ \text{hour}\ 31 \ \text{minutes}.$$

Check: at $t = -1.516$, $u = 70 + 10(0.5)^{-1.516} = 70 + 10(2.86) = 98.6^\circ F$. Correct.

Answer: The murder took place about $1.516$ hours (≈ 1 hour 31 minutes) before 10.00 am, i.e. at approximately 8:29 am.

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Question 3(c) (i) (2 marks)

Question

An inductor of $L = 4$ henrys, a resistor of $R = 24$ ohms and a capacitor of $C = 0.05$ farads are connected in series with an emf of $E$ volts as shown in Figure 3. Given the governing equation for RLC electrical circuit: $L\dfrac{d^2q(t)}{dt^2} + R\dfrac{dq(t)}{dt} + \dfrac{1}{C}q(t) = 2te^{-t}$ where $q(t)$ is the charge of the circuit after $t$ seconds. At $t = 0$, the switch $S$ is closed.

(i) Determine the general complementary solution, $q_c(t)$ of this equation.

Solution

Step 1 — Insert the numerical values.

With $L = 4$, $R = 24$ and $C = 0.05$ (so $\tfrac{1}{C} = 20$):

$$4\frac{d^2q}{dt^2} + 24\frac{dq}{dt} + 20q = 2te^{-t}.$$

Dividing through by 4 puts the equation in standard form:

$$q'' + 6q' + 5q = \frac{1}{2}te^{-t}.$$

Step 2 — Solve the homogeneous (complementary) equation.

For $q'' + 6q' + 5q = 0$ try $q = e^{rt}$, giving the auxiliary (characteristic) equation

$$r^2 + 6r + 5 = 0 \quad\Longrightarrow\quad (r+1)(r+5) = 0 \quad\Longrightarrow\quad r = -1,\ -5 .$$

The roots are real and distinct, so the complementary solution is

$$q_c(t) = c_1e^{-t} + c_2e^{-5t},$$

where $c_1$ and $c_2$ are arbitrary constants.

(Physical reading: the circuit is overdamped, and both natural modes decay exponentially.)

Answer: $q_c(t) = c_1e^{-t} + c_2e^{-5t}$

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Question 3(c) (ii) (3 marks)

Question

(ii) Find the general particular solution, $q_p(t)$ of this equation.

Solution

Method — undetermined coefficients with resonance.

The right-hand side is $\dfrac{1}{2}te^{-t}$. The trial function for $te^{-t}$ would normally be $e^{-t}(At + B)$, but $e^{-t}$ is already a solution of the homogeneous equation (the root $r = -1$ of the auxiliary equation). This is the resonant case, so the trial function must be multiplied by $t$:

$$q_p(t) = e^{-t}\left(At^2 + Bt\right).$$

Step 1 — Derivatives.

$$q_p = e^{-t}\left(At^2 + Bt\right)$$$$q_p' = e^{-t}\left(-At^2 + (2A-B)t + B\right)$$$$q_p'' = e^{-t}\left(At^2 + (B-4A)t + (2A-2B)\right)$$

Step 2 — Substitute into $q'' + 6q' + 5q = \dfrac{1}{2}te^{-t}$.

Collecting the $e^{-t}$ factor:

$$\begin{aligned} q_p'' + 6q_p' + 5q_p &= e^{-t}\Big[\left(A - 6A + 5A\right)t^2 \\ &\qquad + \left((B-4A) + 6(2A-B) + 5B\right)t \\ &\qquad + \left(2A - 2B + 6B\right)\Big] \\[4pt] &= e^{-t}\left(0\cdot t^2 + 8A\,t + \left(2A + 4B\right)\right). \end{aligned}$$

The $t^2$ terms cancel — exactly as expected in the resonant case — leaving

$$q_p'' + 6q_p' + 5q_p = e^{-t}\left(8At + 2A + 4B\right).$$

Step 3 — Match coefficients. This must equal $\dfrac{1}{2}te^{-t}$ for all $t$, so

$$8A = \frac{1}{2} \quad\Longrightarrow\quad A = \frac{1}{16},$$$$2A + 4B = 0 \quad\Longrightarrow\quad 4B = -\frac{2}{16} = -\frac{1}{8} \quad\Longrightarrow\quad B = -\frac{1}{32}.$$

Hence

$$q_p(t) = e^{-t}\left(\frac{1}{16}t^2 - \frac{1}{32}t\right),$$

which agrees with the form quoted in part (iii).

Verification. Substituting $q_p$ into the original equation $4q'' + 24q' + 20q$:

$$4q_p'' + 24q_p' + 20q_p = 4\cdot\frac{1}{2}te^{-t} = 2te^{-t},$$

which is exactly the right-hand side of the governing equation.

Answer: $q_p(t) = e^{-t}\left(\dfrac{1}{16}t^2 - \dfrac{1}{32}t\right) = \dfrac{1}{32}te^{-t}\left(2t - 1\right)$

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Question 3(c) (iii) (3 marks)

Question

(iii) Let $q_c(t) = c_1e^{-t} + c_2e^{-5t}$ and $q_p(t) = e^{(-t)}\left(\dfrac{1}{16}t^2 - \dfrac{1}{32}t\right)$, determine the total solution of charge $q(t)$ of the given initial conditions, $i(0) = 0$ and $q(0) = 0$.

Solution

Method. Form the general solution $q = q_c + q_p$, differentiate to obtain the current $i(t) = \dfrac{dq}{dt}$, and impose the two initial conditions to fix $c_1$ and $c_2$.

Step 1 — General solution.

$$q(t) = c_1e^{-t} + c_2e^{-5t} + e^{-t}\left(\frac{1}{16}t^2 - \frac{1}{32}t\right).$$

Step 2 — Current $i(t) = q'(t)$.

Differentiating the particular part,

$$\frac{d}{dt}\left[e^{-t}\left(\frac{t^2}{16} - \frac{t}{32}\right)\right] = e^{-t}\left(-\frac{t^2}{16} + \frac{5t}{32} - \frac{1}{32}\right),$$

so

$$i(t) = q'(t) = -c_1e^{-t} - 5c_2e^{-5t} + e^{-t}\left(-\frac{t^2}{16} + \frac{5t}{32} - \frac{1}{32}\right).$$

Step 3 — Apply the initial conditions.

$q(0) = 0$:

$$c_1 + c_2 + 0 = 0 \quad\Longrightarrow\quad c_1 = -c_2. \tag{1}$$

$i(0) = 0$:

$$-c_1 - 5c_2 - \frac{1}{32} = 0 \quad\Longrightarrow\quad -c_1 - 5c_2 = \frac{1}{32}. \tag{2}$$

Substituting (1) into (2):

$$c_2 - 5c_2 = \frac{1}{32} \quad\Longrightarrow\quad -4c_2 = \frac{1}{32} \quad\Longrightarrow\quad c_2 = -\frac{1}{128}, \qquad c_1 = \frac{1}{128}.$$

Step 4 — Total solution.

$$q(t) = \frac{1}{128}e^{-t} - \frac{1}{128}e^{-5t} + e^{-t}\left(\frac{t^2}{16} - \frac{t}{32}\right) = e^{-t}\left(\frac{t^2}{16} - \frac{t}{32} + \frac{1}{128}\right) - \frac{1}{128}e^{-5t}.$$

Check.

$$q(0) = \frac{1}{128} - \frac{1}{128} + 0 = 0,$$$$i(0) = -\frac{1}{128} + \frac{5}{128} - \frac{1}{32} = \frac{4}{128} - \frac{4}{128} = 0.$$

Both initial conditions are satisfied.

Answer: $q(t) = \dfrac{1}{128}e^{-t} - \dfrac{1}{128}e^{-5t} + e^{-t}\left(\dfrac{t^2}{16} - \dfrac{t}{32}\right) = e^{-t}\left(\dfrac{t^2}{16} - \dfrac{t}{32} + \dfrac{1}{128}\right) - \dfrac{1}{128}e^{-5t}$ coulomb

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Question 4

Question 4(a) (8 marks)

Question

Solve the following differential equation using Power Series.

$$(1-x)\frac{dy}{dx} - y = 0$$

Solution

Method. Assume a power-series solution about the ordinary point $x = 0$, $y = \sum_{n\ge0}a_nx^n$, substitute, and read off a recurrence relation for the coefficients.

Step 1 — Series and derivatives.

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

Step 2 — Substitute into $(1-x)y' - y = 0$.

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

Step 3 — Re-index so that every sum is in powers of $x^m$.

In the first sum put $m = n-1$, i.e. $n = m+1$, so $\sum_{n\ge1} n a_n x^{n-1} = \sum_{m\ge0}(m+1)a_{m+1}x^m$:

$$\sum_{m=0}^{\infty}(m+1)a_{m+1}x^m - \sum_{m=1}^{\infty} m\,a_m x^m - \sum_{m=0}^{\infty} a_m x^m = 0 .$$

Step 4 — Compare coefficients.

Coefficient of $x^0$ (only the first and third sums contribute):

$$(0+1)a_1 - a_0 = 0 \quad\Longrightarrow\quad a_1 = a_0 .$$

Coefficient of $x^m$ for $m \ge 1$ (all three sums contribute):

$$(m+1)a_{m+1} - m\,a_m - a_m = 0 \quad\Longrightarrow\quad (m+1)a_{m+1} = (m+1)a_m \quad\Longrightarrow\quad a_{m+1} = a_m .$$

Step 5 — Solve the recurrence.

The recurrence $a_{m+1} = a_m$ together with $a_1 = a_0$ gives

$$a_n = a_0 \qquad \text{for every } n \ge 0 .$$

Therefore

$$y = a_0\sum_{n=0}^{\infty} x^n = \frac{a_0}{1-x}, \qquad \lvert x \rvert < 1 .$$

Step 6 — Radius of convergence and check.

By the ratio test, $\displaystyle\lim_{n\to\infty}\left\lvert \frac{a_{n+1}}{a_n}\right\rvert = 1$, so the radius of convergence is $R = 1$ — consistent with the singular point of the differential equation at $x = 1$.

Substituting back:

$$(1-x)\cdot\frac{a_0}{(1-x)^2} - \frac{a_0}{1-x} = \frac{a_0}{1-x} - \frac{a_0}{1-x} = 0 .$$

Answer: $y = \dfrac{a_0}{1-x} = a_0\left(1 + x + x^2 + x^3 + \cdots\right)$, valid for $\lvert x \rvert < 1$, where $a_0 = y(0)$ is an arbitrary constant.

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Question 4(b) (7 marks)

Question

Using Frobenius method, find the indicial roots for the following differential equation:

$$xy'' + y' + 10y = 0$$

Solution

Method. $x = 0$ is a singular point. Substituting the Frobenius series $y = \sum_{n\ge0}a_nx^{n+r}$ with $a_0 \neq 0$ and equating the coefficient of the lowest power of $x$ to zero gives the indicial equation.

Step 1 — Confirm $x = 0$ is a regular singular point.

In standard form $y'' + P(x)y' + Q(x)y = 0$,

$$P(x) = \frac{1}{x}, \qquad Q(x) = \frac{10}{x},$$

so $xP(x) = 1$ and $x^2Q(x) = 10x$ are both analytic at $x = 0$. Hence $x = 0$ is a regular singular point and the Frobenius method applies.

Step 2 — Substitute the Frobenius series.

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

Multiplying $y''$ by $x$,

$$xy'' = \sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{n+r-1}.$$

So the equation becomes

$$\sum_{n=0}^{\infty}\left[(n+r)(n+r-1) + (n+r)\right]a_nx^{n+r-1} + \sum_{n=0}^{\infty}10a_nx^{n+r} = 0,$$$$\sum_{n=0}^{\infty}(n+r)^2a_nx^{n+r-1} + \sum_{n=0}^{\infty}10a_nx^{n+r} = 0 .$$

Step 3 — Indicial equation (coefficient of the lowest power).

The lowest power of $x$ is $x^{r-1}$, which arises only from $n = 0$ in the first sum. Its coefficient must vanish:

$$r^2a_0 = 0 .$$

Since $a_0 \neq 0$ by assumption, the indicial equation is

$$r^2 = 0 \quad\Longrightarrow\quad r = 0 \ \text{(a double root)} .$$

So the indicial roots are

$$r_1 = r_2 = 0 .$$

Step 4 — Recurrence relation (for completeness).

With $r = 0$ the series becomes an ordinary power series. The coefficient of $x^{n-1}$ for $n \ge 1$ is $n^2a_n + 10a_{n-1}$, so

$$n^2a_n + 10a_{n-1} = 0 \quad\Longrightarrow\quad a_n = -\frac{10}{n^2}a_{n-1}, \qquad n \ge 1,$$

which gives

$$a_n = \frac{(-10)^n}{(n!)^2}a_0 .$$

Hence one solution is

$$y_1(x) = a_0\sum_{n=0}^{\infty}\frac{(-10x)^n}{(n!)^2} = a_0\,J_0\!\left(2\sqrt{10x}\right).$$

Because the indicial roots are equal ($r_1 - r_2 = 0$ is an integer), the second linearly independent solution necessarily contains a logarithmic term,

$$y_2(x) = y_1(x)\ln x + \sum_{n=1}^{\infty} b_n x^{n},$$

so the general solution is $y = C_1y_1 + C_2y_2$. (Only the indicial roots were asked for here.)

Answer: $r_1 = r_2 = 0$ — the indicial equation is $r^2 = 0$, so $x = 0$ is a regular singular point with a repeated indicial root $r = 0$.