KIX1001 2019/2020 Semester 1

Original Question Paper (PDF)

Paper details

Academic session 2019/2020, Semester 1 · 4 questions · 6 printed pages · 2 hours. 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) (5 marks)

Question

Find all of the first and second order partial derivative for the function

$$f(a,b) = e^{ab}(2a-b)$$

Solution

Differentiate using the product rule, remembering that

$$\frac{\partial}{\partial a}e^{ab} = b\,e^{ab}, \qquad \frac{\partial}{\partial b}e^{ab} = a\,e^{ab}.$$

First order derivatives.

$$f_a = \frac{\partial}{\partial a}\left[e^{ab}(2a-b)\right] = b\,e^{ab}(2a-b) + e^{ab}(2) = e^{ab}\left[b(2a-b) + 2\right]$$$$f_b = \frac{\partial}{\partial b}\left[e^{ab}(2a-b)\right] = a\,e^{ab}(2a-b) + e^{ab}(-1) = e^{ab}\left[a(2a-b) - 1\right]$$

Second order derivatives. Differentiate $f_a$ and $f_b$ once more:

$$f_{aa} = \frac{\partial}{\partial a}f_a = b\,e^{ab}\left[b(2a-b)+2\right] + e^{ab}(2b) = e^{ab}\left[b(2ab - b^2 + 2) + 2b\right]$$$$f_{bb} = \frac{\partial}{\partial b}f_b = a\,e^{ab}\left[a(2a-b)-1\right] + e^{ab}(-a) = e^{ab}\left[a(2a^2 - ab - 1) - a\right]$$$$f_{ab} = \frac{\partial}{\partial b}f_a = a\,e^{ab}\left[b(2a-b)+2\right] + e^{ab}(2a - 2b) = e^{ab}\left[2a^2b - ab^2 + 2a + 2a - 2b\right]$$

Check. By Clairaut's theorem the mixed derivatives must agree; repeating the calculation from the other order gives

$$f_{ba} = \frac{\partial}{\partial a}f_b = e^{ab}\left(2a^2b - ab^2 + 4a - 2b\right) = f_{ab} \quad \checkmark$$

Answer: $f_a = e^{ab}\left(2ab - b^2 + 2\right)$, $f_b = e^{ab}\left(2a^2 - ab - 1\right)$, $f_{aa} = e^{ab}\left(2ab^2 - b^3 + 4b\right)$, $f_{bb} = e^{ab}\left(2a^3 - a^2b - 2a\right)$, $f_{ab} = f_{ba} = e^{ab}\left(2a^2b - ab^2 + 4a - 2b\right)$.

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Question 1(b) (7 marks)

Question

Suppose a contractor is designing a floorplan for new construction site. The construction site is in a triangle shape where the angle at one vertex is $\theta$ and the lengths of the two sides adjacent to that vertex are $x$ and $y$. Then the area of the construction site can be computed by using the formula as below:

$$\text{Area}(A) = \frac{1}{2}xy\sin\theta$$

Suppose that we measure:

$$x = 40.00 \pm 0.05 \text{ meter}, \qquad y = 30.00 \pm 0.05 \text{ meter}, \qquad \theta = \frac{\pi}{6} \pm 0.01 \text{ radians}$$

Find the area of the construction site and the maximum percentage error of measurement to do the new flooring.

Solution

Use the total differential (small-error propagation). For a function of three measured quantities the maximum absolute error is

$$\delta A \approx \left|\frac{\partial A}{\partial x}\right|\delta x + \left|\frac{\partial A}{\partial y}\right|\delta y + \left|\frac{\partial A}{\partial \theta}\right|\delta \theta .$$

Nominal area. With $x = 40$, $y = 30$ and $\theta = \pi/6$,

$$A = \frac{1}{2}(40)(30)\sin\frac{\pi}{6} = \frac{1}{2}(1200)\left(\frac{1}{2}\right) = 300 \text{ m}^2 .$$

Partial derivatives.

$$\frac{\partial A}{\partial x} = \frac{1}{2}y\sin\theta = \frac{1}{2}(30)\left(\frac{1}{2}\right) = 7.5 \text{ m}$$$$\frac{\partial A}{\partial y} = \frac{1}{2}x\sin\theta = \frac{1}{2}(40)\left(\frac{1}{2}\right) = 10 \text{ m}$$$$\frac{\partial A}{\partial \theta} = \frac{1}{2}xy\cos\theta = \frac{1}{2}(40)(30)\left(\frac{\sqrt{3}}{2}\right) = 300\sqrt{3} \approx 519.62 \text{ m}^2$$

Maximum absolute error. With $\delta x = \delta y = 0.05$ m and $\delta\theta = 0.01$ rad,

$$\delta A \approx (7.5)(0.05) + (10)(0.05) + \left(300\sqrt{3}\right)(0.01) = 0.375 + 0.5 + 3\sqrt{3}$$$$\delta A \approx 0.875 + 5.196 = 6.071 \text{ m}^2 .$$

Maximum percentage error.

$$\frac{\delta A}{A}\times 100\% = \frac{6.071}{300}\times 100\% \approx 2.02\% .$$

The angle measurement dominates the error ($5.196$ of the $6.071 \text{ m}^2$), so the surveyor should refine $\theta$ first if a tighter area tolerance is needed.

Answer: $A = 300 \text{ m}^2$ (exactly $\frac{1}{2}(40)(30)\sin\frac{\pi}{6}$); maximum absolute error $\delta A = \left(\frac{7}{8} + 3\sqrt{3}\right) \approx 6.07 \text{ m}^2$; maximum percentage error $\approx 2.02\%$.

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Question 1(c) (3 marks)

Question

Find the equation of tangent plane and normal line to the surface

$$\ln\left(\frac{x}{y}\right) - z^2(x - 2y) - 3z = 6$$

at the point $(8, 4, -2)$.

Printing check

The printed point does not satisfy the printed surface exactly:

$$\ln\left(\frac{8}{4}\right) - (-2)^2(8 - 2\cdot 4) - 3(-2) = \ln 2 + 6 \approx 6.693 \neq 6 .$$

This is a typo in the paper (the constant is presumably $\ln 2 + 6$). We use the standard reading: the tangent plane and normal line at the given point are those of the level surface of $F$ through that point, i.e. the plane with normal $\nabla F(8,4,-2)$ passing through $(8,4,-2)$. The gradient — and therefore the answer — is unaffected by the constant.

Solution

Write the surface as a level surface $F(x,y,z) = 6$ with

$$F(x,y,z) = \ln x - \ln y - z^2(x - 2y) - 3z - 6 .$$

The gradient $\nabla F$ is normal to the surface, so it is the normal of the tangent plane and the direction vector of the normal line.

Gradient.

$$F_x = \frac{1}{x} - z^2, \qquad F_y = -\frac{1}{y} + 2z^2, \qquad F_z = -2zx + 4zy - 3$$

Evaluate at $(8,4,-2)$.

$$F_x = \frac{1}{8} - (-2)^2 = \frac{1}{8} - 4 = -\frac{31}{8}$$$$F_y = -\frac{1}{4} + 2(-2)^2 = -\frac{1}{4} + 8 = \frac{31}{4}$$$$F_z = -2(-2)(8) + 4(-2)(4) - 3 = 32 - 32 - 3 = -3$$

Tangent plane. A plane with normal $(F_x, F_y, F_z)$ through $(x_0,y_0,z_0)$ is

$$F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0 .$$

Substituting,

$$-\frac{31}{8}(x - 8) + \frac{31}{4}(y - 4) - 3(z + 2) = 0 .$$

Multiply through by $8$:

$$-31(x - 8) + 62(y - 4) - 24(z + 2) = 0$$$$-31x + 248 + 62y - 248 - 24z - 48 = 0$$$$31x - 62y + 24z + 48 = 0 .$$

Normal line. Use the direction vector $\nabla F(8,4,-2)$, scaled by $8$ to clear fractions: $(-31,\ 62,\ -24)$. Hence

$$\frac{x - 8}{-31} = \frac{y - 4}{62} = \frac{z + 2}{-24} = t ,$$

or in parametric form

$$x = 8 - 31t, \qquad y = 4 + 62t, \qquad z = -2 - 24t .$$

Check. At $t = 0$ the line passes through $(8,4,-2)$; and $31(8) - 62(4) + 24(-2) = 248 - 248 - 48 = -48$, so the plane contains the point. $\checkmark$

Answer: Tangent plane $31x - 62y + 24z + 48 = 0$; normal line $\dfrac{x-8}{-31} = \dfrac{y-4}{62} = \dfrac{z+2}{-24}$ (i.e. $x = 8 - 31t$, $y = 4 + 62t$, $z = -2 - 24t$).

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Question 2

Question 2(a) (7 marks)

Question

A project engineer has 200 acre of land for turf installation work with three different types of turfs A, B and C. The cost per acre for installing turf type A, B and C is RM 40, RM 60 and RM 80, respectively. The engineer has a total budget of RM 12,600 available for the installation work. Each acre of turf A requires 20 man-hours, each acre of turf B requires 25 man-hours and each acre of turf C requires 40 man-hours. The engineer has a maximum of 5950 man-hours available. If he wishes to use all of the available land, budget and manpower available, how many acres of each type of turf should be installed? Use Gaussian elimination method to solve the matrix.

Solution

Let $a$, $b$ and $c$ be the number of acres of turf A, B and C. "Use all the land, budget and manpower" gives three equations:

$$a + b + c = 200 \qquad \text{(land)}$$$$40a + 60b + 80c = 12600 \qquad \text{(budget, RM)}$$$$20a + 25b + 40c = 5950 \qquad \text{(manpower, man-hours)}$$

Divide the budget row by $20$ and the manpower row by $5$ to keep the numbers small:

$$2a + 3b + 4c = 630, \qquad 4a + 5b + 8c = 1190 .$$

Augmented matrix and Gaussian elimination.

$$\begin{pmatrix} 1 & 1 & 1 & 200 \\ 2 & 3 & 4 & 630 \\ 4 & 5 & 8 & 1190 \end{pmatrix}$$

Eliminate the first column: $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 4R_1$:

$$\begin{pmatrix} 1 & 1 & 1 & 200 \\ 0 & 1 & 2 & 230 \\ 0 & 1 & 4 & 390 \end{pmatrix}$$

Eliminate the second column: $R_3 \to R_3 - R_2$:

$$\begin{pmatrix} 1 & 1 & 1 & 200 \\ 0 & 1 & 2 & 230 \\ 0 & 0 & 2 & 160 \end{pmatrix}$$

Back-substitution.

$$2c = 160 \implies c = 80$$$$b + 2c = 230 \implies b = 230 - 2(80) = 70$$$$a + b + c = 200 \implies a = 200 - 70 - 80 = 50$$

Check.

$$40(50) + 60(70) + 80(80) = 2000 + 4200 + 6400 = 12600 \text{ RM} \quad \checkmark$$$$20(50) + 25(70) + 40(80) = 1000 + 1750 + 3200 = 5950 \text{ man-hours} \quad \checkmark$$

Answer: $50$ acres of turf A, $70$ acres of turf B and $80$ acres of turf C (so $a = 50$, $b = 70$, $c = 80$).

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Question 2(b) (8 marks)

Question

Company AAA Refinery requires 10 L of gasoline containing 2% additive. The company have drums of the following:

• Gasoline without additive
• Gasoline with 5% additive
• Gasoline with 6% additive

Company AAA Refinery need to use 4 times as much pure gasoline as 5% additive gasoline. How much of each drum is needed to produce gasoline which meets the requirement? [Hint: use total amount of gasoline in each drum as the unknown]

Solution

Let $x$, $y$ and $z$ be the litres taken from the drums of $0\%$, $5\%$ and $6\%$ additive gasoline respectively.

Volume balance (total must be 10 L):

$$x + y + z = 10$$

Additive balance (the blend must contain $2\%$ of 10 L $= 0.2$ L of additive):

$$0.05y + 0.06z = 0.2 \quad\Longleftrightarrow\quad 5y + 6z = 20$$

Mixing ratio (4 times as much pure as 5% additive):

$$x = 4y$$

Solve. Substituting $x = 4y$ into the volume balance:

$$4y + y + z = 10 \implies 5y + z = 10 .$$

Subtract this from the additive balance $5y + 6z = 20$:

$$(5y + 6z) - (5y + z) = 20 - 10 \implies 5z = 10 \implies z = 2 \text{ L}.$$

Then

$$5y = 10 - z = 8 \implies y = 1.6 \text{ L},$$$$x = 4y = 6.4 \text{ L}.$$

Equivalently, by Gaussian elimination on the augmented matrix

$$\begin{pmatrix} 1 & 1 & 1 & 10 \\ 0 & 5 & 6 & 20 \\ 1 & -4 & 0 & 0 \end{pmatrix}$$

Replace row 3 by row 3 minus row 1:

$$\begin{pmatrix} 1 & 1 & 1 & 10 \\ 0 & 5 & 6 & 20 \\ 0 & -5 & -1 & -10 \end{pmatrix}$$

Replace row 3 by row 3 plus row 2:

$$\begin{pmatrix} 1 & 1 & 1 & 10 \\ 0 & 5 & 6 & 20 \\ 0 & 0 & 5 & 10 \end{pmatrix}$$

giving $z = 2$, then $5y = 20 - 12 = 8$ so $y = 1.6$, then $x = 10 - 1.6 - 2 = 6.4$.

Check. $6.4 + 1.6 + 2 = 10$ L $\checkmark$ and the additive content is $0.05(1.6) + 0.06(2) = 0.08 + 0.12 = 0.2$ L $= 2\%$ of 10 L $\checkmark$

Answer: $6.4$ L of pure gasoline, $1.6$ L of 5% additive gasoline and $2$ L of 6% additive gasoline.

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Question 3

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

Question

The water draining process from a tank is governed by the following equation:

$$\frac{dh}{dt} = -\sqrt{2gh}\left(\frac{d}{D}\right)^2$$

where the $g$ is the gravity coefficient (i.e. $9.81 \text{ m/s}$) and $h$ is the height of water level function measured from the ground level. The diameter of the tank, $D$ is $5.6$ m and its outlet's diameter, $d$ is $0.1$ m.

Units in the printed paper

The paper writes the gravitational acceleration as "$9.81$ m/s" in both Q3(a) and Q3(b), which is a velocity unit; the correct unit is m/s$^{2}$. The value $g = 9.81$ m/s$^{2}$ is used throughout the solutions below.

(i) Given the initial height, $h_0$ of the water level is $4.0$ m, suggest a suitable strategy and use it to calculate the height of water level function in the tank over time, $h(t)$ as shown in Figure Q3(a).

Solution

Strategy: separation of variables. The right-hand side depends only on $h$ and the left-hand side only on $t$, so the ODE is separable (the equation is autonomous with $\sqrt{2g}\,(d/D)^2$ a constant).

Write $k = \sqrt{2g}\left(\frac{d}{D}\right)^2$ so that $\dfrac{dh}{dt} = -k\sqrt{h}$, and separate:

$$\frac{dh}{\sqrt{h}} = -k\,dt .$$

Integrate both sides:

$$\int h^{-1/2}\,dh = -k\int dt \implies 2\sqrt{h} = -kt + C .$$

Apply the initial condition $h(0) = h_0 = 4.0$ m:

$$2\sqrt{4.0} = C \implies C = 4 .$$

Hence

$$\sqrt{h} = 2 - \frac{k}{2}t = 2 - \frac{1}{2}\sqrt{2g}\left(\frac{d}{D}\right)^2 t ,$$

and squaring gives the height as a function of time:

$$h(t) = \left[\,2 - \frac{1}{2}\sqrt{2g}\left(\frac{d}{D}\right)^2 t\,\right]^2 .$$

Numerical form. With $g = 9.81$, $d = 0.1$, $D = 5.6$:

$$\sqrt{2g} = \sqrt{19.62} \approx 4.4294, \qquad \left(\frac{d}{D}\right)^2 = \left(\frac{0.1}{5.6}\right)^2 = \frac{1}{3136} \approx 3.1888\times10^{-4}$$$$\frac{1}{2}\sqrt{2g}\left(\frac{d}{D}\right)^2 = \frac{1}{2}(4.4294)(3.1888\times10^{-4}) \approx 7.0624\times10^{-4}\ \text{m}^{1/2}\,\text{s}^{-1}$$$$h(t) \approx \left(2 - 7.0624\times10^{-4}\,t\right)^2 \text{ m} .$$

Verification. Differentiating, $\dfrac{dh}{dt} = -2\left(7.0624\times10^{-4}\right)\left(2 - 7.0624\times10^{-4}t\right) = -\sqrt{2g}\left(\frac{d}{D}\right)^2\sqrt{h}$, which is the original ODE. Also $h(0) = 4.0$ m. $\checkmark$

Answer: $h(t) = \left[\,2 - \dfrac{1}{2}\sqrt{2g}\left(\dfrac{d}{D}\right)^2 t\,\right]^2$, i.e. $h(t) \approx \left(2 - 7.062\times10^{-4}\,t\right)^2$ m (valid for $0 \le t \le 2832$ s).

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

Question

Calculate the time (in second) required to empty the tank.

Solution

The tank is empty when $h(t) = 0$. From

$$h(t) = \left[\,2 - \frac{1}{2}\sqrt{2g}\left(\frac{d}{D}\right)^2 t\,\right]^2 = 0$$

the bracket must vanish:

$$2 = \frac{1}{2}\sqrt{2g}\left(\frac{d}{D}\right)^2 t \implies t = \frac{4}{\sqrt{2g}}\left(\frac{D}{d}\right)^2 = \frac{4D^2}{d^2\sqrt{2g}} .$$

Substitute the data:

$$t = \frac{4(5.6)^2}{(0.1)^2\sqrt{19.62}} = \frac{125.44}{(0.01)(4.4294)} = \frac{12544}{4.4294} \approx 2831.96 \text{ s} .$$

That is $2831.96/60 \approx 47.2$ minutes.

Answer: $t = \dfrac{4D^2}{d^2\sqrt{2g}} \approx 2832$ s (about $47.2$ minutes).

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Question 3(b) (8 marks)

Question

The landing-gear unit of an airplane is examined under drop test to ensure a safety and comfort landing, as shown in Figure Q3(b). The vibration at the airplane body, $x(t)$ due to the dropping height, $h$ can be governed by the simplified equation:

$$m\ddot{x} + c\dot{x} + kx = F(t)$$

where $m$ is the equivalent mass of the airplane, $c$ and $k$ are the damping and stiffness coefficients respectively of the landing-gear unit. $x$, $\dot{x}$ and $\ddot{x}$ are the displacement, velocity and acceleration of the mass respectively. $F(t) = mg$ is the forcing function where $g$ is the gravity coefficient (i.e. $9.81 \text{ m/s}$).

Given the dropping height of the testing is $h = 5.10$ m, determine the impact velocity at ground and the total vibration, $x(t)$ experienced by the airplane in this condition.

$$60000\ddot{x} + 120000\dot{x} + 3000000x = 60000g, \qquad x(0) = 0, \qquad \dot{x}(0) = \sqrt{2gh}$$

Hint: Euler's Formula: $e^{\pm ix} = \cos x \pm i(\sin x)$.

Solution

Impact velocity. The body falls freely through $h = 5.10$ m from rest, so at the moment of impact

$$v_0 = \sqrt{2gh} = \sqrt{2(9.81)(5.10)} = \sqrt{100.062} \approx 10.003 \text{ m/s} .$$

This is exactly the initial condition $\dot{x}(0) = \sqrt{2gh}$ supplied in the question.

Reduce the ODE. Divide the whole equation by $m = 60000$:

$$\ddot{x} + 2\dot{x} + 50x = g = 9.81 .$$

This is a linear second-order ODE with constant coefficients (forced, underdamped).

Complementary function. The auxiliary equation is

$$r^2 + 2r + 50 = 0 \implies r = \frac{-2 \pm \sqrt{4 - 200}}{2} = -1 \pm \sqrt{-49} = -1 \pm 7i .$$

Using Euler's formula $e^{(-1 \pm 7i)t} = e^{-t}\left(\cos 7t \pm i\sin 7t\right)$, the complementary function is

$$x_c(t) = e^{-t}\left(A\cos 7t + B\sin 7t\right),$$

with damped natural frequency $\omega_d = 7 \text{ rad/s}$ and damping ratio $\zeta = c/(2\sqrt{mk}) = 1/\sqrt{50} \approx 0.141$ (underdamped).

Particular integral. The forcing is the constant $mg$, so try $x_p = X$ constant:

$$0 + 0 + 50X = 9.81 \implies X = \frac{9.81}{50} = 0.1962 .$$

General solution.

$$x(t) = 0.1962 + e^{-t}\left(A\cos 7t + B\sin 7t\right) .$$

Apply the initial conditions. First, $x(0) = 0$:

$$0.1962 + A = 0 \implies A = -0.1962 .$$

Differentiate:

$$\dot{x}(t) = -e^{-t}\left(A\cos 7t + B\sin 7t\right) + e^{-t}\left(-7A\sin 7t + 7B\cos 7t\right) .$$

At $t = 0$, using $v_0 = 10.003$:

$$\dot{x}(0) = -A + 7B = 0.1962 + 7B = 10.003 \implies 7B = 9.8068 \implies B = 1.4010 .$$

Total vibration.

$$x(t) = 0.1962 + e^{-t}\left(-0.1962\cos 7t + 1.4010\sin 7t\right) \text{ m} .$$

Equivalently, combining the two harmonic terms into a single sine with amplitude $R = \sqrt{A^2 + B^2} = \sqrt{0.1962^2 + 1.4010^2} \approx 1.4147$ m and phase $\delta = \arctan(A/B) \approx -0.139$ rad:

$$x(t) \approx 0.1962 + 1.4147\,e^{-t}\sin\left(7t - 0.139\right) \text{ m} .$$

Check. $x(0) = 0.1962 - 0.1962 = 0$ $\checkmark$ and $\dot{x}(0) = 0.1962 + 7(1.4010) = 10.003$ m/s $\checkmark$

Answer: Impact velocity $v_0 = \sqrt{2gh} = \sqrt{100.062} \approx 10.00$ m/s; total vibration $x(t) = 0.1962 + e^{-t}\left(-0.1962\cos 7t + 1.401\sin 7t\right)$ m $\approx 0.1962 + 1.415\,e^{-t}\sin\left(7t - 0.139\right)$ m.

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Question 4

Question

Given two ordinary differential equations as follow:

$$2x^2y'' + 3xy' - (x^2 + 1)y = 0 \qquad \text{(4a)}$$ $$y'' - xy' - x^2y = 0 \qquad \text{(4b)}$$

For each equation, determine the characteristic of the point $x = 0$. Based on the results obtained, find either (a) or (b) for each equation.

• (a) Two linearly independent solutions using the Power Series Method at the point $x = 0$ (9 marks)
• (b) The indicial roots using the Frobenius Method at the point $x = 0$ plus determine whether two linearly independent Frobenius solutions can be found (6 marks)

Classification of $x = 0$.

Write each equation in the standard form $y'' + P(x)y' + Q(x)y = 0$ and test whether $xP(x)$ and $x^2Q(x)$ are analytic at $x = 0$.

Equation (4a). Dividing by $2x^2$,

$$P(x) = \frac{3}{2x}, \qquad Q(x) = -\frac{x^2 + 1}{2x^2} .$$

$P$ and $Q$ are singular at $x = 0$, but

$$xP(x) = \frac{3}{2}, \qquad x^2Q(x) = -\frac{x^2 + 1}{2}$$

are both analytic (indeed polynomial) at $x = 0$. Hence $x = 0$ is a regular singular point of (4a), so the Frobenius method (part b) applies.

Equation (4b). Here

$$P(x) = -x, \qquad Q(x) = -x^2$$

are both analytic at $x = 0$ (they are polynomials), so $x = 0$ is an ordinary point of (4b) and the power series method (part a) applies.

Question 4(a) (9 marks)

Question

Two linearly independent solutions using the Power Series Method at the point $x = 0$.

From the classification above, this applies to equation (4b):

$$y'' - xy' - x^2y = 0 .$$

Solution

Since $x = 0$ is an ordinary point, seek a power series solution

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

Re-index $y''$ and $x^2y$ onto powers of $x^n$:

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

Substitute into the ODE.

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

Collecting the coefficient of $x^n$:

$$2a_2 + 6a_3x + \sum_{n=2}^{\infty}\Big[(n+2)(n+1)a_{n+2} - n a_n - a_{n-2}\Big]x^n = 0 .$$

Every coefficient must vanish, so:

• $n = 0$: $2a_2 = 0 \implies a_2 = 0$
• $n = 1$: $6a_3 - a_1 = 0 \implies a_3 = \dfrac{a_1}{6}$
• $n \ge 2$: $(n+2)(n+1)a_{n+2} = n a_n + a_{n-2}$, i.e. the recurrence relation

$$a_{n+2} = \frac{n\,a_n + a_{n-2}}{(n+2)(n+1)}, \qquad n \ge 2 .$$

First solution (take $a_0 = 1$, $a_1 = 0$). Then $a_2 = a_3 = 0$ and all odd coefficients vanish. Applying the recurrence:

$$a_4 = \frac{2a_2 + a_0}{4\cdot3} = \frac{1}{12}, \qquad a_6 = \frac{4a_4 + a_2}{6\cdot5} = \frac{1}{90},$$$$a_8 = \frac{6a_6 + a_4}{8\cdot7} = \frac{3}{1120}, \qquad a_{10} = \frac{8a_8 + a_6}{10\cdot9} = \frac{41}{113400}$$$$y_1(x) = 1 + \frac{x^4}{12} + \frac{x^6}{90} + \frac{3x^8}{1120} + \frac{41x^{10}}{113400} + \cdots$$

Second solution (take $a_0 = 0$, $a_1 = 1$). Then $a_2 = 0$ and all even coefficients vanish:

$$a_3 = \frac{1}{6}, \qquad a_5 = \frac{3a_3 + a_1}{5\cdot4} = \frac{3}{40}, \qquad a_7 = \frac{5a_5 + a_3}{7\cdot6} = \frac{13}{1008}, \qquad a_9 = \frac{7a_7 + a_5}{9\cdot8} = \frac{119}{51840}$$$$y_2(x) = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{13x^7}{1008} + \frac{119x^9}{51840} + \cdots$$

Linear independence. $y_1$ is even with $y_1(0) = 1$, $y_2$ is odd with $y_2'(0) = 1$, so the Wronskian at the origin is

$$W(y_1,y_2)(0) = y_1(0)y_2'(0) - y_1'(0)y_2(0) = (1)(1) - (0)(0) = 1 \neq 0 .$$

Hence $y_1$ and $y_2$ are linearly independent, and the general solution is $y = c_1y_1 + c_2y_2$ with $c_1$, $c_2$ arbitrary constants. Because $P(x) = -x$ and $Q(x) = -x^2$ are entire, both series converge for all $x$.

Check (substitution of the first few terms). Substituting the truncated series back into $y'' - xy' - x^2y$ leaves $-\frac{571}{90720}x^{10} - \frac{41}{113400}x^{12} + \cdots$, i.e. terms of order $x^{10}$ and higher — the first two terms beyond those the recurrence determines — confirming the recurrence.

Answer: $y = c_1y_1 + c_2y_2$ where

$$y_1(x) = 1 + \frac{x^4}{12} + \frac{x^6}{90} + \frac{3x^8}{1120} + \frac{41x^{10}}{113400} + \cdots$$ $$y_2(x) = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{13x^7}{1008} + \frac{119x^9}{51840} + \cdots$$

These are linearly independent since $W(y_1,y_2)(0) = 1 \neq 0$.

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Question 4(b) (6 marks)

Question

The indicial roots using the Frobenius Method at the point $x = 0$ plus determine whether two linearly independent Frobenius solutions can be found.

From the classification above, this applies to equation (4a):

$$2x^2y'' + 3xy' - (x^2 + 1)y = 0 .$$

Solution

Since $x = 0$ is a regular singular point, apply the Frobenius method: seek

$$y = x^r\sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+r}, \qquad a_0 \neq 0 .$$

Then

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

Substitute into the ODE.

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

Re-index the $x^{n+r+2}$ sum with $n \to n-2$ and collect the coefficient of $x^{n+r}$:

$$\Big[2(n+r)(n+r-1) + 3(n+r) - 1\Big]a_n - a_{n-2} = 0, \qquad a_{-2} = a_{-1} = 0 .$$

Write the indicial polynomial

$$\phi(r) = 2r(r-1) + 3r - 1 = 2r^2 + r - 1 = (2r - 1)(r + 1) .$$

Indicial equation. Setting $n = 0$ (so $a_{-2} = 0$) and $a_0 \neq 0$:

$$\phi(r)a_0 = 0 \implies 2r^2 + r - 1 = 0 \implies (2r - 1)(r + 1) = 0$$$$r_1 = \frac{1}{2}, \qquad r_2 = -1 .$$

Recurrence relation. For $n \ge 1$,

$$a_n = \frac{a_{n-2}}{\phi(n+r)} = \frac{a_{n-2}}{2(n+r)^2 + (n+r) - 1} .$$

First root $r_1 = \frac{1}{2}$. Here

$$\phi\left(n + \tfrac{1}{2}\right) = 2\left(n+\tfrac{1}{2}\right)^2 + \left(n+\tfrac{1}{2}\right) - 1 = 2n^2 + 3n = n(2n+3),$$

so with $a_0 = 1$:

$$a_2 = \frac{1}{2\cdot 7} = \frac{1}{14}, \qquad a_4 = \frac{a_2}{4\cdot 11} = \frac{1}{616}, \qquad a_6 = \frac{a_4}{6\cdot 15} = \frac{1}{55440}, \qquad a_1 = a_3 = a_5 = 0$$$$y_1(x) = x^{1/2}\left(1 + \frac{x^2}{14} + \frac{x^4}{616} + \frac{x^6}{55440} + \cdots\right)$$

Second root $r_2 = -1$. Here

$$\phi(n - 1) = 2(n-1)^2 + (n-1) - 1 = 2n^2 - 3n = n(2n-3),$$

so with $a_0 = 1$:

$$a_2 = \frac{1}{2\cdot 1} = \frac{1}{2}, \qquad a_4 = \frac{a_2}{4\cdot 5} = \frac{1}{40}, \qquad a_6 = \frac{a_4}{6\cdot 9} = \frac{1}{2160}, \qquad a_1 = a_3 = a_5 = 0$$$$y_2(x) = x^{-1}\left(1 + \frac{x^2}{2} + \frac{x^4}{40} + \frac{x^6}{2160} + \cdots\right)$$

Can two linearly independent Frobenius solutions be found?

The two indicial roots are

$$r_1 - r_2 = \frac{1}{2} - (-1) = \frac{3}{2} .$$

Since the difference of the indicial roots is not an integer, Frobenius theory guarantees that the two series above are genuinely linearly independent (neither is a constant multiple of the other, and no logarithmic term is required). Their Wronskian is non-zero for $x > 0$, so the general solution on $x > 0$ is

$$y = C_1\,x^{1/2}\left(1 + \frac{x^2}{14} + \frac{x^4}{616} + \cdots\right) + C_2\,x^{-1}\left(1 + \frac{x^2}{2} + \frac{x^4}{40} + \cdots\right) .$$

Check. Substituting the truncated series into $2x^2y'' + 3xy' - (x^2+1)y$ leaves only the first omitted term ($-x^{17/2}/55440$ for $r_1$ and $-x^{7}/2160$ for $r_2$), confirming both recurrences.

Answer: Indicial roots $r_1 = \dfrac{1}{2}$ and $r_2 = -1$ (from $2r^2 + r - 1 = 0$). Because $r_1 - r_2 = \dfrac{3}{2}$ is not an integer, two linearly independent Frobenius solutions can be found:

$$y_1(x) = x^{1/2}\left(1 + \frac{x^2}{14} + \frac{x^4}{616} + \frac{x^6}{55440} + \cdots\right),$$ $$y_2(x) = x^{-1}\left(1 + \frac{x^2}{2} + \frac{x^4}{40} + \frac{x^6}{2160} + \cdots\right), \qquad x > 0 .$$

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