FAD1014: Rapid-Fire Drill Pack — Leak Topics

Objective: Master the 3 leak-highlighted topics for the final exam through high-volume mechanical drilling.
Target: 2–3 min per problem. If stalled >3 min, skip and mark for review.
Total problems: 45
Estimated time: ~100 min

Cheat Sheet (Memorize First)

Maclaurin & Taylor Series

Maclaurin (Taylor at $x=0$): $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$

Taylor (about $x = a$): $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$

Function Series Valid for
$e^x$ $\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ All $x$
$\sin x$ $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ All $x$
$\cos x$ $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ All $x$
$\ln(1+x)$ $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ $-1 < x \leq 1$
$\sinh x$ $\displaystyle \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$ All $x$
$\cosh x$ $\displaystyle \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots$ All $x$
$(1+x)^n$ $\displaystyle \sum_{r=0}^\infty \binom{n}{r} x^r = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots$ $\vert x \vert < 1$

Ellipse

Feature Horizontal ($a > b$) Vertical ($b > a$)
Standard (origin) $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
Centre $(h,k)$ $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$ $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$
Vertices $(h \pm a, k)$ $(h, k \pm b)$
Foci $(h \pm c, k)$, $c^2 = a^2 - b^2$ $(h, k \pm c)$, $c^2 = b^2 - a^2$
Major axis length $2a$ $2b$
Minor axis length $2b$ $2a$
Latus rectum $\dfrac{2b^2}{a}$ $\dfrac{2a^2}{b}$
General form $Ax^2 + By^2 + Cx + Dy + K = 0$, $A \neq B$, same sign

Differential Equations — Quick ID Table

Type Form Method
Separable $\dfrac{dy}{dx} = f(x)g(y)$ Separate: $\int \frac{dy}{g(y)} = \int f(x),dx$
First-order linear $\dfrac{dy}{dx} + P(x)y = Q(x)$ Integrating factor $I = e^{\int P,dx}$, then $yI = \int Q I,dx + C$
Bernoulli $\dfrac{dy}{dx} + P(x)y = Q(x)y^n$ ($n\neq0,1$) Substitute $v = y^{1-n}$ to reduce to linear
Exact $M,dx + N,dy = 0$, $\partial M/\partial y = \partial N/\partial x$ Find $F$ with $\partial F/\partial x = M$, $\partial F/\partial y = N$; solution $F = C$
Non-homogeneous 2nd order $ay'' + by' + cy = f(x)$ Complementary function (solve $ay''+by'+cy=0$) + particular integral

Part A: Maclaurin & Taylor Series

Target: 2–3 min per problem.

Set A1 — Standard Expansions from Memory (5 problems)

Write the Maclaurin series for each function up to and including the $x^5$ term.

  1. $e^{2x}$
  2. $\sin(3x)$
  3. $\cos(x^2)$
  4. $\ln(1 - x)$
  5. $(1 + 2x)^{1/2}$

Score: ___/5

Set A2 — Derivation from Definition (4 problems)

Derive the Maclaurin series using $f(x) = \sum f^{(n)}(0)x^n/n!$ up to the indicated order.

  1. $f(x) = \tan x$ up to $x^3$
  2. $f(x) = e^x \sin x$ up to $x^3$ (use multiplication of known series)
  3. $f(x) = \dfrac{\sin x}{x}$ up to $x^4$ (use division of known series)
  4. $f(x) = \dfrac{e^x - 1}{x}$ up to $x^4$

Score: ___/4

Set A3 — Taylor Series about $x = a$ (3 problems)

Find the Taylor series about the given point up to the indicated order.

  1. $f(x) = \sin x$ about $x = \pi/4$ up to $(x - \pi/4)^3$
  2. $f(x) = \sqrt{x}$ about $x = 4$ up to $(x - 4)^3$
  3. $f(x) = \ln x$ about $x = 2$ up to $(x - 2)^3$

Score: ___/3

Set A4 — Term-by-Term Integration (3 problems)

Use the Maclaurin series of the integrand to evaluate the indefinite integral as a power series. Write at least the first 3 non-zero terms.

  1. $\displaystyle \int \frac{\sin x}{x},dx$
  2. $\displaystyle \int e^{-x^2},dx$
  3. $\displaystyle \int \frac{1 - \cos x}{x^2},dx$

Score: ___/3

Part B: Ellipse

Target: 2–3 min per problem.

Set B1 — Standard Form to Features (4 problems)

For each ellipse in standard form, find: (i) centre, (ii) vertices, (iii) foci, (iv) length of major and minor axes, (v) length of latus rectum.

  1. $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$
  2. $\dfrac{(x-2)^2}{16} + \dfrac{(y+3)^2}{25} = 1$
  3. $\dfrac{(x+1)^2}{9} + \dfrac{(y-4)^2}{4} = 1$
  4. $4x^2 + 9y^2 = 36$ (Hint: divide to get 1 on RHS)

Score: ___/4

Set B2 — General to Standard (Completing Square) (4 problems)

Convert each general form to standard form by completing the square, then find centre, $a$, $b$, and $c$.

  1. $x^2 + 4y^2 + 6x - 16y + 21 = 0$
  2. $9x^2 + 4y^2 - 36x + 24y + 36 = 0$
  3. $x^2 + 9y^2 + 2x - 18y + 1 = 0$
  4. $4x^2 + y^2 - 8x + 4y - 8 = 0$

Score: ___/4

Set B3 — Equation from Given Data (4 problems)

Write the ellipse equation in standard form given the following information.

  1. Centre $(0,0)$, vertices $(\pm 5, 0)$, foci $(\pm 3, 0)$
  2. Centre $(0,0)$, vertices $(0, \pm 7)$, foci $(0, \pm 3)$
  3. Centre $(3, -2)$, vertices $(3 \pm 4, -2)$, foci $(3 \pm 2, -2)$
  4. Foci $(\pm 4, 0)$ and latus rectum length $= 18$ (centre at origin, horizontal)

Score: ___/4

Set B4 — Sketch & Identify (2 problems)

For each equation, determine orientation (horizontal/vertical), find centre, vertices, and foci, then describe the rough sketch.

  1. $\dfrac{(x+3)^2}{36} + \dfrac{(y-1)^2}{16} = 1$
  2. $25x^2 + 16y^2 = 400$ (Hint: divide first)

Score: ___/2

Part C: Differential Equations

Target: 2–3 min per problem.

Set C1 — Type Identification (5 problems)

Identify each DE as one of: separable / first-order linear / Bernoulli / exact / non-homogeneous 2nd order. Do NOT solve — just classify.

  1. $\dfrac{dy}{dx} = \dfrac{x^2}{1 + y^2}$
  2. $\dfrac{dy}{dx} + 2xy = x e^{-x^2}$
  3. $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2 y^3$
  4. $(2x + y),dx + (x - 3y),dy = 0$
  5. $y'' - 4y' + 3y = e^{2x}$

Score: ___/5

Set C2 — Separable & Linear DEs (5 problems)

Solve each DE. Find the general solution unless an initial condition is given.

  1. $\dfrac{dy}{dx} = \dfrac{x}{y}$, $y(0) = 2$
  2. $\dfrac{dy}{dx} = 3x^2 e^{-y}$
  3. $\dfrac{dy}{dx} + 3y = 6$, $y(0) = 1$
  4. $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$
  5. $x \dfrac{dy}{dx} - 2y = x^3$, $y(1) = 2$

Score: ___/5

Set C3 — Bernoulli & Exact DEs (3 problems)

Solve each DE.

  1. $\dfrac{dy}{dx} + \dfrac{y}{x} = x y^2$ (Bernoulli, $n=2$)
  2. $\dfrac{dy}{dx} - y = e^{2x} y^3$ (Bernoulli, $n=3$)
  3. $(2xy - 3),dx + (x^2 + 4y),dy = 0$

Score: ___/3

Set C4 — Non-Homogeneous 2nd Order (3 problems)

Find the general solution (complementary function + particular integral) for each.

  1. $y'' - 3y' + 2y = e^{3x}$
  2. $y'' + 4y = \sin 2x$ (Hint: particular integral takes form $x(A\cos 2x + B\sin 2x)$ due to resonance)
  3. $y'' - 2y' - 3y = 3x$

Score: ___/3

Final Scorecard

Part Sets Problems Raw Score
A — Maclaurin & Taylor Series A1, A2, A3, A4 15 ___/15
B — Ellipse B1, B2, B3, B4 14 ___/14
C — Differential Equations C1, C2, C3, C4 16 ___/16
TOTAL 45 ___/45

Proficiency Benchmarks

  • 32/45 (71%) — Proficient. You can handle standard exam problems.
  • 38/45 (84%) — Solid. Fast and accurate.
  • 42/45 (93%) — Exam-ready. Any mistake is a careless slip.

Speed Benchmarks

  • <75 min: Excellent mechanical fluency.
  • 75–100 min: Good. Review missed patterns.
  • >100 min: Drill the specific sets you scored lowest on again tomorrow.

Error Log Template

After grading, list every wrong problem number with a one-word reason:

Problem Reason
e.g. 5 forgot binomial sign

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Set A1 — Standard Expansions from Memory

  1. $e^{2x} = 1 + 2x + \dfrac{(2x)^2}{2!} + \dfrac{(2x)^3}{3!} + \dfrac{(2x)^4}{4!} + \dfrac{(2x)^5}{5!} + \cdots = 1 + 2x + 2x^2 + \dfrac{4}{3}x^3 + \dfrac{2}{3}x^4 + \dfrac{4}{15}x^5 + \cdots$

  2. $\sin(3x) = 3x - \dfrac{(3x)^3}{3!} + \dfrac{(3x)^5}{5!} + \cdots = 3x - \dfrac{27}{6}x^3 + \dfrac{243}{120}x^5 + \cdots = 3x - \dfrac{9}{2}x^3 + \dfrac{81}{40}x^5 + \cdots$

  3. $\cos(x^2) = 1 - \dfrac{(x^2)^2}{2!} + \dfrac{(x^2)^4}{4!} - \cdots = 1 - \dfrac{x^4}{2} + \dfrac{x^8}{24} - \cdots$

  4. $\ln(1 - x) = \ln(1 + (-x)) = (-x) - \dfrac{(-x)^2}{2} + \dfrac{(-x)^3}{3} - \dfrac{(-x)^4}{4} + \dfrac{(-x)^5}{5} + \cdots = -x - \dfrac{x^2}{2} - \dfrac{x^3}{3} - \dfrac{x^4}{4} - \dfrac{x^5}{5} - \cdots$

  5. $(1 + 2x)^{1/2} = 1 + \dfrac{1}{2}(2x) + \dfrac{\frac{1}{2}(-\frac{1}{2})}{2!}(2x)^2 + \dfrac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{3!}(2x)^3 + \dfrac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{4!}(2x)^4 + \dfrac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2})}{5!}(2x)^5 + \cdots$ $$= 1 + x - \dfrac{1}{2}x^2 + \dfrac{1}{2}x^3 - \dfrac{5}{8}x^4 + \dfrac{7}{8}x^5 + \cdots$$

Set A2 — Derivation from Definition

  1. $f(x) = \tan x$, $f(0)=0$, $f'(x)=\sec^2 x$, $f'(0)=1$, $f''(x)=2\sec^2 x\tan x$, $f''(0)=0$, $f'''(x)=2(2\sec^2 x\tan^2 x + \sec^4 x)$, $f'''(0)=2$ $$\tan x = 0 + 1\cdot x + 0\cdot x^2 + \dfrac{2}{3!}x^3 + \cdots = x + \dfrac{x^3}{3} + \cdots$$

  2. $e^x \sin x = \left(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\right)\left(x - \dfrac{x^3}{3!} + \cdots\right)$ $$= x + x^2 + \left(-\dfrac{1}{6} + \dfrac{1}{2}\right)x^3 + \cdots = x + x^2 + \dfrac{1}{3}x^3 + \cdots$$

  3. $\dfrac{\sin x}{x} = \dfrac{1}{x}\left(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots\right) = 1 - \dfrac{x^2}{6} + \dfrac{x^4}{120} - \cdots$

  4. $\dfrac{e^x - 1}{x} = \dfrac{1}{x}\left(\left(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \cdots\right) - 1\right) = \dfrac{1}{x}\left(x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \cdots\right)$ $$= 1 + \dfrac{x}{2} + \dfrac{x^2}{6} + \dfrac{x^3}{24} + \cdots$$

Set A3 — Taylor Series about $x = a$

  1. $f(x)=\sin x$, $a=\pi/4$. $f(\pi/4)=\sqrt{2}/2$, $f'(\pi/4)=\sqrt{2}/2$, $f''(\pi/4)=-\sqrt{2}/2$, $f'''(\pi/4)=-\sqrt{2}/2$ $$\sin x = \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}\left(x-\dfrac{\pi}{4}\right) - \dfrac{\sqrt{2}}{2\cdot 2!}\left(x-\dfrac{\pi}{4}\right)^2 - \dfrac{\sqrt{2}}{2\cdot 3!}\left(x-\dfrac{\pi}{4}\right)^3 + \cdots$$ $$= \dfrac{\sqrt{2}}{2}\left[1 + \left(x-\dfrac{\pi}{4}\right) - \dfrac{1}{2}\left(x-\dfrac{\pi}{4}\right)^2 - \dfrac{1}{6}\left(x-\dfrac{\pi}{4}\right)^3 + \cdots\right]$$

  2. $f(x)=\sqrt{x}$, $a=4$. $f(4)=2$, $f'(x)=1/(2\sqrt{x})$, $f'(4)=1/4$, $f''(x)=-1/(4x^{3/2})$, $f''(4)=-1/32$, $f'''(x)=3/(8x^{5/2})$, $f'''(4)=3/256$ $$\sqrt{x} = 2 + \dfrac{1}{4}(x-4) - \dfrac{1}{32\cdot 2!}(x-4)^2 + \dfrac{3}{256\cdot 3!}(x-4)^3 + \cdots$$ $$= 2 + \dfrac{1}{4}(x-4) - \dfrac{1}{64}(x-4)^2 + \dfrac{1}{512}(x-4)^3 + \cdots$$

  3. $f(x)=\ln x$, $a=2$. $f(2)=\ln 2$, $f'(x)=1/x$, $f'(2)=1/2$, $f''(x)=-1/x^2$, $f''(2)=-1/4$, $f'''(x)=2/x^3$, $f'''(2)=1/4$ $$\ln x = \ln 2 + \dfrac{1}{2}(x-2) - \dfrac{1}{4\cdot 2!}(x-2)^2 + \dfrac{1}{4\cdot 3!}(x-2)^3 + \cdots$$ $$= \ln 2 + \dfrac{1}{2}(x-2) - \dfrac{1}{8}(x-2)^2 + \dfrac{1}{24}(x-2)^3 + \cdots$$

Set A4 — Term-by-Term Integration

  1. $\dfrac{\sin x}{x} = 1 - \dfrac{x^2}{3!} + \dfrac{x^4}{5!} - \dfrac{x^6}{7!} + \cdots = 1 - \dfrac{x^2}{6} + \dfrac{x^4}{120} - \dfrac{x^6}{5040} + \cdots$ $$\displaystyle\int \dfrac{\sin x}{x},dx = \int \left(1 - \dfrac{x^2}{6} + \dfrac{x^4}{120} - \cdots\right)dx = C + x - \dfrac{x^3}{18} + \dfrac{x^5}{600} - \cdots$$

  2. $e^{-x^2} = 1 - x^2 + \dfrac{x^4}{2!} - \dfrac{x^6}{3!} + \cdots = 1 - x^2 + \dfrac{x^4}{2} - \dfrac{x^6}{6} + \cdots$ $$\displaystyle\int e^{-x^2},dx = \int \left(1 - x^2 + \dfrac{x^4}{2} - \dfrac{x^6}{6} + \cdots\right)dx = C + x - \dfrac{x^3}{3} + \dfrac{x^5}{10} - \dfrac{x^7}{42} + \cdots$$

  3. $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \cdots$, so $1 - \cos x = \dfrac{x^2}{2} - \dfrac{x^4}{24} + \dfrac{x^6}{720} - \cdots$ $$\dfrac{1 - \cos x}{x^2} = \dfrac{1}{2} - \dfrac{x^2}{24} + \dfrac{x^4}{720} - \cdots$$ $$\displaystyle\int \dfrac{1 - \cos x}{x^2},dx = \int \left(\dfrac{1}{2} - \dfrac{x^2}{24} + \dfrac{x^4}{720} - \cdots\right)dx = C + \dfrac{x}{2} - \dfrac{x^3}{72} + \dfrac{x^5}{3600} - \cdots$$

Set B1 — Standard Form to Features

  1. $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$. $a^2=25$, $b^2=9$, $a=5$, $b=3$. Horizontal ($a>b$). $c^2=25-9=16$, $c=4$.

    • Centre: $(0,0)$
    • Vertices: $(\pm 5, 0)$
    • Foci: $(\pm 4, 0)$
    • Major axis: $10$, Minor axis: $6$
    • Latus rectum: $2b^2/a = 2(9)/5 = 18/5 = 3.6$

  2. $\dfrac{(x-2)^2}{16} + \dfrac{(y+3)^2}{25} = 1$. $a^2=16$, $b^2=25$, so $b>a$ (vertical). $b=5$, $a=4$. $c^2=25-16=9$, $c=3$.

    • Centre: $(2, -3)$
    • Vertices: $(2, -3 \pm 5) = (2, 2)$ and $(2, -8)$
    • Foci: $(2, -3 \pm 3) = (2, 0)$ and $(2, -6)$
    • Major axis: $10$, Minor axis: $8$
    • Latus rectum: $2a^2/b = 2(16)/5 = 32/5 = 6.4$

  3. $\dfrac{(x+1)^2}{9} + \dfrac{(y-4)^2}{4} = 1$. $a^2=9$, $b^2=4$, $a=3$, $b=2$. Horizontal ($a>b$). $c^2=9-4=5$, $c=\sqrt{5}$.

    • Centre: $(-1, 4)$
    • Vertices: $(-1 \pm 3, 4) = (2, 4)$ and $(-4, 4)$
    • Foci: $(-1 \pm \sqrt{5}, 4)$
    • Major axis: $6$, Minor axis: $4$
    • Latus rectum: $2b^2/a = 2(4)/3 = 8/3$

  4. $4x^2 + 9y^2 = 36 \Rightarrow \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$. $a^2=9$, $b^2=4$, $a=3$, $b=2$. Horizontal. $c^2=9-4=5$, $c=\sqrt{5}$.

    • Centre: $(0,0)$
    • Vertices: $(\pm 3, 0)$
    • Foci: $(\pm \sqrt{5}, 0)$
    • Major axis: $6$, Minor axis: $4$
    • Latus rectum: $2b^2/a = 2(4)/3 = 8/3$

Set B2 — General to Standard (Completing Square)

  1. $x^2 + 4y^2 + 6x - 16y + 21 = 0$ $$(x^2 + 6x) + 4(y^2 - 4y) = -21$$ $$(x^2 + 6x + 9) + 4(y^2 - 4y + 4) = -21 + 9 + 16$$ $$(x+3)^2 + 4(y-2)^2 = 4$$ $$\dfrac{(x+3)^2}{4} + \dfrac{(y-2)^2}{1} = 1$$ Centre $(-3, 2)$, $a=2$, $b=1$, $c^2=4-1=3$, $c=\sqrt{3}$. Horizontal.

  2. $9x^2 + 4y^2 - 36x + 24y + 36 = 0$ $$9(x^2 - 4x) + 4(y^2 + 6y) = -36$$ $$9(x^2 - 4x + 4) + 4(y^2 + 6y + 9) = -36 + 36 + 36$$ $$9(x-2)^2 + 4(y+3)^2 = 36$$ $$\dfrac{(x-2)^2}{4} + \dfrac{(y+3)^2}{9} = 1$$ Centre $(2, -3)$, $a=2$, $b=3$, $c^2=9-4=5$, $c=\sqrt{5}$. Vertical (since $b>a$).

  3. $x^2 + 9y^2 + 2x - 18y + 1 = 0$ $$(x^2 + 2x) + 9(y^2 - 2y) = -1$$ $$(x^2 + 2x + 1) + 9(y^2 - 2y + 1) = -1 + 1 + 9$$ $$(x+1)^2 + 9(y-1)^2 = 9$$ $$\dfrac{(x+1)^2}{9} + \dfrac{(y-1)^2}{1} = 1$$ Centre $(-1, 1)$, $a=3$, $b=1$, $c^2=9-1=8$, $c=2\sqrt{2}$. Horizontal.

  4. $4x^2 + y^2 - 8x + 4y - 8 = 0$ $$4(x^2 - 2x) + (y^2 + 4y) = 8$$ $$4(x^2 - 2x + 1) + (y^2 + 4y + 4) = 8 + 4 + 4$$ $$4(x-1)^2 + (y+2)^2 = 16$$ $$\dfrac{(x-1)^2}{4} + \dfrac{(y+2)^2}{16} = 1$$ Centre $(1, -2)$, $a=2$, $b=4$, $c^2=16-4=12$, $c=2\sqrt{3}$. Vertical.

Set B3 — Equation from Given Data

  1. Centre $(0,0)$, vertices $(\pm 5, 0)$ so $a=5$, foci $(\pm 3, 0)$ so $c=3$. $b^2 = a^2 - c^2 = 25 - 9 = 16$, $b=4$. $$\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$$

  2. Centre $(0,0)$, vertices $(0, \pm 7)$ so $b=7$, foci $(0, \pm 3)$ so $c=3$. $a^2 = b^2 - c^2 = 49 - 9 = 40$, $a=2\sqrt{10}$. $$\dfrac{x^2}{40} + \dfrac{y^2}{49} = 1$$

  3. Centre $(3, -2)$, vertices $(3 \pm 4, -2)$ so $a=4$, foci $(3 \pm 2, -2)$ so $c=2$. $b^2 = a^2 - c^2 = 16 - 4 = 12$, $b=2\sqrt{3}$. $$\dfrac{(x-3)^2}{16} + \dfrac{(y+2)^2}{12} = 1$$

  4. Foci $(\pm 4, 0)$ so $c=4$ and centre $(0,0)$, horizontal. Latus rectum $= 2b^2/a = 18$. Also $c^2 = a^2 - b^2 = 16$. So $b^2 = a^2 - 16$. Sub into latus rectum: $2(a^2-16)/a = 18 \Rightarrow 2a^2 - 32 = 18a \Rightarrow 2a^2 - 18a - 32 = 0 \Rightarrow a^2 - 9a - 16 = 0$. $a = \dfrac{9 \pm \sqrt{81+64}}{2} = \dfrac{9 \pm \sqrt{145}}{2}$. Take positive: $a = \dfrac{9 + \sqrt{145}}{2}$. Then $b^2 = a^2 - 16$. $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2-16} = 1$ where $a = \dfrac{9 + \sqrt{145}}{2}$.

Set B4 — Sketch & Identify

  1. $\dfrac{(x+3)^2}{36} + \dfrac{(y-1)^2}{16} = 1$. Centre $(-3, 1)$. $a=6$, $b=4$. Horizontal ($a>b$). $c^2=36-16=20$, $c=2\sqrt{5} \approx 4.47$. Vertices: $(-3 \pm 6, 1) = (3,1)$ and $(-9,1)$. Foci: $(-3 \pm 2\sqrt{5}, 1)$. Major axis $12$ horizontal, minor axis $8$ vertical. Co-vertices at $(-3, 1\pm 4) = (-3,5)$ and $(-3,-3)$.

  2. $25x^2 + 16y^2 = 400 \Rightarrow \dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$. Centre $(0,0)$. $a=4$, $b=5$. Vertical ($b>a$). $c^2=25-16=9$, $c=3$. Vertices: $(0, \pm 5)$. Foci: $(0, \pm 3)$. Major axis $10$ vertical, minor axis $8$ horizontal. Co-vertices at $(\pm 4, 0)$.

Set C1 — Type Identification

  1. $\dfrac{dy}{dx} = \dfrac{x^2}{1 + y^2}$ — Separable (can write as $(1+y^2),dy = x^2,dx$)

  2. $\dfrac{dy}{dx} + 2xy = x e^{-x^2}$ — First-order linear (form $dy/dx + P(x)y = Q(x)$ with $P=2x$, $Q=xe^{-x^2}$)

  3. $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2 y^3$ — Bernoulli ($n=3$, RHS has $y^3$)

  4. $(2x + y),dx + (x - 3y),dy = 0$ — Exact (check: $\partial M/\partial y = 1$, $\partial N/\partial x = 1$, they are equal)

  5. $y'' - 4y' + 3y = e^{2x}$ — Non-homogeneous 2nd order (2nd derivative present, RHS $\neq 0$)

Set C2 — Separable & Linear DEs

  1. $\dfrac{dy}{dx} = \dfrac{x}{y}$, $y(0) = 2$. $$y,dy = x,dx \Rightarrow \int y,dy = \int x,dx \Rightarrow \dfrac{y^2}{2} = \dfrac{x^2}{2} + C \Rightarrow y^2 = x^2 + 2C$$ $y(0)=2 \Rightarrow 4 = 0 + 2C \Rightarrow C=2$. So $y^2 = x^2 + 4$, $y = \sqrt{x^2 + 4}$ (positive root since $y(0)>0$).

  2. $\dfrac{dy}{dx} = 3x^2 e^{-y}$. $$e^y,dy = 3x^2,dx \Rightarrow \int e^y,dy = \int 3x^2,dx \Rightarrow e^y = x^3 + C$$ $$y = \ln(x^3 + C)$$

  3. $\dfrac{dy}{dx} + 3y = 6$, $y(0) = 1$. Integrating factor: $I = e^{\int 3,dx} = e^{3x}$. $$y e^{3x} = \int 6 e^{3x},dx = 2e^{3x} + C \Rightarrow y = 2 + Ce^{-3x}$$ $y(0)=1 \Rightarrow 1 = 2 + C \Rightarrow C = -1$. So $y = 2 - e^{-3x}$.

  4. $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$. Integrating factor: $I = e^{\int (1/x),dx} = e^{\ln x} = x$. $$y \cdot x = \int x^2 \cdot x,dx = \int x^3,dx = \dfrac{x^4}{4} + C$$ $$y = \dfrac{x^3}{4} + \dfrac{C}{x}$$

  5. $x \dfrac{dy}{dx} - 2y = x^3$, $y(1) = 2$. Rewrite: $\dfrac{dy}{dx} - \dfrac{2}{x}y = x^2$. Integrating factor: $I = e^{\int (-2/x),dx} = e^{-2\ln x} = x^{-2}$. $$y \cdot x^{-2} = \int x^2 \cdot x^{-2},dx = \int 1,dx = x + C$$ $$y = x^2(x + C) = x^3 + Cx^2$$ $y(1)=2 \Rightarrow 2 = 1 + C \Rightarrow C = 1$. So $y = x^3 + x^2$.

Set C3 — Bernoulli & Exact DEs

  1. $\dfrac{dy}{dx} + \dfrac{y}{x} = x y^2$ (Bernoulli, $n=2$). Let $v = y^{1-2} = y^{-1}$. Then $\dfrac{dv}{dx} = -y^{-2}\dfrac{dy}{dx}$. Multiply original by $-y^{-2}$: $-y^{-2}\dfrac{dy}{dx} - \dfrac{1}{x}y^{-1} = -x$ $$\dfrac{dv}{dx} - \dfrac{1}{x}v = -x$$ Integrating factor: $I = e^{\int (-1/x),dx} = e^{-\ln x} = 1/x$. $$v \cdot \dfrac{1}{x} = \int (-x) \cdot \dfrac{1}{x},dx = \int -1,dx = -x + C$$ $$v = -x^2 + Cx \Rightarrow y^{-1} = Cx - x^2 \Rightarrow y = \dfrac{1}{Cx - x^2}$$

  2. $\dfrac{dy}{dx} - y = e^{2x} y^3$ (Bernoulli, $n=3$). Let $v = y^{1-3} = y^{-2}$. Then $\dfrac{dv}{dx} = -2y^{-3}\dfrac{dy}{dx}$. Multiply original by $-2y^{-3}$: $-2y^{-3}\dfrac{dy}{dx} + 2y^{-2} = -2e^{2x}$ $$\dfrac{dv}{dx} + 2v = -2e^{2x}$$ Integrating factor: $I = e^{\int 2,dx} = e^{2x}$. $$v e^{2x} = \int -2e^{2x} \cdot e^{2x},dx = \int -2e^{4x},dx = -\dfrac{1}{2}e^{4x} + C$$ $$v = -\dfrac{1}{2}e^{2x} + Ce^{-2x} \Rightarrow y^{-2} = Ce^{-2x} - \dfrac{1}{2}e^{2x}$$ $$y^2 = \dfrac{1}{Ce^{-2x} - \frac{1}{2}e^{2x}}$$

  3. $(2xy - 3),dx + (x^2 + 4y),dy = 0$. $M = 2xy - 3$, $N = x^2 + 4y$. $\partial M/\partial y = 2x$, $\partial N/\partial x = 2x$. They are equal, so exact. Find $F$: $\partial F/\partial x = M = 2xy - 3 \Rightarrow F = \int (2xy - 3),dx = x^2 y - 3x + g(y)$. $\partial F/\partial y = x^2 + g'(y) = N = x^2 + 4y \Rightarrow g'(y) = 4y \Rightarrow g(y) = 2y^2$. So $F = x^2 y - 3x + 2y^2$. Solution: $x^2 y - 3x + 2y^2 = C$.

Set C4 — Non-Homogeneous 2nd Order

  1. $y'' - 3y' + 2y = e^{3x}$. Auxiliary: $r^2 - 3r + 2 = 0 \Rightarrow (r-1)(r-2)=0 \Rightarrow r=1, 2$. Complementary: $y_c = C_1 e^{x} + C_2 e^{2x}$. Particular: Try $y_p = A e^{3x}$. Then $y_p' = 3A e^{3x}$, $y_p'' = 9A e^{3x}$. Substitute: $9A e^{3x} - 9A e^{3x} + 2A e^{3x} = 2A e^{3x} = e^{3x} \Rightarrow A = 1/2$. $y_p = \dfrac{1}{2}e^{3x}$. General: $y = C_1 e^{x} + C_2 e^{2x} + \dfrac{1}{2}e^{3x}$.

  2. $y'' + 4y = \sin 2x$. Auxiliary: $r^2 + 4 = 0 \Rightarrow r = \pm 2i$. Complementary: $y_c = C_1 \cos 2x + C_2 \sin 2x$. Resonance: $\sin 2x$ is in CF, so try $y_p = x(A\cos 2x + B\sin 2x)$. $$y_p' = (A\cos 2x + B\sin 2x) + x(-2A\sin 2x + 2B\cos 2x)$$ $$y_p'' = 2(-2A\sin 2x + 2B\cos 2x) + x(-4A\cos 2x - 4B\sin 2x)$$ $$= -4A\sin 2x + 4B\cos 2x - 4Ax\cos 2x - 4Bx\sin 2x$$ Substitute into $y'' + 4y$: $$(-4A\sin 2x + 4B\cos 2x - 4Ax\cos 2x - 4Bx\sin 2x) + 4x(A\cos 2x + B\sin 2x) = \sin 2x$$ $$-4A\sin 2x + 4B\cos 2x = \sin 2x$$ So $4B = 0 \Rightarrow B=0$, and $-4A = 1 \Rightarrow A = -1/4$. $y_p = -\dfrac{x}{4}\cos 2x$. General: $y = C_1 \cos 2x + C_2 \sin 2x - \dfrac{x}{4}\cos 2x$.

  3. $y'' - 2y' - 3y = 3x$. Auxiliary: $r^2 - 2r - 3 = 0 \Rightarrow (r-3)(r+1)=0 \Rightarrow r=3, -1$. Complementary: $y_c = C_1 e^{3x} + C_2 e^{-x}$. Particular: Try $y_p = Ax + B$. Then $y_p' = A$, $y_p'' = 0$. Substitute: $0 - 2A - 3(Ax + B) = -3Ax - (2A + 3B) = 3x$. So $-3A = 3 \Rightarrow A = -1$, and $-(2(-1) + 3B) = 0 \Rightarrow 2 - 3B = 0 \Rightarrow B = 2/3$. $y_p = -x + \dfrac{2}{3}$. General: $y = C_1 e^{3x} + C_2 e^{-x} - x + \dfrac{2}{3}$.

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