Tutorials

Tutorial 1: Review of Algebra and Calculus

Outcomes: At the end of this tutorial you should be able to

- Understand the rules of the natural exponent and logarithmic functions.
- Solve equations containing these functions.
- Differentiate and integrate functions using the associated rules.
- Differentiate and integrate functions using the associated methods

Logarithmic Equations

Solve each of the following logarithmic equations for \(x\):

  1. \(\ln{(5x+4)}=0\)

  2. \(\ln{x}+\ln{0.2}=\ln{e}\)

  3. \(\ln{4x^2}-\ln{16}=\ln{4}-\ln{2x}\)

Revision of Differentiation and Integration Techniques

Differentiate the following functions:

  1. \(x(7x+8)^{2}\)

  2. \(\dfrac{5x^{2}}{(2x^{3}+4)^{4}}\)

Integrate the following functions:

  1. \((2x+1)\ln(x+1)\)

  2. \(x e^{x^{2}}\)

Multiple Choice Questions

  1. If \(f(x)=(x^2-3x)^6(4-x)^5\), then \(f'(x)\) is equal to:

\(\text{(A) } (x^2-3x)^5(4-x)^4(5x^2-21x+24)\)

\(\text{(B) } -(x^2-3x)^5(4-x)^4(7x^2-51x+72)\)

\(\text{(C) } -(x^2-3x)^5(4-x)^4(17x^2-81x+72)\)

\(\text{(D) } 6(x^2-3x)^5(2x-3)(-5)(4-x)^4\)

\(\text{(E) none of these}\)

2) If \(y=e^x \ln{x}\), then \(\dfrac{dy}{dx}\) is equal to:

\(\text{(A) } \dfrac{e^x(1+x\ln{x})}{x}\)

\(\text{(B) } e^x(x+\ln{x})\)

\(\text{(C) } xe^x\)

\(\text{(D) } \dfrac{x}{e^x}\)

\(\text{(E) none of these}\)

3) If \(\dfrac{500}{12+5e^{-0.5x}}\), then \(\dfrac{dy}{dx}\) is equal to:

\(\text{(A) } 500(-1)(12+5e^{-0.5x})^{-2}\)

\(\text{(B) } 500(-1)(12+5e^{-0.5x})^{-2}(12+5e^{-0.5x})\)

\(\text{(C) } 500(-1)(12+5e^{-0.5x})^{-2}(-2.5e^{-0.5x})\)

\(\text{(D) } 500(-1)(5(0.5)e^{-0.5x})^{-2}\)

\(\text{(E) none of these}\)

4) \({\displaystyle \int \dfrac{x^2+4x-\sqrt{x}}{x^2} dx}\), expressed in terms of an arbitrary constant \(c\) is equal to:

\(\text{(A) } x+\ln{4x}-\frac{2}{3}x^{\frac{3}{2}}+c\)

\(\text{(B) } x+\ln{4x}+\dfrac{2}{\sqrt{x}}+c\)

\(\text{(C) } x+4\ln{x}+\dfrac{1}{2\sqrt{x}}+c\)

\(\text{(D) } x+4\ln{x}+\dfrac{2}{\sqrt{x}}+c\)

\(\text{(E) none of these}\)

5) \({\displaystyle \int \left(\dfrac{3x}{2}-\dfrac{9}{4}\right) e^{x(x-3)} dx}\), expressed in terms of an arbitrary constant \(c\) is equal to:

\(\text{(A) } \dfrac{3}{4}e^{x^2-3x}+c\)

\(\text{(B) } e^{x^2-3x}+c\)

\(\text{(C) } \dfrac{9}{4}e^{x^2-3x}+c\)

\(\text{(D) } \dfrac{3}{2}e^{x^2-3x}+c\)

\(\text{(E) } \dfrac{4}{3}e^{x^2-3x}+c\)

6) \({\displaystyle \int (3x+1)^{2} e^{-2x} dx}\), expressed in terms of an arbitrary constant \(c\) is equal to:

\(\text{(A) } -0.5(3x+1) e^{-2x}+1.5(3x+1)^{2} e^{-2x}-2.25 e^{-2x} + c\)

\(\text{(B) } -0.5(3x+1)^{2} e^{-2x}+1.5(3x+1) e^{-2x}+2.25 e^{-2x} + c\)

\(\text{(C) } -0.5(3x+1)^{2} e^{-2x}-1.5(3x+1) e^{-2x}-2.25 e^{-2x} + c\)

\(\text{(D) }-0.5(3x+1)^{2} e^{-2x}+1.5(3x+1) e^{-2x}-2.25 e^{-2x} + c\)

\(\text{(E) none of these}\)

Tutorial 2: Classification, Solutions, and Direction Fields

Outcomes: At the end of this tutorial you should be able to

- Explain what an ordinary differential equation is, and why classifying one is a useful first step.
- Classify a given differential equation by its order, linearity, homogeneity, and coefficients, and justify the classification.
- Verify by substitution whether a given function is a solution of a differential equation.
- Sketch a family of solution curves.
- Sketch and interpret a direction field for a first-order equation, and relate it to the family of solution curves without solving the equation.

Conceptual Questions

  1. Explain the ways in which one can classify a differential equation, and why classification matters for what comes after it.
  2. What does the degree of a differential equation inform us about its linearity?
  3. In your own words: what does a single lineal element at a point \((x,y)\) tell you, and where does that information come from?

Classification of ODEs

Classify the following differential equations by checking first if they are ODEs, and then in terms of their order, linearity, homogeneity, and coefficients. Give reasons for your answers.

  1. \(t \dfrac{d^3 x}{dt^3}-2\left(\dfrac{d x}{dt}\right) ^4+x=0\)
  2. \(y y'+2y=1+x^2\), where \(y=y(x)\)
  3. \(y''+9y=\sin y\), where \(y=y(x)\)
  4. \(\dfrac{d^2 R}{dt^2}=\dfrac{\kappa}{R^2}\), where \(\kappa\) (read: kappa) is a constant
  5. \(x^5\dfrac{d^4 y}{dx^4}-x^3 \dfrac{d^3 y}{dx^3}+6y=0\)
  6. \(\dfrac{d^2 x}{dt^2}=\sqrt{1+\left( \dfrac{d x}{dt}\right) ^2}\)

Solutions of ODEs

  1. In the questions below, verify that the indicated function is an explicit solution of the given differential equation, where \(y=y(x)\).
      1. \(2y'+y=0; \quad y= e^{-x/2}\)
      1. \(y''+y=\tan x; \quad y=-(\cos x)\ln(\sec x+\tan x)\)

  2. In the question below, verify that the indicated family of functions is a solution of the given differential equation.         \(x^3\dfrac{\textrm{d}^3 y}{\textrm{d}x^3}+2x^2 \dfrac{\textrm{d}^2 y}{\textrm{d}x^2}-x\dfrac{\textrm{d} y}{\textrm{d}x}+y=12x^2; \quad y=c_1x^{-1} +c_2x+c_3x \ln x+4x^2\),         where \(c_1\), \(c_2\), \(c_3\) are arbitrary constants.

  3. Draw a rough sketch or use a graphing calculator (like Desmos) to find the family of solution curves for the differential equation \(y'=\cos x; \quad y=\sin x + c\), where \(y=y(x)\).

Direction Fields

  1. Consider the differential equation \(\dfrac{dy}{dx} = y - x\).
      1. Compute the slope \(f(x,y)\) at each of the nine grid points \(x, y \in \{-1, 0, 1\}\), and record your values in a small table.
      1. Using your table, sketch the direction field on this grid by hand.
      1. On your sketch, lightly trace the solution curve that passes through the point \((0, 0)\), following the lineal elements. Do the same through \((0, 1)\). Describe, in a sentence, how the two curves behave differently as \(x\) increases.

  2. Without computing a single slope, match each differential equation below to the description of its direction field. Justify each choice in one sentence.
      1. \(\dfrac{dy}{dx} = 2\)
      1. \(\dfrac{dy}{dx} = x\)
      1. \(\dfrac{dy}{dx} = y\)
      1. \(\dfrac{dy}{dx} = -\dfrac{x}{y}\)

    1. The lineal elements are identical along every horizontal line, flat on the \(x\)-axis, and steepen as you move away from it.
    2. The lineal elements everywhere lie tangent to circles centred at the origin.
    3. Every lineal element in the entire plane has the same slope.
    4. The lineal elements are identical along every vertical line, flat on the \(y\)-axis, and steepen as you move away from it.

  3. Consider the differential equation \(\dfrac{dy}{dx} = 1 - y\).
      1. Sketch the direction field by hand on the region \(x \in [-2, 2]\), \(y \in [-1, 3]\) (a grid spacing of \(1\), or \(0.5\) if you are feeling thorough, is fine).
      1. There is one constant function that solves this differential equation. Read it off your direction field, and verify it by substitution.
      1. Using your field, describe the long-run behaviour (as \(x \to \infty\)) of the solution starting at \((0, 3)\), and of the solution starting at \((0, -1)\). What single feature of the field makes both answers obvious?
      1. Check your sketch and your answers using the interactive direction field widget in the notes (the equation \(y(1-y)\) in the widget’s menu behaves similarly near its constant solutions — explore both).

  4. Connecting the ideas. The differential equation \(y' = \cos x\) from Question 3 of the Solutions of ODEs section has the family of solutions \(y = \sin x + c\).
      1. What special structure does the direction field of \(y' = \cos x\) have, given that the slope does not depend on \(y\)?
      1. Use that structure to explain, in one or two sentences, why every member of the family \(y = \sin x + c\) is just a vertical shift of every other member — and why the “+ \(c\)” from integration was therefore inevitable.

  5. True or false, with a reason: if the direction field of a first-order differential equation \(\frac{dy}{dx} = f(x,y)\) (with \(f\) well-behaved) shows a lineal element of slope zero at the point \((a, b)\), then the constant function \(y = b\) must be a solution of the differential equation.