Maths with David

    IBDP. AI. HL. DEs. Solving DEs

    When we solve equations, we find a value. However, when we solve differential equations, we find a function, for instance, a solution to \frac{dy}{dx} = y^2 is y = - \frac{1}{x}

    We can find the general solution or a particular solution of differential equations.

    The general solution includes a constant of integration, e.g. \frac{dy}{dx} = y^2 has general solution y = - \frac{1}{x - c} . It therefore represents a whole family of solutions.

    The particular solution does not include a constant of integration. Further information is required about eg. Initial conditions (i.e. what x or \frac{dx}{dt} equals when t=0) or boundary conditions, in order to find a particular solution.

    Worked Example. Particular Solution of Differential Equation

    • Show that T = ce^{ \frac{-t}{4}} + 4 is a solution to the differential equations for any constant c.
    • Sketch the solution curves for c = 6, c = 26 and c = 56.
    • Given that the initial water temperature was 80ºC, find the particular solution.

    A glass of water has temperature TºC. It is placed in the refrigerator and its temperature decreases according to the differential equations \frac{dT}{dt} + \frac{1}{4}T = 1 , where t is time in minutes.

    Exercise

    1. Verify that
      • a. y = x^4 is a solution to \frac{dy}{dx} = 4x^3
      • b. y = 5e^{2x} is a solution to \frac{dy}{dx} = 2y
      • c. y = \sqrt{ x^2 + 1 } is a solution to \frac{dy}{dx} = \frac{x}{y}
      • d. y = - \frac{1}{x} is a solution to \frac{dy}{dx} = y^2
      • e. y = 3e^ { \frac{x^2}{2} +x} is a solution to \frac{dy}{dx} - y = xy

    2. Match each differential equations with the correct solution:

    a. \frac{dy}{dx} = \frac{1}{2y}b. \frac{dy}{dx} = \frac{-2y}{x}c. \frac{dy}{dx} = \frac{2x}{3y^2}
    A. y = \sqrt[3]{x^2+1} B. y = \sqrt{x+3} C. y = \frac{1}{x^2}

    3. Verify that for any constant c:

    • a. y = x^3 + c is a solution to \frac{dy}{dx} = 3x^2
    • b. y = ce^{-x} is a solution to \frac{dy}{dx} = -y
    • c. y = \frac{-2}{x^2 + c} is a solution to \frac{dy}{dx} = xy^2

    4. The population of guinea pigs P grows according to the model \frac{dP}{dt} - \frac{P}{6} = 0 , where time is in months.

    • a. Show that P = ce^{ \frac{t}{6} } is a solution to the differential equation for any constant c.
    • b. Sketch the solution curves for c = -2, c = 0, c = 2, c = 6, and c = 10.
    • c. Explain why c must be positive for the model to be realistic.
    • d. Given that there were initially 30 guinea pigs in the field, find the particular solution.

    5. Consider the algae growth model \frac{dG}{dt} = k \sqrt{G} , k > 0.

    • a. Show that G(t) = ( \frac{kt}{2} + \sqrt{a} )^2 is a solution for any constant a.
    • b. Explain the significance of a by calculating G(0).
    • c. Show that G(t) is a quadratic function whose vertex occurs for some t < 0.
    • d. Sketch G(t) for t > 0.

    6. Consider the differential equation \frac{dy}{dx} = 4x

    • a. Show that y = 2x^2 + c is  a solution to the differential equation for any constant c.
    • b. Sketch the solution curves for c = 0, \pm 1 , \pm 2 .
    • c. Find the particular solution which passes through (1,5).
    • d.Find the equation of the tangent to the particular solution at (1, g).

    7. Consider the differential equation \frac{dy}{dx} = 2x - y

    • a. Show that y = 2x - 2 + ce^{-x} is  a solution to the differential equation for any constant c.
    • b. Sketch the solution curves for c = 0, \pm 1, \pm 2 .
    • c. Find the particular solution which passes through (0, 1).

    8. For the frictionless mass on a spring described at the start of the previous section, we can write the differential equation as \frac{d^2x}{dx^2} + \omega^2 x = 0 , where \omega^2 = \frac{k}{m} . This is a particularly important equation called the Simple Harmonic Motion (“SHM”) equation.

    • a. Verify that each of the following are solutions to the SHM equation:
      • x = sin \omega t
      • x = sin ( \omega t + b)
      • x = cos ( \omega t + b)
    • b. Prove that x = e^{ \lambda t} is a solution to the SHM equation provided \lambda is purely imaginary. Find \lambda in this case.
    • c. Prove that if x_1(t) and x_2(t) are both solutions to the SHM equation then x = Ax_1(t) + Bx_2(t) is also a solution.

    Answers

    Screenshot
    Screenshot