Second Order Differential Equations (DP IB Applications & Interpretation (AI): HL): Revision Note

Euler's method: second order

What is a second order differential equation?

• A second order differential equation is a differential equation containing one or more second derivatives

• In this course you consider second order differential equations of the form

  $\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}=f\left(x,\frac{\mathrm{d}x}{\mathrm{d}t},t\right)$

• Some examples include;

  • $\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}=x\frac{\mathrm{d}x}{\mathrm{d}t}+2x^2+3t+5$

  • $\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}=2\frac{\mathrm{d}x}{\mathrm{d}t}+3$

How do I apply Euler’s method to second order differential equations?

• You need to turn the second order differential equation into a pair of coupled first order differential equations

• Use the substitution $y=\frac{dx}{dt}$

  • Then $\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}$

• The pair of equations then becomes the system

$$\begin{gathered} \frac{dx}{dt}=y \\ \frac{dy}{dt}=f\left(x,y,t\right) \end{gathered}$$

• You can then use Euler's method in the same way

• Write down the recursion equations using the formulae from the exam formula booklet:

  • $x_{n+1}=x_n+h\times f_1\left(x_n, y_n, t_n\right)$

  • $y_{n+1}=y_n+h\times f_1\left(x_n, y_n, t_n\right)$

  • $t_{n+1}=t_n+h$ 

    • h in those equations is the step size

    • The exam question will usually tell you the correct value of h to use

• Use the recursion feature on your GDC to calculate the Euler’s method approximation over the correct number of steps

  • The values for $x_0$, $y_0$ and $t_0$ will come from the boundary conditions given in the question

  • Frequently you will be given an initial condition

    • Look out for terms like ‘initially’ or ‘at the start’

    • In this case $t_0=0$

Exact solutions of second order differential equations

How can I find the exact solution for a second order differential equation?

• You might be asked to find exact solutions to second order differential equations of the form
  

  $\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}+a\frac{\mathrm{d}x}{\mathrm{d}t}+bx=0$
  

  •  where $a$ and $b$ are constants

• Use the substitution $y=\frac{dx}{dt}$ to turn it into a coupled differential equation

$$\begin{gathered} \frac{dx}{dt}=y \\ \frac{dy}{dt}=-bx-ay \end{gathered}$$

• This can be written in the form $\dot{\mathit{x}}=\mathit{M}\mathit{x}$

  • $\dot{\mathit{x}}=\left(\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right)$

  • $\mathit{x}=\left(\begin{array}{c}x\\ y\end{array}\right)$

  • $\mathit{M}=\left(\begin{array}{cc}0& 1\\ -b& -a\end{array}\right)$

• You can then investigate the solutions of the coupled differential equations

  • The characteristic equation is ${\lambda }^2+a\lambda +b=0$

• If the eigenvalues ${\lambda }_1$ and ${\lambda }_2$ are real, distinct, non-zero

  • The exact solution is of the form $x=C{\mathrm{e}}^{{\lambda }_{1}t}+D{\mathrm{e}}^{{\lambda }_{2}t}$

    • Where $C$ and $D$ are constants

  • You can find the value of the constants given initial or boundary conditions

    • Remember that $y=\frac{dx}{dt}$

    • This helps if you are given the initial value of $\frac{dx}{dt}$

• For example, consider $\frac{d^{2}x}{d{t}^2}-4\frac{dx}{dt}+3x=0$

  • The eigenvalues of $\left(\begin{array}{cc}0& 1\\ -3& 4\end{array}\right)$ are 1 and 3

  • The general solution is $x=C{\mathrm{e}}^t+D{\mathrm{e}}^{3t}$