Math Insight

Forward Euler and linear approximations

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  1. For the following differential equation \begin{align*} \diff{ w }{ t } &= 1.1 w + 5.1\\ w(0) & = -4.3 \end{align*} calculate a Forward Euler approximation to $w(1.8)$ using a time step of $\Delta t = 0.6$. In this problem, we guide your through the steps of the calculation. It may easier to understand the steps by writing the differential equation using an explicit argument of $t$ \begin{align*} \diff{ w }{ t }(t) &= 1.1 w{\left (t \right )} + 5.1. \end{align*}

    1. The following steps will guide you in calculating the Forward Euler approximation after one step of size $\Delta t = 0.6$, yielding an estimate of $w(0.6)$.

      What is the initial condition? $w(0) = $

      Based on that result and the differential equation, what is the slope of $w(t)$ at time $t=0$?
      $\diff{ w }{t}(0) = 1.1 w{\left (0 \right )} + 5.1 =$

      Those two numbers are enough information to make a linear approximation at $t=0$.
      $L(t) = w(0) + \diff{ w }{t}(0) (t-0) =$

      We use this linear approximation to approximate the value of $w(t)$ at $t=\Delta t = 0.6$.
      $w(0.6) \approx L(0.6) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 0.6$, yielding an estimate of $w(1.2)$.

      Given the value of $w(0.6)$ estimated above and the differential equation, what is the slope of $w(t)$ at time $t=\Delta t = 0.6$?
      $\diff{ w }{t}(0.6) = 1.1 w{\left (0.6 \right )} + 5.1 =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 0.6$.
      $L(t) = w(0.6) + \diff{ w }{t}(0.6) (t-0.6) =$

      We use this linear approximation to approximate the value of $w(t)$ at $t=2\Delta t = 1.2$.
      $w(1.2) \approx L(1.2) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 0.6$, yielding an estimate of $w(1.8)$.

      Given the value of $w(1.2)$ estimated above and the differential equation, what is the slope of $w(t)$ at time $t=2\Delta t = 1.2$?
      $\diff{ w }{t}(1.2) = 1.1 w{\left (1.2 \right )} + 5.1 =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 1.2$.
      $L(t) = w(1.2) + \diff{ w }{t}(1.2) (t-1.2) =$

      We use this linear approximation to approximate the value of $w(t)$ at $t=3\Delta t = 1.8$.
      $w(1.8) \approx L(1.8) =$

  2. For the following logistic differential equation \begin{align*} \diff{ z }{ t } &= 0.5 z \left(- \frac{z}{240} + 1\right)\\ z(0) & = 96 \end{align*} calculate a Forward Euler approximation to $z(6.0)$ using a time step of $\Delta t = 2.0$. In this problem, we guide your through the steps of the calculation. It may easier to understand the steps by writing the differential equation using an explicit argument of $t$ \begin{align*} \diff{ z }{ t }(t) &= 0.5 \left(- \frac{1}{240} z{\left (t \right )} + 1\right) z{\left (t \right )}. \end{align*}

    1. The following steps will guide you in calculating the Forward Euler approximation after one step of size $\Delta t = 2.0$, yielding an estimate of $z(2.0)$.

      What is the initial condition? $z(0) = $

      Based on that result and the differential equation, what is the slope of $z(t)$ at time $t=0$?
      $\diff{ z }{t}(0) = 0.5 \left(- \frac{1}{240} z{\left (0 \right )} + 1\right) z{\left (0 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=0$.
      $L(t) = z(0) + \diff{ z }{t}(0) (t-0) =$

      We use this linear approximation to approximate the value of $z(t)$ at $t=\Delta t = 2.0$.
      $z(2.0) \approx L(2.0) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 2.0$, yielding an estimate of $z(4.0)$.

      Given the value of $z(2.0)$ estimated above and the differential equation, what is the slope of $z(t)$ at time $t=\Delta t = 2.0$?
      $\diff{ z }{t}(2.0) = 0.5 \left(- \frac{1}{240} z{\left (2.0 \right )} + 1\right) z{\left (2.0 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 2.0$.
      $L(t) = z(2.0) + \diff{ z }{t}(2.0) (t-2.0) =$

      We use this linear approximation to approximate the value of $z(t)$ at $t=2\Delta t = 4.0$.
      $z(4.0) \approx L(4.0) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 2.0$, yielding an estimate of $z(6.0)$.

      Given the value of $z(4.0)$ estimated above and the differential equation, what is the slope of $z(t)$ at time $t=2\Delta t = 4.0$?
      $\diff{ z }{t}(4.0) = 0.5 \left(- \frac{1}{240} z{\left (4.0 \right )} + 1\right) z{\left (4.0 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 4.0$.
      $L(t) = z(4.0) + \diff{ z }{t}(4.0) (t-4.0) =$

      We use this linear approximation to approximate the value of $z(t)$ at $t=3\Delta t = 6.0$.
      $z(6.0) \approx L(6.0) =$

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Math 201, Spring 22