Math Insight

Forward Euler and linear approximations

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  1. For the following differential equation \begin{align*} \diff{ v }{ t } &= - 1.8 v - 5.4\\ v(0) & = -5.9 \end{align*} calculate a Forward Euler approximation to $v(1.2)$ using a time step of $\Delta t = 0.4$. 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{ v }{ t }(t) &= - 1.8 v{\left (t \right )} - 5.4. \end{align*}

    1. The following steps will guide you in calculating the Forward Euler approximation after one step of size $\Delta t = 0.4$, yielding an estimate of $v(0.4)$.

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

      Based on that result and the differential equation, what is the slope of $v(t)$ at time $t=0$?
      $\diff{ v }{t}(0) = - 1.8 v{\left (0 \right )} - 5.4 =$

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

      We use this linear approximation to approximate the value of $v(t)$ at $t=\Delta t = 0.4$.
      $v(0.4) \approx L(0.4) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 0.4$, yielding an estimate of $v(0.8)$.

      Given the value of $v(0.4)$ estimated above and the differential equation, what is the slope of $v(t)$ at time $t=\Delta t = 0.4$?
      $\diff{ v }{t}(0.4) = - 1.8 v{\left (0.4 \right )} - 5.4 =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 0.4$.
      $L(t) = v(0.4) + \diff{ v }{t}(0.4) (t-0.4) =$

      We use this linear approximation to approximate the value of $v(t)$ at $t=2\Delta t = 0.8$.
      $v(0.8) \approx L(0.8) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 0.4$, yielding an estimate of $v(1.2)$.

      Given the value of $v(0.8)$ estimated above and the differential equation, what is the slope of $v(t)$ at time $t=2\Delta t = 0.8$?
      $\diff{ v }{t}(0.8) = - 1.8 v{\left (0.8 \right )} - 5.4 =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 0.8$.
      $L(t) = v(0.8) + \diff{ v }{t}(0.8) (t-0.8) =$

      We use this linear approximation to approximate the value of $v(t)$ at $t=3\Delta t = 1.2$.
      $v(1.2) \approx L(1.2) =$

  2. For the following logistic differential equation \begin{align*} \diff{ y }{ t } &= 0.1 y \left(- \frac{y}{960} + 1\right)\\ y(0) & = 864 \end{align*} calculate a Forward Euler approximation to $y(30.0)$ using a time step of $\Delta t = 10.0$. 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{ y }{ t }(t) &= 0.1 \left(- \frac{1}{960} y{\left (t \right )} + 1\right) y{\left (t \right )}. \end{align*}

    1. The following steps will guide you in calculating the Forward Euler approximation after one step of size $\Delta t = 10.0$, yielding an estimate of $y(10.0)$.

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

      Based on that result and the differential equation, what is the slope of $y(t)$ at time $t=0$?
      $\diff{ y }{t}(0) = 0.1 \left(- \frac{1}{960} y{\left (0 \right )} + 1\right) y{\left (0 \right )} =$

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

      We use this linear approximation to approximate the value of $y(t)$ at $t=\Delta t = 10.0$.
      $y(10.0) \approx L(10.0) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 10.0$, yielding an estimate of $y(20.0)$.

      Given the value of $y(10.0)$ estimated above and the differential equation, what is the slope of $y(t)$ at time $t=\Delta t = 10.0$?
      $\diff{ y }{t}(10.0) = 0.1 \left(- \frac{1}{960} y{\left (10.0 \right )} + 1\right) y{\left (10.0 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 10.0$.
      $L(t) = y(10.0) + \diff{ y }{t}(10.0) (t-10.0) =$

      We use this linear approximation to approximate the value of $y(t)$ at $t=2\Delta t = 20.0$.
      $y(20.0) \approx L(20.0) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 10.0$, yielding an estimate of $y(30.0)$.

      Given the value of $y(20.0)$ estimated above and the differential equation, what is the slope of $y(t)$ at time $t=2\Delta t = 20.0$?
      $\diff{ y }{t}(20.0) = 0.1 \left(- \frac{1}{960} y{\left (20.0 \right )} + 1\right) y{\left (20.0 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 20.0$.
      $L(t) = y(20.0) + \diff{ y }{t}(20.0) (t-20.0) =$

      We use this linear approximation to approximate the value of $y(t)$ at $t=3\Delta t = 30.0$.
      $y(30.0) \approx L(30.0) =$

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Math 1241, Fall 2020