Math Insight

Forward Euler and linear approximations

Name:
Group members:
Section:
  1. For the following differential equation \begin{align*} \diff{ y }{ t } &= 1.6 y + 3\\ y(0) & = -0.8 \end{align*} calculate a Forward Euler approximation to $y(1.2)$ 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{ y }{ t }(t) &= 1.6 y{\left (t \right )} + 3. \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 $y(0.4)$.

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

      Based on that result and the differential equation, what is the slope of $y(t)$ at time $t=0$?
      $\diff{ y }{t}(0) = 1.6 y{\left (0 \right )} + 3 =$

      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 $t=\Delta t = 0.4$.
      $y(0.4) \approx L(0.4) =$

    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 $y(0.8)$.

      Given the value of $y(0.4)$ estimated above and the differential equation, what is the slope of $y(t)$ at time $t=\Delta t = 0.4$?
      $\diff{ y }{t}(0.4) = 1.6 y{\left (0.4 \right )} + 3 =$

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

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

    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 $y(1.2)$.

      Given the value of $y(0.8)$ estimated above and the differential equation, what is the slope of $y(t)$ at time $t=2\Delta t = 0.8$?
      $\diff{ y }{t}(0.8) = 1.6 y{\left (0.8 \right )} + 3 =$

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

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

  2. For the following logistic differential equation \begin{align*} \diff{ v }{ t } &= 0.1 v \left(- \frac{v}{80} + 1\right)\\ v(0) & = 64 \end{align*} calculate a Forward Euler approximation to $v(30.0)$ 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{ v }{ t }(t) &= 0.1 \left(- \frac{1}{80} v{\left (t \right )} + 1\right) v{\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 $v(10.0)$.

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

      Based on that result and the differential equation, what is the slope of $v(t)$ at time $t=0$?
      $\diff{ v }{t}(0) = 0.1 \left(- \frac{1}{80} v{\left (0 \right )} + 1\right) v{\left (0 \right )} =$

      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 $t=\Delta t = 10.0$.
      $v(10.0) \approx L(10.0) =$

    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 $v(20.0)$.

      Given the value of $v(10.0)$ estimated above and the differential equation, what is the slope of $v(t)$ at time $t=\Delta t = 10.0$?
      $\diff{ v }{t}(10.0) = 0.1 \left(- \frac{1}{80} v{\left (10.0 \right )} + 1\right) v{\left (10.0 \right )} =$

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

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

    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 $v(30.0)$.

      Given the value of $v(20.0)$ estimated above and the differential equation, what is the slope of $v(t)$ at time $t=2\Delta t = 20.0$?
      $\diff{ v }{t}(20.0) = 0.1 \left(- \frac{1}{80} v{\left (20.0 \right )} + 1\right) v{\left (20.0 \right )} =$

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

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

Thread navigation

Math 1241, Fall 2020