Math Insight

Forward Euler and linear approximations

Name:
Group members:
Section:
  1. For the following logistic differential equation \begin{align*} \diff{ y }{ t } &= 0.3 y \left(- \frac{y}{10} + 1\right)\\ y(0) & = 5 \end{align*} calculate a Forward Euler approximation to $y(9.9)$ using a time step of $\Delta t = 3.3$. 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) &= 0.3 \left(- \frac{1}{10} 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 $\Delta t = 3.3$, yielding an estimate of $y(3.3)$.

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

      Based on that result and the differential equation, what is the slope of $y(t)$ at time $t=0$?
      $\diff{ y }{t}(0) = 0.3 \left(- \frac{1}{10} y{\left (0 \right )} + 1\right) y{\left (0 \right )} =$

      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 $t=\Delta t = 3.3$.
      $y(3.3) \approx L(3.3) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 3.3$, yielding an estimate of $y(6.6)$.

      Given the value of $y(3.3)$ estimated above and the differential equation, what is the slope of $y(t)$ at time $t=\Delta t = 3.3$?
      $\diff{ y }{t}(3.3) = 0.3 \left(- \frac{1}{10} y{\left (3.3 \right )} + 1\right) y{\left (3.3 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 3.3$.
      $L(t) = y(3.3) + \diff{ y }{t}(3.3) (t-3.3) =$

      We use this linear approximation to approximate the value of $y(t)$ at $t=2\Delta t = 6.6$.
      $y(6.6) \approx L(6.6) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 3.3$, yielding an estimate of $y(9.9)$.

      Given the value of $y(6.6)$ estimated above and the differential equation, what is the slope of $y(t)$ at time $t=2\Delta t = 6.6$?
      $\diff{ y }{t}(6.6) = 0.3 \left(- \frac{1}{10} y{\left (6.6 \right )} + 1\right) y{\left (6.6 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 6.6$.
      $L(t) = y(6.6) + \diff{ y }{t}(6.6) (t-6.6) =$

      We use this linear approximation to approximate the value of $y(t)$ at $t=3\Delta t = 9.9$.
      $y(9.9) \approx L(9.9) =$

  2. For the following differential equation \begin{align*} \diff{ x }{ t } &= - 1.7 x + 0.7\\ x(0) & = -3.3 \end{align*} calculate a Forward Euler approximation to $x(1.2)$ 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{ x }{ t }(t) &= - 1.7 x{\left (t \right )} + 0.7. \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 $x(0.4)$.

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

      Based on that result and the differential equation, what is the slope of $x(t)$ at time $t=0$?
      $\diff{ x }{t}(0) = - 1.7 x{\left (0 \right )} + 0.7 =$

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

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

    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 $x(0.8)$.

      Given the value of $x(0.4)$ estimated above and the differential equation, what is the slope of $x(t)$ at time $t=\Delta t = 0.4$?
      $\diff{ x }{t}(0.4) = - 1.7 x{\left (0.4 \right )} + 0.7 =$

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

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

    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 $x(1.2)$.

      Given the value of $x(0.8)$ estimated above and the differential equation, what is the slope of $x(t)$ at time $t=2\Delta t = 0.8$?
      $\diff{ x }{t}(0.8) = - 1.7 x{\left (0.8 \right )} + 0.7 =$

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

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

Thread navigation

Math 1241, Fall 2020