Applications of Series — Question 7

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Question 7

Derive one forward Euler step for y′=f(t,y)y'=f(t,y) from Taylor’s theorem. Identify the discarded term. Then apply one step with h=0.1h=0.1 to y′=yy'=y, y(0)=1y(0)=1, and bound its local error.

Original worksheet page 1: question and worked solution for 4-17-007
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Question 7 – Solution

Step 1: Expand the exact solution in time.

Taylor’s theorem about tnt_n gives y(tn+h)=y(tn)+hy′(tn)+h22y″(ξn)y(t_n+h)=y(t_n)+h y'(t_n)+\frac{h^2}{2}y''(\xi_n) for some ξn\xi_n between tnt_n and tn+ht_n+h.

Step 2: Replace the derivative using the ODE.

Since y′(tn)=f(tn,y(tn))y'(t_n)=f(t_n,y(t_n)), dropping the quadratic remainder produces Euler’s update yn+1=yn+hf(tn,yn).\boxed{y_{n+1}=y_n+h f(t_n,y_n)}. The one-step local truncation error is h2y″(ξn)/2h^2y''(\xi_n)/2, hence is order h2h^2.

Step 3: Apply the method.

For y′=yy'=y, y0=1y_0=1, and h=0.1h=0.1, y1=1+0.1(1)=1.1.y_1=1+0.1(1)=\boxed{1.1}. The exact value is e0.1e^{0.1}. On [0,0.1][0,0.1], y″(t)=et≤e0.1y''(t)=e^t\le e^{0.1}, so |e0.1−1.1|≤e0.1(0.1)22<0.00553.|e^{0.1}-1.1|\le\frac{e^{0.1}(0.1)^2}{2}<0.00553.

Original worksheet page 2: question and worked solution for 4-17-007

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