18.03SC Practice Problems 3 Euler’s method Solution Suggestions dy 1. Use Euler’s method to estimate the value at x = 1.5 of the solution of dx = y� = F ( x, y) = y2 − x2 for which y(0) = −1. Use step size h = 0.5. Recall the notation x0 = 0, y0 = −1, xn+1 = xn + h, yn+1 = yn + mn h, mn = F ( xn , yn ). Make a table with columns n, xn , yn , mn , mn h. Draw the Euler polygon. We are estimating y(1.5) using Euler’s method with step size 0.5. This takes 3 steps, as outlined in the following table. n 0 1 2 3 xn 0 0.5 1.0 1.5 yn -1 -0.5 -0.5 -0.875 mn 1 0 -0.75 1.5−0 0.5 = mn h 0.5 0 -0.375 Thus, Euler’s method gives the estimate y(1.5) ≈ y3 = −0.875. The corresponding Euler polygon for this estimation is Euler polygon and actual integral curve for Question 1. 2. Is the estimate found in Question 1 likely to be too large or too small? It is likely to be too large. One way to see this is to use the second derivative test to check the concavity of solutions around the inital point. That is, ﬁnd taking the derivative of the differential equation y�� = d d 2 ( F ( x, y)) = (y − x2 ) = 2yy� − 2x, dx dx d2 y dx2 = y�� by 18.03SC Practice Problems 3 OCW 18.03SC and evaluate the second derivative at the initial point ( x0 , y0 ) to get y�� |(0,−1) = 2(−1)(1) − 2(0) = −2 < 0. This means the solution that goes through the initial point is concave down. The tangent to a concave down function lies above the function in a small neighborhood, so the Euler estimate for one step is likely to overshoot. Running the same check for the next two endpoints shows that the second derivative is negative at each endpoint of the Euler polygon, so each of the three steps is likely to overshoot. This suggests the estimate found is likely to be greater than the actual value of our solution y = y( x ) at x = 1.5. 2