2. Consider the IVP for 0 t6. y=-3.8y, y(0) = 1 The exact solution of this...

 

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2. Consider the IVP for 0 ≤t≤6. y=-3.8y, y(0) = 1 The exact solution of this IVP is y = e−3.8t The goal of this exercise is to visualize how Euler's method is related to the slope field of the differ- ential equation. In order to do this we will plot the direction field together with the approximations and the exact solution. (a) To plot the slope field we will use the MATLAB commands meshgrid and quiver. Enter the following commands: t = 0:0.4:6; y = -11:2.4:13; %define a grid in t & y directions = [T, Y] meshgrid (t, y); dT = ones (size (T)); dY -3.8*Y; quiver (T, Y, dT, dY) axis tight hold on % create 2d matrices of points in ty-plane %dt 1 for all points %dy = -3.8*y; this is the ODE % draw arrows (t, y)->(t+dt, t+dy) %adjust look After running these commands you should get the graph of the slope field. (b) Use linspace to generate a vector of 100 t-values between 0 and 6. Evaluate the (ex- act) solution y at these t-values and plot it in black together with the direction field (use 'linewidth', 2). (c) Enter the function defining the ODE as anonymous. Use euler.m with N = 10 to determine the approximation to the solution. Plot the approxi- mated points in red (use 'ro-', 'linewidth',2 as line-style in your plot), together with the exact solution and the direction field. You should get Figure 2. The legend is not required, but to produce it, you can use legend ('Slope Field', 'Exact Sltn', 'Approx. Sltn (N=10)', 'location', 'northwest'); Based on the slope field and the geometrical meaning of Euler's method explain why the ap- proximations are so inaccurate for this particular value of the stepsize. Sope Fil for E-35 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ + ++ + THT T PT TT TT TT TTTTTTT D T 2 3 Figure 2: Euler's method applied to y' = -3.8y, y(0) = 1, N = 10 compared to the exact solution. (d) Open a new figure by typing figure. Plot again the direction field but in a different window: t = 0:0.4:6; y = -0.6:0.2:1.3; Repeat part (b) and repeat part (c) but this time with N = 20. You should get Figure 3. Because of the different scaling of the y-axis, the exact solution (black curve) looks a bit different, but it is the same curve as in the previous figure. Comment on the result from a geometrical point of view. DE 04 02 == Figure 3: Euler's method applied to y' = -3.8y, y(0) = 1, N = 20, compared to the exact solution. 2. Consider the IVP for 0 ≤t≤6. y=-3.8y, y(0) = 1 The exact solution of this IVP is y = e−3.8t The goal of this exercise is to visualize how Euler's method is related to the slope field of the differ- ential equation. In order to do this we will plot the direction field together with the approximations and the exact solution. (a) To plot the slope field we will use the MATLAB commands meshgrid and quiver. Enter the following commands: t = 0:0.4:6; y = -11:2.4:13; %define a grid in t & y directions = [T, Y] meshgrid (t, y); dT = ones (size (T)); dY -3.8*Y; quiver (T, Y, dT, dY) axis tight hold on % create 2d matrices of points in ty-plane %dt 1 for all points %dy = -3.8*y; this is the ODE % draw arrows (t, y)->(t+dt, t+dy) %adjust look After running these commands you should get the graph of the slope field. (b) Use linspace to generate a vector of 100 t-values between 0 and 6. Evaluate the (ex- act) solution y at these t-values and plot it in black together with the direction field (use 'linewidth', 2). (c) Enter the function defining the ODE as anonymous. Use euler.m with N = 10 to determine the approximation to the solution. Plot the approxi- mated points in red (use 'ro-', 'linewidth',2 as line-style in your plot), together with the exact solution and the direction field. You should get Figure 2. The legend is not required, but to produce it, you can use legend ('Slope Field', 'Exact Sltn', 'Approx. Sltn (N=10)', 'location', 'northwest'); Based on the slope field and the geometrical meaning of Euler's method explain why the ap- proximations are so inaccurate for this particular value of the stepsize. Sope Fil for E-35 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ + ++ + THT T PT TT TT TT TTTTTTT D T 2 3 Figure 2: Euler's method applied to y' = -3.8y, y(0) = 1, N = 10 compared to the exact solution. (d) Open a new figure by typing figure. Plot again the direction field but in a different window: t = 0:0.4:6; y = -0.6:0.2:1.3; Repeat part (b) and repeat part (c) but this time with N = 20. You should get Figure 3. Because of the different scaling of the y-axis, the exact solution (black curve) looks a bit different, but it is the same curve as in the previous figure. Comment on the result from a geometrical point of view. DE 04 02 == Figure 3: Euler's method applied to y' = -3.8y, y(0) = 1, N = 20, compared to the exact solution.