Question

Osts 10 Problem 3 Numerically integrate the 2nd order linear differential equation on the interval y(t) = 2e" - 2e-41 and compare it to the solution a) Plot the numerical solution and the true solution for y(t) (20 pts) b) Plot the numerical solution and the true solution for dy/dt (10 pts)

only using matlab

Answer 1

Program Screen Shot:

function dy = ode_fun (t,Y) x driver_file. m ode_fun.m X { from: y' + 5y + 4y = 0, y(0) = 0, y'(0) = 6 let: y' = x meaning two lst order ode's are formed as: x' = -5x - 4y, x(0) = y'(0) = 6 y' = x, y(0) = 0 84 % extract x and y in the same order initial conditions were set x = Y(1); y = Y(2); 8t compute the derivatives dx = -5*x - 4*y; $for x' dy = x; $for y' dY = (dx;dy]; end

clc, clear, close all X 2 driver_file.m ode_fun. m x from: y'' + 5y + 4y = 0, y(0) = 0, y' (0) = 6 let: y' = x meaning two lst order ode's are formed as: x' = -5x - 4y, x(0) = y'(0) = 6 y' = x, y(0) = 0 La %% set the initial conditions and time interval x0 = 6; yo = 0; YO = (x0;y0]; & initial conditions tspan = [010]; $ time interval 8% solve numerically [t,Y] = ode45 (Code_fun, tspan, Y); using ode 45 %% extract the numerical solution y_num = Y(:,2); $ 2nd colum contains the solution y dydt_num= Y(:,1); $ 1st column contains the solution dydt 8% compute the true y and dydt solutions v_true = 2.*exp(-t) - 2.*exp(-4*t); dydt_true = -2.*exp(-1) + 8. *exp(-4*t); 8graph the results figure, plot(t,v_num,'o b',t,v_true,'- r'),xlabel('t'), ylabel('y(t)') legend('y_n_u_m', 'y_t_r_u_e'), grid on figure, plot(t, dydt_num,'ob',t, dydt_true,'- r'),xlabel('t'),ylabel('dydt') legend ('dydt_n_u_m', 'dydt_t_r_u_e'),grid on

Sample Output:

num ее 0-0— 0 eeeeeeeeeeeeeeeeeeeeeeeee y(t) 00 ecoco ооооооооо 4 5 6 оооооооооооо 1 2 3 7

odyctum dych grue dydt 0 1 2 3 4 5 6 7 8 9 10

Program Code to Copy:

function dY = ode_fun(t,Y)
%%
%{
from: y'' + 5y + 4y = 0, y(0) = 0, y'(0) = 6
let: y' = x
meaning two 1st order ode's are formed as:
x' = -5x - 4y, x(0) = y'(0) = 6
y' = x, y(0) = 0
%}
%
%% % extract x and y in the same order initial conditions were set
x = Y(1); y = Y(2);
%
%% compute the derivatives
dx = -5*x - 4*y; % for x'
dy = x; % for y'
%
dY = [dx;dy];

end

______________________________________________

% driver_file.m

clc,clear,close all
%%
%{
from: y'' + 5y + 4y = 0, y(0) = 0, y'(0) = 6
let: y' = x
meaning two 1st order ode's are formed as:
x' = -5x - 4y, x(0) = y'(0) = 6
y' = x, y(0) = 0
%}
%% set the initial conditions and time interval
x0 = 6; y0 = 0;
Y0 = [x0;y0]; % initial conditions
tspan = [0 10]; % time interval
%% solve numerically
[t,Y] = ode45(@ode_fun,tspan,Y0); % using ode45
%% extract the numerical solution
y_num = Y(:,2); % 2nd colum contains the solution y
dydt_num = Y(:,1); % 1st column contains the solution dydt
%% compute the true y and dydt solutions
y_true = 2.*exp(-t) - 2.*exp(-4*t);
dydt_true = -2.*exp(-t) + 8.*exp(-4*t);
%% graph the results
figure,plot(t,y_num,'o b',t,y_true,'- r'),xlabel('t'),ylabel('y(t)')
legend('y_n_u_m','y_t_r_u_e'),grid on
%
figure,plot(t,dydt_num,'o b',t,dydt_true,'- r'),xlabel('t'),ylabel('dydt')
legend('dydt_n_u_m','dydt_t_r_u_e'),grid on

%%

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