Question

a = 1

b = 2

c= 3

Let a, b and c be the three integers previously assigned to you. Let q be the three-digit number abc. The shape, y(x), of a certain structure is governed by the following differential equation. 1 dy dx? x dx Suppose that y(0.5)=0.71459 y(3)=4 +

Answer 1

ANSWER:

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• I have provided the output image of the code so you can easily cross-check for the correct output of the code.
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CODE TEXT

% defining given variables

a = 1; b = 2; c = 3;

% q = abc, i.e.

q = a*100 + b*10 + c;

% 1. function required for ode45

% for eqn, y'' = -y'/x + (1/x^2 - 1)*y,

% puting, y' = y(2)

% hence, y'' = A - 2*a*w*y(2) - w^2*y(1)

% defining inline function vdp1

vdp1 = @(x,y) [y(2); -y(2)/x + (1/x^2 - 1)*y(1)];

% defining error tollerance to least four significant figures

% i.e 10^-4

opts = odeset('AbsTol',1e-4);

% using vdp1 , initial values and tollerance

% solving eqn by ode45, it returns x [y y']

[x,Y] = ode45(vdp1,[0.5 3],[0.7145*q; q],opts);

% 2. Ploting curve, x vs y

plot(x,Y(:,1));

xlabel('X');

ylabel('y');

title("Curve for y'' = -y'/x + (1/x^2 - 1)*y");

grid on;

% 3. slopes i.e y'(x) at end points

y_diff = Y(:,2);

fprintf("End point slopes on intervals [0.5 , 3] is: [%.2f, %.2f]", ...

y_diff(1),y_diff(end));

% 4.

% shooting method is good method can find values with low error

% results shown in graph

CODE IMAGE

1 2 a = % defining given variables 1; b = 2; C = 3; % q = abc, i.e. 9 a*100 + b*10 + C; 3 4 - 5 6 7 8 9 10 11 - % 1. function required for ode45 % for eqn, y'' -y'/x + (1/x^2 - 1)*y, ko puting, y' = y(2) % hence, y' = A - 2*a*w*y(2) W^2*y(1) % defining inline function vdp1 vdp1 = @(x,y) [y(2); -y(2)/x + (1/x^2 - 1)*y(1)]; % defining error tollerance to least four significant figures % i.e 10^-4 opts = odeset('AbsTol',1e-4); % using vdp1 , initial values and tollerance % solving eqn by ode45, it returns x [y y'] [x,Y] = ode45 (vdp1,[0.5 3],[0.7145*q; q],opts); 12 3 14 15 16 % 2. Ploting curve, X vs y plot(X,Y(:,1)); xlabel('X'); ylabel('y'); title("Curve for y'' = -y'/x + (1/x^2 - 1)*y"); grid on; - N 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 % 3. slopes i.e y'(x) at end points y_diff = Y(:,2); fprintf("End point slopes on intervals [0.5 , 3] is: [%.2f, %.2f]", y_diff(1),y_diff(end)); % 4. % shooting method is good method can find values with low error % results shown in graph

OUTPUT IMAGE

Curve for y" = -y'lx + (1/x2 - 1)*y 200 180 160 >140 120 100 80 0.5 1 1.5 2 2.5 3 X