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MTH643 Final Term Mega File All in 1 2022 by Everblue

Introduction to MATLAB PDF 16.8 MBUpdated 23 Oct 2026
Final term past paper MTH643 Final Term Mega File All in 1 2022 by Everblue PDF 16.8 MB
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Final term past paper
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MTH643 Introduction to MATLAB
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PDF 16.8 MB
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23 Oct 2026
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Text from pages 2, 3, 4 of 168. Download the file for all 168 pages.

yy=eval(vectorize(y)); zz=eval(vectorize(z)); plot(t,xx,t,yy,t,zz)

title(' Graph of solution curves of DE') grid

using Euler’s Formula for n=10, function [xvalues,yvalues]=euler(f,x0,y0,xn,n);

x0=0; y0=pi; xn=1; n=10; x(1)=x0; y(1)=y0; xn=1; dx=(xn-x0)/n; for k=1:n

y(k+1)=y(k)+sin(x(k)*y(k))*dx; end xvalues=x' yvalues=y'

[0; 1] using Euler’s method for n = 20.

%using Euler’s method for n = 20. dy/dx=(y-x)/(y+x) ,y(0)=0.5 function [xvalues,yvalues]=euler(f,x0,y0,xn,n);

x0=0; y0=0.5; xn=1; n=20; x(1)=x0; y(1)=y0; xn=1; dx=(xn-x0)/n; for k=1:n

y(k+1)=y(k)+((y(k)-x(k))/(y(k)+x(k)))*dx; end xvalues=x' yvalues=y' [0; 1] using improved Euler’s Formula for n=10,

(P#231-235 – LECT#08)

%using improved Euler’s Formula for n=10, dy/dx=Sin(xy),y(0)=? function [xvalues,yvalues]=impeuler(f,x0,y0,xn,n);

x0=0; y0=pi; xn=1; n=10; x(1)=x0; y(1)=y0; xn=1; dx=(xn-x0)/n; for k=1:n

y(k+1)=y(k)+dx/2*(f(x(k),y(k))+f(x(k),y(k)+dx*y(k))); end xvalues=x' yvalues=y'

[0,1] using Improved Euler’s method for n = 20.

%using Improved Euler’s method for n = 20. dy/dx=(y-x)/(y+x) ,y(0)=0.5 function [xvalues,yvalues]=impeuler(f,x0,y0,xn,n);

x0=0; y0=1; xn=0.5; n=20; x(1)=x0; y(1)=y0; xn=1; dx=(xn-x0)/n; for k=1:n

y(k+1)=y(k)+dx/2*(f(x(k),y(k))+f(x(k),y(k)+dx*y(k))); end xvalues=x' yvalues=y' L08_Q44: Find the numerical solution of the following initial value problem on the interval [0,1] using improved Euler’s method by taking step size ∆𝑡= 0.01, Moreover, plot the graph of the approximated solution curve. Note: The iterative form of the improved Euler’s method is given below;

%Find the numerical solution of the following initial value problem on the interval [0,1] %using improved Euler’s method by taking step size ∆t=0.01, %Moreover, plot the graph of the approximated solution curve. dy/dt=(y-t)/(y+t) ,y(0)=0.5 function [yvalues,tvalues]=impeuler(f,y0,t0,yn,n);

y0=0; t0=1; yn=0.5; n=20; y(1)=y0; t(1)=t0; yn=1; dt=0.01; for k=1:n y(k+1)=y(k)+dt; t(k+1)=t(k)+dt/2*(f(y(k),t(k))+f(y(k),t(k)+dt*t(k))); end yvalues=y' tvalues=t' plot(yvalues,tvalues)

dy dx = Sin xy , y 0 = π

%using Runge–Kutta method of fourth order for n=10, dy/dx=Sin(xy),y(0)=?

x0=0; y0=pi; xn=1; n=10; x(1)=x0; y(1)=y0; dx=(xn-x0)/n; for k=1:n

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