Question

answer in matlab code

Employ the bvp4c command to find the approximate solution of the boundary value problem governed by the second-order nonhomogeneous differential equation, 9. with the boundary conditions of y(0) 5 and y(1)-2. Plot to compare the approximate solution with the exact solution obtained by using the dsolve command.

Answer 1

nt we. have 2- 2 HeMce 2

Matlab code for solving ode using bvp4c clear all close all %solution using bvp4c solinit - bvpinit(linspace (0,1),1 0]) sol- bvp4c (twoode, êtwobc, solinit); xx-linspace(0,1) yy deval (sol, xx) plot (xx, yy (1,:), 'linewidth', 2) 용号号응용음음응용음음응용음음용용음음용용음음용용음음용용음응용용융응용용응응용용융응용욤 solution using dsolve syms y(x) eqndiff (y,x,2)+2 diff (y, x, 1)+y(x) x 2; cond- [y(0)-#5, y(1) #21: ySol ()dsolve (eqn, cond) fprint f ('solution using dsolve is y(x)-.) disp ySol(x)) for i-1:length (xx) end hold on plot (xx,y ext, o', linewidth',2) title( 'Y (x) vs. x plot') xlabl'x') ylabel'y(x) ') legend('Using bvp4c,'using dsolve') function dydxtwoode (x,y) dydx- [Y (2) (x.2-2"y (2)-y(1)) function res-twobc (ya,yb) res[ya(1)-5; yb(1)-2; end 용음음응용음음응용음음응용음음용용음음용용음음용용음음용용음응용용음응용욤응응용용융응용욤 solution using dsolve is y(x)-x"2 -4*x x* exp (-x)*(exp (1) exp(-x) - - - 1) +6

Y(x) vs. x plot Using bvp4c dsolve 0 4.5 3.5 2.5 0.1 0 03 0.5 06 0700.91 Published with MATLAB R2018a

%Matlab code for solving ode using bvp4c
clear all
close all

%solution using bvp4c

solinit = bvpinit(linspace(0,1),[1 0]);
sol= bvp4c(@twoode,@twobc,solinit);

xx=linspace(0,1);
yy=deval(sol,xx);
plot(xx,yy(1,:),'linewidth',2)

    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%solution using dsolve
syms y(x)
eqn = diff(y,x,2)+2*diff(y,x,1)+y(x) == x^2;
cond = [y(0)==5, y(1)==2];
ySol(x) = dsolve(eqn,cond);
fprintf('Solution using dsolve is y(x)=')
disp(ySol(x))

for i=1:length(xx)
    y_ext(i)=ySol(xx(i));
end

hold on
plot(xx,y_ext,'o','linewidth',2)
title('Y(x) vs. x plot')
xlabel('x')
ylabel('y(x)')
legend('Using bvp4c','using dsolve')

    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function dydx = twoode(x,y)

    dydx= [y(2); (x.^2-2*y(2)-y(1))];
  
end

function res =twobc(ya,yb)
    res = [ya(1)-5; yb(1)-2];
end

    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%