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+// example 7.11
+// solve the boundary value problem      u''=u'+1;
+// u(0)=1;   u(x=1)=2(%e-1);          h=1/3;
+
+
+// we know;     u''=(u(j-1)-2*u(j)+u(j+1))/h^2;
+// we know;      u'=(u(j+1)-u(j-1))/2h;
+
+// 1) second order method;
+ x=0:1/3:1;                                  
+ 
+ u=[u0 u1 u2 u3 ];
+// hence;
+disp('(u(j-1)-2*u(j)+u(j+1))/h^2=((u(j+1)-u(j-1))/2h)+1')           // for j=1,2;
+
+
+disp('for j=1            (7/6)*u0-2*u1+(5/6)*u2=(1/9)')
+
+disp('for j=2            (7/6)*u1-2*u2+(5/6)*u3=(1/9)')
+
+
+// hence eliminating u1!
+// solving for u1,u2, 
+u0=1;
+u3=2*(%e-1);
+u1=1.454869;
+u2=2.225019;
+
+disp(x);
+disp(u);
+
+