3 files changed, 112 insertions, 116 deletions
diff --git a/50/CH7/EX7.11/ex_7_11.sce b/50/CH7/EX7.11/ex_7_11.sce
index 2e785541b..9120eacfa 100755
--- a/50/CH7/EX7.11/ex_7_11.sce
+++ b/50/CH7/EX7.11/ex_7_11.sce
@@ -1,32 +1,30 @@
-// 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);
-
-
+// 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= rand(1,4);;
+// 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);
+\ No newline at end of file
diff --git a/50/CH7/EX7.5/ex_7_5.sce b/50/CH7/EX7.5/ex_7_5.sce
index 8d6eed3f7..dc8b9e584 100755
--- a/50/CH7/EX7.5/ex_7_5.sce
+++ b/50/CH7/EX7.5/ex_7_5.sce
@@ -1,52 +1,52 @@
-// example 7.5
-// solve the boundary value problem      u''=u+x;
-// u(x=0)=u(0)=0;   u(x=1)=u(4)=0;          h=1/4;
-
-
-// we know;     u''=(u(j-1)-2*u(j)+u(j+1))/h^2;
-
-// 1) second order method;
- x=0:1/4:1;                                  
- u0=0;
- u4=0;
- u=[u0 u1 u2 u3 u4];
-// hence;
-disp('(u(j-1)-2*u(j)+u(j+1))/h^2=u(j)+x(j)')           // for j=1,2,3;
-
-disp('for j=1            -16*u0+33*u1-16*u2=-.25')
-
-disp('for j=2            -16*u1+33*u2-16*u3=-.50')
-
-disp('for j=3            -16*u2+33*u3-16*u4=-.75')
-
-// hence solving for u1,u2,u3)   , 
-u1=-.034885;
-u2=-.056326;
-u3=-.050037;
-
-disp(x);
-disp(u);
-
-// 2) numerov method;
- x=0:1/4:1;                                  
- u0=0;
- u4=0;
- u=[u0 u1 u2 u3 u4];
-// since according to numerov method we get the following system of equations;
-disp('(191*u(j-1)-394*u(j)+191*u(j+1)=x(j-1)+10*x(j)+x(j+1)')           // for j=1,2,3;
-
-disp('for j=1            191*u0-394*u1+191*u2=3')
-
-disp('for j=2            191*u1-394*u2+191*u3=6')
-
-disp('for j=3            191*u2-394*u3+191*u4=9')
-
-// hence solving for u1,u2,u3   , 
-u1=-.034885
-u2=-.056326
-u3=-.050037
-
-
-disp(x);
-disp(u);
-
+// example 7.5
+// solve the boundary value problem      u''=u+x;
+// u(x=0)=u(0)=0;   u(x=1)=u(4)=0;          h=1/4;
+
+
+// we know;     u''=(u(j-1)-2*u(j)+u(j+1))/h^2;
+
+// 1) second order method;
+ x=0:1/4:1;                                  
+ u0=0;
+ u4=0;
+u1_3 = rand(1,3)
+ u=[u0 u1_3 u4];
+// hence;
+disp('(u(j-1)-2*u(j)+u(j+1))/h^2=u(j)+x(j)')           // for j=1,2,3;
+
+disp('for j=1            -16*u0+33*u1-16*u2=-.25')
+
+disp('for j=2            -16*u1+33*u2-16*u3=-.50')
+
+disp('for j=3            -16*u2+33*u3-16*u4=-.75')
+
+// hence solving for u1,u2,u3)   , 
+u1=-.034885;
+u2=-.056326;
+u3=-.050037;
+
+disp(x);
+disp(u);
+
+// 2) numerov method;
+ x=0:1/4:1;                                  
+ u0=0;
+ u4=0;
+ u=[u0 u1 u2 u3 u4];
+// since according to numerov method we get the following system of equations;
+disp('(191*u(j-1)-394*u(j)+191*u(j+1)=x(j-1)+10*x(j)+x(j+1)')           // for j=1,2,3;
+
+disp('for j=1            191*u0-394*u1+191*u2=3')
+
+disp('for j=2            191*u1-394*u2+191*u3=6')
+
+disp('for j=3            191*u2-394*u3+191*u4=9')
+
+// hence solving for u1,u2,u3   , 
+u1=-.034885
+u2=-.056326
+u3=-.050037
+
+
+disp(x);
+disp(u);
+\ No newline at end of file
diff --git a/50/CH7/EX7.6/ex_7_6.sce b/50/CH7/EX7.6/ex_7_6.sce
index 04a28e398..1f7ca0460 100755
--- a/50/CH7/EX7.6/ex_7_6.sce
+++ b/50/CH7/EX7.6/ex_7_6.sce
@@ -1,32 +1,30 @@
-// example 7.6
-// solve the boundary value problem      u''=u*x;
-// u(0)+u'(0)=1;   u(x=1)=0;          h=1/3;
-
-
-// we know;     u''=(u(j-1)-2*u(j)+u(j+1))/h^2;
-
-// 1) second order method;
- x=0:1/3:1;                                  
- 
- u3=1;
- u=[u0 u1 u2 u3 ];
-// hence;
-disp('(u(j-1)-2*u(j)+u(j+1))/h^2=u(j)*x(j)')           // for j=0,1,2,3;
-
-disp('for j=0            u1!-2*u0+u1=0')            // u1!=u(-1)
-
-disp('for j=1            u0-2*u1+u2=(1/27)u1')
-
-disp('for j=2            u1-2*u2+u3=(2/27)u2')
-
-// we know;      u'=(u(j+1)-u(j-1))/2h
-// hence eliminating u1!
-// solving for u0,u1,u2,u3 , 
-u0=-.9879518;
-u1=-.3253012;
-u2=-.3253012;
-
-disp(x);
-disp(u);
-
-
+// example 7.6
+// solve the boundary value problem      u''=u*x;
+// u(0)+u'(0)=1;   u(x=1)=0;          h=1/3;
+
+
+// we know;     u''=(u(j-1)-2*u(j)+u(j+1))/h^2;
+
+// 1) second order method;
+ x=0:1/3:1;                                  
+ u1_2 = rand(1,3)
+ u3=1;
+ u=[u1_2 u3];
+// hence;
+disp('(u(j-1)-2*u(j)+u(j+1))/h^2=u(j)*x(j)')           // for j=0,1,2,3;
+
+disp('for j=0            u1!-2*u0+u1=0')            // u1!=u(-1)
+
+disp('for j=1            u0-2*u1+u2=(1/27)u1')
+
+disp('for j=2            u1-2*u2+u3=(2/27)u2')
+
+// we know;      u'=(u(j+1)-u(j-1))/2h
+// hence eliminating u1!
+// solving for u0,u1,u2,u3 , 
+u0=-.9879518;
+u1=-.3253012;
+u2=-.3253012;
+
+disp(x);
+disp(u);
+\ No newline at end of file