100 Chapter 2
SOLUTION: We consider two possibilities for values of λ. We first consider λ = 0. For this
case, the system basis vectors are
> restart:y1(x):=1;y2(x):=x;
y1(x) := 1
y2(x) := x (2.79)
General solution
> y(x):=C1*y1(x)+C2*y2(x);
y(x) := C1 +C2x (2.80)
Substituting into the boundary conditions yields
> eval(subs(x=0,y(x)))=0;
C1 = 0 (2.81)
> eval(subs(x=b,diff(y(x),x)+h*y(x)))=0;
C2 +h(C1 +C2b) = 0 (2.82)
The only solution to the preceding is the trivial solution. We next consider λ>0. We set
λ = μ
2
, and for this case, the system basis vectors are
> y1(x):=sin(mu*x);y2(x):=cos(mu*x);
y1(x) := sin(μx)
y2(x) := cos(μx) (2.83)
General solution
> y(x):=C1*y1(x)+C2*y2(x);
y(x) := C1 sin(μx) +C2 cos(μx) (2.84)
Substituting into the boundary conditions yields
> eval(subs(x=0,y(x)))=0;
C2 = 0 (2.85)
> eval(subs(x=b,diff(y(x),x)+h*y(x)))=0;
C1 cos(μb)μ −C2 sin(μb)μ +h(C1 sin(μb) +C2 cos(μb)) = 0 (2.86)