#let y = e^lx.eq (1)
#Then substitution and omission of common factor e^lx gives the following equation where l is lamda
#the given eq then becomes
reset()
y = var('y')
l = var('l')
expand(l^4 - 5*l^2 +4 == 0) # eq(2)
#this is a quadratic equation in u = l^2, eq(2) becomes

u = var('u')
#u = l^2
expand(u^2 - 5*u +4 == 0)
print solve([u^2 - 5*u +4 == 0],u)
#as u = l^2 and u =1,4 we have

l = sqrt(1);print l
l = -sqrt(1);print l
l = sqrt(4);print l
l = -sqrt(4);print l
# substituting these values of l in eq(1) we get
var('c1,c2,c3,c4')

y = c1*exp(-2*x) + c2*exp(-x) + c3*exp(x) + c4*exp(2*x)
expand(y)

Kreyszig-2.13-3 (last edited 2010-12-17 11:39:36 by 172)