A solution to a possible QotD
Since y´=(ex/3)/y we can
Separate: y dy=ex/3dx
Integrate: ∫y dy=∫ex/3dx
The general solution is in implicit form as
[y2/2]=3ex/3+C
Initial condition leads to constant of integration: so when x=0,
then y=2, and therefore:
[22/2]=3e0/3+C which is 2=3+C, and C=–1.
The transition from implicit description to explicit description
[y2/2]=3ex/3–1 and so
y2=6ex/3–2 and
y=+/–sqrt(6ex/3–2). Which sign should be used? Since
y(0)=2>0, I'll take the + sign, and the
solution in explicit form is
y=sqrt(6ex/3–2).
Note If I had as initial condition y(0)=–2 then the
solution would be
y=–sqrt(6ex/3–2). Really -- the sign is
important.
Maintained by
greenfie@math.rutgers.edu and last modified 3/23/2009.