Let u (x, t) = ∑ t (t) sin (n π x/l), and n =1.∞

Substituting into the normalized equation, the initial conditions are as follows

∑[T'(t)+(nπa/l)？ T(t)]sin(nπx/l)=2sin(2x/l)

U(x，t)=∑T(0)sin(nπx/l)=0，T(0)=0。

Using the orthogonality of the function system {sin (n π x/l) | n = 0 ... on (0, l), there are

T'(t)+(nπa/l)？ t(t)=(4/l)∫(0 ~ l)sin(nπx/l)sin(2x/l)dx

=(- 1)^n(4nπsin2)/[4-(nπ)？ ]=q(n) (remember)

So T(t)=exp[-(nπa/l)? t]∫(0~t)q(n)exp[(nπa/l)？ t]dt

=(l/nπa)？ q(n){ 1-exp[-(nπa/l)？ t]}

The definite solution u(x, t)=∑T(t)sin(nπx/l)

=∑(l/nπa)？ q(n){ 1-exp[-(nπa/l)？ t]}sin(nπx/l)，n= 1..∞