v(x,2)x0=0v(xD)x=04 0 只求特解 t=0 傅立叶级数法 u(x)=2O)m"7 nZA 2G nZX 2G nAIl 4G cOS 0 (2k+1)丌 T2k+()+[ (2k+1) 4G )2-b72k+1(1)= 2k+1)丌 241(O)=072k+1(O)=0 拉普拉斯变换: 2k+(p){p2+[( (2k+1)ma 4G )2-b]}= p(2k+1)ru(x,t) x=0 = 0 ux (x,t) x=l = 0 0 u t=0 = 只求特解 傅立叶级数法   = = 0 ( , ) ( )sin n n l n x u x t T t   = l n dx l n x l G G 0 sin 2    = = 0 sin n n l n x G G  l l n x n G 0 cos 2   = − (2 1) 4 + = k G   (2 1) 4 ) ] ( ) (2 1) ''( ) [( 2 1 2 2 1 + − = + + + + k G b T t l k a T t k k T2k+1 (0) = 0 T2k+1 '(0) = 0 拉普拉斯变换：   (2 1) 4 ) ]} (2 1) ( ){ [( ~ 2 2 2 1 + − = + + + p k G b l k a T k p p