3、用初始条件定特解: 由方程<2>可得: f1(x)+f2(x)=(x)<5> 由方程〈3>可得: dfi(+at)d(x+at d(x+ at) dt df,(x-at d(x-at) y(x) d(x-at) dt t=03、用初始条件定特解： 由方程<2>可得： f 1 (x) + f 2 (x) = j (x) < 5 > 由方程<3>可得： ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 0 1 x dt d x at d x at df x at dt d x at d x at df x at t t = y - - - + + + + = =