则方程<1化为: l(,n)=0 50n 2、求通解 选择 5 (x+at) [n=(x-ar 即:x=012X5+n) t=(1/2a)(-m) 0vdn=C1(2)
则方程<1>化为： ( , ) 0 2 = ¶ ¶ ¶ x h x h u 2、求通解： 选择 î í ì = - = + ( ) ( ) x at x at h x î í ì = - = + (1/ 2 )( ) (1/ 2)( ) x h x h t a x 即： ò = ¶ ¶ = ¶ ¶ \ 0 ( ) 1 h x h h x x d C u u
a5=c1(5) 0l=C1(5)+f2(n) 即(2,n)=f(2)+(m)<4> 故通解为 u(x,y)=f(x+ at)+f2(x-at
ò = + ¶ ¶ = ¶ ¶ ( ) ( ) ( ) 1 x 1 x 2 h x x x d C f u C u 即 u(x ,h) = f 1 (x ) + f 2 (h) < 4 > 故通解为： ( , ) ( ) ( ) u x y = f 1 x + at + f 2 x - at
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=0
3、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 - - - + + + + = =
0f1(x)-af2(x)=V(x) 即 f(x)-(x)= V()dc+C<6> 5式加上<6式再除2,可得: f(x)=70(m、r y(ada 2a
( ) ( ) ( ) 1 2 af x af x = y x ¢ - ¢ 即 + < > - = ò ( ) 6 1 ( ) ( ) 0 1 2 x x d C a f x f x y a a <5>式加上<6>式再除2，可得： 2 ( ) 2 1 ( ) 2 1 ( ) 0 1 C d a f x x x x = + + ò j y a a