高等数学试用于华东理工大学的学生
有二阶连续导数，证明： 自测 13/三.设 , ψ 有二阶连续导数，证明：函数 三1 1 x + at z ( x , t ) = [ ( x + at ) + ( x at )] + ∫ x atψ (u)du 2 2a
证明： 证明：
2z 2z 2 =a 满足关系式(偏微分方程) 满足关系式(偏微分方程) 2 2 . t x z 1 1 ′ ( x + at ) + ′ ( x at )] + [ψ ( x + at ) ψ ( x at )] = [ 2a x 2 1 1 2z 1 = [ ′′( x + at ) + ′′( x at ) + ψ ′( x + at ) ψ ′( x at )] a a x 2 2 1 z 1 ′( x + at ) a ′( x at )] + [aψ ( x + at ) + aψ ( x at )] = [ a 2a t 2
1 2z a2 ′′( x + at ) + ′′( x at )] + [aψ ′( x + at ) aψ ′( x at )] = [ 2 2 a t 19