【设变换为u=x-2y、v=x+ay,可把方程damp;#178;z/dxamp;#178;+damp;#178;z/(dxdy)-damp;#178;z/dyamp;#178;=0化简为damp;#178;z/(dudv)=0且damp;#178;z/(dudv)=damp;#178;z/(dvdu)求常数a可把方程6damp;#178;z/dxamp;#178;+damp;#178;z/(dx
<p>问题：【设变换为u=x-2y、v=x+ay,可把方程damp;#178;z/dxamp;#178;+damp;#178;z/(dxdy)-damp;#178;z/dyamp;#178;=0化简为damp;#178;z/(dudv)=0且damp;#178;z/(dudv)=damp;#178;z/(dvdu)求常数a可把方程6damp;#178;z/dxamp;#178;+damp;#178;z/(dxdy)-damp;#178;z/dyamp;#178;=0化简为damp;#178;z/(du】<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">吕婷婷的回答：<div class="content-b">网友采纳 dz/dx=dz/du*1+dz/dv*1=dz/du+dz/dv;dz/dy=dz/du*(-2)+dz/dv*(a)=adz/dv-2dz/du;于是d^2z/dx^2=d(dz/du+dz/dv)/dx=d^2z/du^2+d^2z/dudv+d^2z/dvdu+d^2z/dv^2=d^2z/du^2+2d^2z/dudv+d^2z/dv^2；d^2z/dxdy=d(dz/du+dz/...<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">化建宁的回答：<div class="content-b">网友采纳 我2B了。。。少了个系数6<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">吕婷婷的回答：<div class="content-b">网友采纳 前面做的都准确，最后一步代入得(12+a-2+4a)d^2z/dudv+(6+a-a^2)d^2z/dv^2=0，于是6+a-a^2=0,12+a-2+4a不等于0即可。解得a=3。
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