meili 发表于 2022-10-27 15:47:35

设函数z=f(u,v)具有二阶连续偏导数,且满足方程∂2f∂u2+∂2f∂v2=0,证明函数z=f(exsiny,excosy)满足方程∂2z∂x2+∂2z∂y2=0.

<p>问题:设函数z=f(u,v)具有二阶连续偏导数,且满足方程∂2f∂u2+∂2f∂v2=0,证明函数z=f(exsiny,excosy)满足方程∂2z∂x2+∂2z∂y2=0.
<p>答案:↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">李声威的回答:<div class="content-b">网友采纳  设u=exsiny,v=excosy,则∂z∂x=∂f∂uexsiny+∂f∂vexcosy,∂z∂y=∂f∂uexcosy−∂f∂vexsiny∂2z∂x2=∂∂x[∂f∂uexsiny]+∂∂x[∂f∂vexcosy]=[∂2f∂u2exsiny+∂2f∂u∂vexcosy]exsiny+∂f∂u exsin...
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查看完整版本: 设函数z=f(u,v)具有二阶连续偏导数,且满足方程∂2f∂u2+∂2f∂v2=0,证明函数z=f(exsiny,excosy)满足方程∂2z∂x2+∂2z∂y2=0.