【f(x)具有二阶连续导数,f(0)=1,f#39;(0)=-1,且[xy(x+y)-f(x)y]dx+[f#39;(x)+x^2y]dy为某二元函数的全微分,求f(x)】
<p>问题：【f(x)具有二阶连续导数,f(0)=1,f#39;(0)=-1,且dx+dy为某二元函数的全微分,求f(x)】<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">何海燕的回答：<div class="content-b">网友采纳 设该二元函数为g(x,y),则g'x(x,y)=xy(x+y)-f(x)y两边对x求积分g(x,y)=x³y/3+x²y²/2-y∫f(x)dxg'y(x,y)=f'(x)+x²y两边对y求积分g(x,y)=f'(x)y+x²y²/2+C∴x³y/3-y∫f(x)dx=f'(x)y+...
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