• 2022-06-03
    4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$,并且$z(x,0)=x,z(0,y)={{y}^{2}}$,则$z(x,y)=$( )
    A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$
    B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$
    C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$
    D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$