设 $z=f(x, y)$ 满足 $\frac{\partial^{2} f}{\partial y^{2}}=2 x, f(x, 1)=0, \frac{\partial f(x, 0)}{\partial y}=\sin {x}$,
求$f(x, y)$.
解
\[\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)=2 x \Longleftrightarrow \frac{\partial f}{\partial y}=2 x y+\varphi(x), \varphi(x)为x 的任意函数 \\f(x, y)=x y^{2}+\varphi(x) y+\phi(x)\ \ \psi(x) 也是 x 的任意函数. \\由 \frac{\partial f(x, 0)}{\partial y}=\sin x, \\得 \left.[2 x y+\varphi(x)]\right|_{y=0}=\sin x, \\即 \varphi(x)=\sin x. \\由 f(x, 1)=0, 得 \left.\left[x y^{2}+\varphi(x) y+\psi(x)\right]\right|_{y=1}=x+\sin x+\psi(x)=0 \\即 \psi(x)=-x-\sin x \\因此, f(x, y)=x y^{2}+y \sin x-x-\sin x.\]