Find the solution z=z(x,y) of (∂^2 z)/(∂ x ∂ y) = x+y…
Find the solution $z=z(x,y)$ of \[ \frac{\partial^2 z}{\partial x\,\partial y} = x+y \] satisfying $z(x,0)=x$ and $z(0,y)=y^2$.
Explanation of Method
Integrate the mixed partial twice in succession — once with respect to $y$, once with respect to $x$ — introducing an arbitrary function of the "other" variable at each stage. Then pin those two arbitrary functions down using the two given boundary curves.
Step 1 · Integrate with respect to $y$
Treating $x$ as a parameter, \[ \frac{\partial z}{\partial x} = xy+\frac{y^2}{2}+\varphi(x), \] where $\varphi(x)$ is an arbitrary function of $x$.
Step 2 · Integrate with respect to $x$
\[ z = \frac{x^2y}{2}+\frac{xy^2}{2}+\Phi(x)+\psi(y), \] where $\Phi(x)=\int\varphi(x)\,dx$ and $\psi(y)$ are arbitrary functions.
Step 3 · Apply the boundary conditions
Setting $y=0$: \[ x = \Phi(x)+\psi(0) \implies \Phi(x)=x-C_0,\qquad C_0:=\psi(0). \] Setting $x=0$: \[ y^2 = \Phi(0)+\psi(y) \implies \psi(y)=y^2-\Phi(0)=y^2+C_0. \] Substituting $\Phi(x)=x-C_0$ and $\psi(y)=y^2+C_0$ back into the general solution, the constant $C_0$ cancels: \[ z = \frac{x^2y}{2}+\frac{xy^2}{2}+x+y^2. \]
Verification
With $z=\dfrac{x^2y}{2}+\dfrac{xy^2}{2}+x+y^2$: \[ z_x = xy+\frac{y^2}{2}+1, \qquad z_{xy}=x+y. \checkmark \] \[ z(x,0)=x, \qquad z(0,y)=y^2. \checkmark \]
Result
The solution of the mixed boundary value problem is: