We are tasked with solving the partial differential equation:

$\frac{\partial^2 z}{\partial x \partial y} = x^2 y$

Subject to the conditions:

• $z(x,0) = x^2$
• $z(1,y) = \cos y$

Step 1: Integrate with respect to $y$

First, integrate $\frac{\partial^2 z}{\partial x \partial y} = x^2 y$ with respect to $y$:

$\frac{\partial z}{\partial x} = \int x^2 y \, dy = x^2 \frac{y^2}{2} + f(x),$

where $f(x)$ is an arbitrary function of $x$ (since the integration is with respect to $y$).

Step 2: Integrate with respect to $x$

Next, integrate $\frac{\partial z}{\partial x} = x^2 \frac{y^2}{2} + f(x)$ with respect to $x$:

$z = \int \left( x^2 \frac{y^2}{2} \right) dx + \int f(x) dx = \frac{x^3 y^2}{6} + F(x) + g(y),$

where $F(x)$ is the antiderivative of $f(x)$ (i.e., $F'(x) = f(x)$) and $g(y)$ is an arbitrary function of $y$ (since the integration is with respect to $x$).

Thus, the general solution is:

$z = \frac{x^3 y^2}{6} + F(x) + g(y).$

Step 3: Apply the boundary conditions

Condition 1: $z(x, 0) = x^2$

Substitute $y = 0$ into $z = \frac{x^3 y^2}{6} + F(x) + g(y)$:

$z(x, 0) = \frac{x^3 (0)^2}{6} + F(x) + g(0) = F(x) + g(0).$

Since $z(x, 0) = x^2$, we have:

$F(x) + g(0) = x^2.$

Let $g(0) = C$, where $C$ is a constant. Then $F(x) = x^2 - C$.

Condition 2: $z(1, y) = \cos y$

Substitute $x = 1$ into $z = \frac{x^3 y^2}{6} + F(x) + g(y)$:

$z(1, y) = \frac{(1)^3 y^2}{6} + F(1) + g(y) = \frac{y^2}{6} + F(1) + g(y).$

Since $z(1, y) = \cos y$, we have:

$\frac{y^2}{6} + F(1) + g(y) = \cos y.$

From the first condition, $F(x) = x^2 - C$, so $F(1) = 1^2 - C = 1 - C$. Substituting this:

$\frac{y^2}{6} + (1 - C) + g(y) = \cos y.$

Simplify:

$g(y) = \cos y - \frac{y^2}{6} - 1 + C.$

Final Solution:

Substitute $F(x)$ and $g(y)$ back into the general solution:

$z = \frac{x^3 y^2}{6} + (x^2 - C) + \left( \cos y - \frac{y^2}{6} - 1 + C \right).$

Simplify:

$z = \frac{x^3 y^2}{6} + x^2 - C + \cos y - \frac{y^2}{6} - 1 + C.$

Combine like terms:

$z = \frac{x^3 y^2}{6} + x^2 + \cos y - \frac{y^2}{6} - 1.$

Final Answer:

$z = \frac{x^3 y^2}{6} - \frac{y^2}{6} + x^2 + \cos y - 1.$

Would you like further clarifications or details?

Here are 5 related questions for further exploration:

1. How can you verify that the solution satisfies the given partial differential equation?
2. What changes if one boundary condition is altered (e.g., $z(1, y) = \sin y$)?
3. How would the solution differ if the partial derivative equation involved $\frac{\partial^2 z}{\partial x^2}$?
4. Can this method be extended to higher-order partial differential equations?
5. How does the arbitrary function separation method work in cases where multiple variables appear nonlinearly?

Tip: Always substitute the final solution back into the original PDE and boundary conditions to confirm its correctness!