Partial Differential Equations 1

I'm going to show a method to solve the PDE without knowing the form of the actual solution given (i.e. solving from scratch):

We have:

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

$z(x,0)=x^2$ , $z(1,y)=\cos(y)$

Integrate with respect to $x$ to obtain:

$\frac{\partial z}{\partial y} = \frac{1}{3} x^3 y + f(y)$

Where $f(y)$ is some arbitrary function. Integrate again with respect to $y$ this time, obtaining:

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

Where $g(x)$ is arbitrary and $f(y)$ remains the same since the integral of an arbitrary function of $y$ with respect to $y$ is again an arbitrary function. Using boundary conditions, we have:

$z(x,0) = x^2 = f(0) +g(x)$

$z(1,y) = \frac{1}{6} y^2 +f(y) +g(1) =\cos(y)$

We may now find $f(y)$ and $g(x)$ :

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

$g(x) = x^2 +g(1) -1$

Returning to our original expression for $z$ we have:

$z(x,y) = \frac{1}{6} x^3y^2 +\cos(y) -\frac{1}{6}y^2 -g(1) +x^2 +g(1) -1$

From which it immediately follows that: $A+B+C+D = \frac{7}{3}$

It is given that: $\quad z(x,y) = Ax^3y^2+B\cos{y} - Cy^2+Dx^2-1$

$\Rightarrow z(x,0) = 0 + B - 0 + Dx^2 - 1 = x^2\quad \Rightarrow B = D = 1$

$\Rightarrow z(1,y) = Ay^2+\cos{y} - Cy^2+1-1 = \cos{y} \quad \Rightarrow A = C$

$\Rightarrow z(x,y) = Ay^2(x^3-1)+\cos{y} +x^2-1$

Therefore,

$\dfrac {\partial z} {\partial x} = 3Ax^2y^2 + 2x$

$\dfrac {\partial ^2 z} {\partial x \partial y} = 6Ax^2y = x^2y \quad \Rightarrow A = \frac {1}{6}$

Therefore, $A+B+C+D = \frac {1}{6} + 1 + \frac {1}{6} + 1 = \boxed {2\frac{1}{3}}$