Question: $\frac{\partial^2z}{\partial x^2}=a^2z\,$ given that when $x=0, \frac{\partial z}{\partial x}=asinx\;\text{and}\;\frac{\partial z}{\partial y}=a$
To solve this I assumed $z=Ae^{ax}+Be^{-ax}$ where A and B are constants or functions of y.
Assuming this, I get $$\frac{\partial z}{\partial x}=aAe^{ax}-aBe^{-ax}\qquad(1)$$ $$\frac{\partial z}{\partial y}=A'e^{ax}+B'e^{-ax}\qquad(2)$$
Inserting $x=0\,$ here gives $$\text{From eq (1)}\rightarrow A=B$$ $$\text{From eq (2)}\rightarrow A'+B'=a$$
Solving for A and B gives, $A=B=\frac{ay}{2}+c$, thus $$z=\left(\frac{ay}{2}+c\right)(e^{ax}+e^{-ax})$$
The procedure I used seems correct to me but since the question gave $\frac{\partial z}{\partial x}=asinx$, am I supposed to find the solution in terms of trigonometric functions only. Moreover, is directly assuming z to be a particular type of function a valid step.
Also, please tell the formula of sine or cosine in terms of e. Say can Euler's formula be used only for complex powers or is it applicable for real powers as well
$$\frac{\partial^2z}{\partial x^2}=a^2z\,$$ You don't "assume" that the solution is $z=Ae^{ax}+Be^{-ax}$ insofar you first prove that $e^{ax}$ and $e^{-ax}$ are particular solutions. The general solution is the linear combination of two independent solutions. So, this is not an assumption this is a factual result.
Your calculus is correct, leading to $$z=\left(\frac{ay}{2}+c\right)(e^{ax}+e^{-ax}) ,$$ which satisfies both boundary conditions specified as : When $x=0, \frac{\partial z}{\partial x}=a\sin(x)\;\text{and}\;\frac{\partial z}{\partial y}=a$
Note that the first condition (at $x=0$ ) is : $$\frac{\partial z}{\partial x}=0\quad\text{because}\quad\sin(x)=0$$ There is no sinusoidal function in the condition. So, you are not supposed to find the solution in terms of trigonometric functions only.
Nevertheless this manner to specify the condition appears strange. Are you sure that there is no typo in the wording ?