1. **Stating the problem:** Solve the partial differential equation $$\frac{\partial^2 z}{\partial x^2} + z = 0$$ with initial conditions at $$x=0$$: $$z = e^y$$ and $$\frac{\partial z}{\partial x} = 1$$. 2. **Understanding the equation:** This is a second-order linear PDE in variable $$x$$ with parameter $$y$$ treated as a constant during differentiation with respect to $$x$$. 3. **Rewrite the PDE as an ODE in $$x$$:** For fixed $$y$$, $$z$$ is a function of $$x$$ satisfying $$\frac{d^2 z}{d x^2} + z = 0$$. 4. **Characteristic equation:** Assume solution $$z = e^{m x}$$, then $$m^2 + 1 = 0 \Rightarrow m = \pm i$$. 5. **General solution:** $$z(x,y) = A(y) \cos x + B(y) \sin x$$ where $$A(y)$$ and $$B(y)$$ are functions of $$y$$. 6. **Apply initial conditions at $$x=0$$:** - $$z(0,y) = A(y) = e^y$$ - $$\frac{\partial z}{\partial x} = -A(y) \sin x + B(y) \cos x$$, so at $$x=0$$: $$\frac{\partial z}{\partial x}(0,y) = B(y) = 1$$. 7. **Final solution:** $$ z(x,y) = e^y \cos x + \sin x $$ This satisfies the PDE and given initial conditions.