Question

The solution of $\frac{\partial^2 z}{\partial x^2} + z = 0$, satisfying $z(0,y) = e^y$, $\left(\frac{\partial z}{\partial x}\right)_{x=0} = 1$ is $z(x,y) = $}

• A. \(e^y \sin x + \cos x\)
• B. \(\sin x + e^y e^x \cos x\)
• C. \(e^y \cos x + \sin x\)
• D. \(e^y \cos x + y \sin x\)

Hint:

You can quickly eliminate incorrect options by directly evaluating the conditions: - Checking $x=0 \Rightarrow z = e^y$ eliminates options (A) and (B) since they give values of $1$ and $1+e^y$. - Differentiating remaining choices eliminates option (D) based on the second condition.

Answer 1

The Correct Option is C

Concept: The equation $\frac{\partial^2 z}{\partial x^2} + z = 0$ is a partial differential equation where the differentiation is done exclusively with respect to $x$. Hence, it can be treated as an ordinary linear differential equation with respect to $x$, while treating $y$ as an independent parameter. The auxiliary equation for $\frac{d^2z}{dx^2} + z = 0$ is: \[ m^2 + 1 = 0 \quad \Rightarrow \quad m = \pm i \] The general solution for imaginary roots contains arbitrary functions of $y$ instead of standard constants: \[ z(x,y) = f(y)\cos x + g(y)\sin x \] Using the provided boundary conditions, we isolate the unknown functions $f(y)$ and $g(y)$.

Step 1: Set up the general solution form.
Given the differential equation structure, the solution must take the form: \[ z(x,y) = f(y)\cos x + g(y)\sin x \quad \cdots (1) \]

Step 2: Apply the first boundary condition $z(0,y) = e^y$.
Substitute $x = 0$ into equation (1): \[ z(0,y) = f(y)\cos(0) + g(y)\sin(0) \] Since $\cos(0) = 1$ and $\sin(0) = 0$: \[ z(0,y) = f(y) \cdot 1 + g(y) \cdot 0 = f(y) \] We are explicitly given that $z(0,y) = e^y$, therefore: \[ f(y) = e^y \]

Step 3: Differentiate the general solution with respect to $x$ to use the second condition.
Differentiating equation (1) partially with respect to $x$: \[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}[f(y)\cos x + g(y)\sin x] = -f(y)\sin x + g(y)\cos x \]

Step 4: Apply the second boundary condition $\left(\frac{\partial z}{\partial x}\right)_{x=0 = 1$.}
Substitute $x = 0$ into the expression for the derivative: \[ \left(\frac{\partial z}{\partial x}\right)_{x=0} = -f(y)\sin(0) + g(y)\cos(0) = g(y) \] We are given that this derivative value equals $1$ at $x=0$, thus: \[ g(y) = 1 \]

Step 5: Assemble the complete explicit solution.
Substitute $f(y) = e^y$ and $g(y) = 1$ back into the structural equation (1): \[ z(x,y) = e^y \cos x + 1 \cdot \sin x = e^y \cos x + \sin x \] This matches Option (C).