Question:medium

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) = $}

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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.
Updated On: Jun 25, 2026
  • \(e^y \sin x + \cos x\)
  • \(\sin x + e^y e^x \cos x\)
  • \(e^y \cos x + \sin x\)
  • \(e^y \cos x + y \sin x\)
Show Solution

The Correct Option is C

Solution and Explanation

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