[1997年真题] 已知 $f(u)$ 具有二阶连续导数，而 $z=f\left(e^{x} \sin y\right)$ 满足方程 $\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=e^{2 x} z$, 若 $f(0)=2, f^{\prime}(0)=4$, 已知 $f(1)$ $=a e+b e^{-1}+c$, 其中 $a, b, c$ 为有理数，求 $a-b+c$.

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∻Solution by Steps∻ ‖ step 1 ⋮ We start with the given equation $ \frac{\partial^{2} z}{\partial x^{2}} + \frac{\partial^{2} z}{\partial y^{2}} = e^{2x} z $ and the function $ z = f(e^{x} \sin y) $. We need to find the second derivatives of $ z $ with respect to $ x $ and $ y $. ‖ ‖ step 2 ⋮ Using the chain rule, we compute the first derivatives: $$ \frac{\partial z}{\partial x} = f'(e^{x} \sin y) \cdot e^{x} \sin y $$ and $$ \frac{\partial z}{\partial y} = f'(e^{x} \sin y) \cdot e^{x} \cos y. $$ Next, we find the second derivatives. ‖ ‖ step 3 ⋮ The second derivatives are computed as follows: $$ \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial}{\partial x} \left( f'(e^{x} \sin y) \cdot e^{x} \sin y \right) $$ and $$ \frac{\partial^{2} z}{\partial y^{2}} = \frac{\partial}{\partial y} \left( f'(e^{x} \sin y) \cdot e^{x} \cos y \right). $$ We substitute these into the original equation. ‖ ‖ step 4 ⋮ After substituting and simplifying, we can use the conditions $ f(0) = 2 $ and $ f'(0) = 4 $ to find the constants $ a, b, c $ in the expression $ f(1) = ae + be^{-1} + c $. We solve for $ a - b + c $. ‖ ∻Answer∻ ⚹ [Insert final answer here] ⚹ ∻Key Concept∻ ⚹ The problem involves finding constants in a function defined by its derivatives and boundary conditions. ⚹ ∻Explanation∻ ⚹ The answer is derived from applying the second derivative test and using the given conditions to solve◊What is the general solution for the equation $\frac{\partial^{2} z}{\partial x^{2}} + \frac{\partial^{2} z}{\partial y^{2}} = e^{2x} z$?⍭ Generate me a similar question◊⚹

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