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\begin{aligned} & \Rightarrow \mathrm{dE}=\frac{\mathrm{Adl}}{\mathrm{r}} \cdot 2 \sin \left(\frac{\omega \mathrm{t}-\mathrm{kr}+\mathrm{k} \Delta+\omega \mathrm{t}-\mathrm{kr}-\mathrm{k} \Delta}{2}\right) \cos \left(\frac{\omega \mathrm{t}-\mathrm{kr}+\mathrm{k} \Delta-\omega \mathrm{t}+\mathrm{kr}+\mathrm{k} \Delta}{2}\right) \\ & \Rightarrow \mathrm{dE}=\frac{2 \mathrm{Adl}}{\mathrm{r}} \sin \left(\frac{2 \omega \mathrm{t}-2 \mathrm{kr}}{2}\right) \cos \left(\frac{2 \mathrm{k} \Delta}{2}\right) \\ & \Rightarrow \mathrm{dE}=\frac{2 \mathrm{Adl}}{\mathrm{r}} \sin (\omega \mathrm{t}-\mathrm{kr}) \cos (\mathrm{k} \Delta) \\ & \Rightarrow \mathrm{dE}=\frac{2 \mathrm{~A}}{\mathrm{r}} \sin (\omega \mathrm{t}-\mathrm{kr}) \cos (\mathrm{kl} \sin \theta) \mathrm{dl} As the wave eq is given by: $$ \frac{\partial^2 y}{\partial x^2}=\frac{1}{v^2}\left(\frac{\partial^2 y}{\partial x^2}\right) $$ We want to find out the solution of this equation. We know that general formula for traveling wave is $$ y(x, t)=f(x \pm v t) $$ Let us suppose that equation (2) is the solution of (1). Suppose $x \pm v t=z \quad----(A)$ Therefore, equation (2) can be written as $$ y(x, t)=f(z)--^{-}-(B) $$ Differetiating (B) w.r.t $\mathrm{x}$ $$ \begin{aligned} & \frac{\partial y}{\partial x}=\frac{\partial f(z)}{\partial x} \\ & \Rightarrow \frac{\partial y}{\partial x}=\frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial x} \\ & \Rightarrow \frac{\partial y}{\partial x}=\frac{\partial f}{\partial z} \end{aligned} $$ Again differentiating yith respect to $x$ we get: $$ \begin{aligned} & \frac{\partial^2 y}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial z}\right) \\ & \Rightarrow \frac{\partial^2 y}{\partial x^2}=\frac{\partial}{\partial z}\left(\frac{\partial f}{\partial x}\right) \\ & \frac{\partial^2 y}{\partial x^2}=\frac{\partial}{\partial z} \cdot \frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial x} \quad \because \frac{\partial z}{\partial x}=\frac{\partial(x \pm v t)}{\partial x}=1 \\ & \Rightarrow \frac{\partial^2 y}{\partial x^2}=\frac{\partial^2 f}{\partial z^2}----(3) \end{aligned} $$ Similarly differentiating (B) with respect to $t$, we have: $$ \frac{\partial y}{\partial t}=\frac{\partial f(z)}{\partial t}=\frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial t} $$

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