# Transmission Line Differential Equation

I was reading the transmission line chapter of Engineering Electromagnetics by Buck and William H. Hayt. I don't understand why the following is true: $$\frac{\partial{^{2}V}}{\partial{z}^2}=LC\frac{\partial{^{2}V}}{\partial{t}^2} +(LG+RC)\frac{\partial{V}}{\partial{t}} +RGV \ \textbf{(1)} \Rightarrow \frac{d^{2}\ \boldsymbol{V_{s}}}{dz ^{2}}=-\omega^{2}LC\ \boldsymbol {V_{s}}+j\omega(LG+RC) \boldsymbol {V_{s}}+RGV_{s} \ \textbf{(2)}$$

                          Figure 1: Transmission Line Model


I understand that we can get equation (1) from the basic KCL and KCL. Nevertheless, according to the definition of the textbook： $$V = \frac{1}{2}|V_{o}| e^{j \phi}e^{-j\beta z}e^{j \omega t} + \frac{1}{2}|V_{o}| e^{-j \phi}e^{j\beta z}e^{-j \omega t} = \frac12 \boldsymbol{V_s}e^{j \omega t}+\frac12\boldsymbol{V_s^*}e^{-j \omega t} \ \textbf{(3)}$$ where $$\V\$$ is the instantaneous value of the voltage whereas $$\\boldsymbol{V_s}\$$ is the phasor form of the complex voltage without $$\e^{j\omega t}\$$: $$\boldsymbol{V_{s}}=|V_{o}| e^{-j \phi}e^{j\beta z} \ \textbf{(4)}$$

Therefore, when we are solving eq(1), we have to deal with the conjugate term shown in eq(3).

For me to get the result shown on the textbook eq(2), I need to do the following: $$\text{Define} \ V = \alpha + \alpha^{*} \text{where} \ \alpha = \frac{1}{2} \boldsymbol{V_{s}} e^{j \omega t}$$ LHS： $$\frac{\partial{^{2}V}}{\partial{z}^2} = \frac{\partial{^{2}(\alpha + \alpha^{*})}}{\partial{z}^2}$$ RHS:   $$LC\frac{\partial{^{2}}(\alpha + \alpha^{*})}{\partial{t}^2} +(LG+RC)\frac{\partial(\alpha + \alpha^{*})}{\partial{t}} +RG(\alpha + \alpha^{*})$$

Here is my question: Can I simply disregard all the conjugates $$\\alpha^{*}\$$ that appear in the equation since it will allow me to get the correct answer? If so, what is the mathematical reasoning behind this? Is it because it is a linear combination? I would like to know the exact law/theorem. If not, where did I do wrong? Sorry, I am kind of rusty in math. $$\frac{\partial{^{2}(\alpha) }}{\partial{z}^2} = LC\frac{\partial{^{2}}(\alpha )}{\partial{t}^2} +(LG+RC)\frac{\partial(\alpha )}{\partial{t}} +RG(\alpha )$$ $$\frac{\partial{^{2} (\frac{1}{2} \boldsymbol{V_{s}} e^{j \omega t}) }}{\partial{z}^2} = LC\frac{\partial{^{2}}( \frac{1}{2} \boldsymbol{V_{s}} e^{j \omega t})}{\partial{t}^2} +(LG+RC)\frac{\partial( \frac{1}{2} \boldsymbol{V_{s}} e^{j \omega t})}{\partial{t}} +RG( \frac{1}{2} \boldsymbol{V_{s}} e^{j \omega t} )$$ $$\frac{d^{2}{\boldsymbol{V_{s}}}}{dt^{2}}=-\omega^{2}LC\boldsymbol{V_{s}}+j\omega(LG+RC)\boldsymbol{V_{s}}+RGV_{s} \ \textbf{(2)}$$