Semi-infinite cable

In this case, we assume that at x=0 the voltage is clamped to V=V₀. The boundedness of the voltage as $x\to\infty$ means that we must take  

$$V(x) = V_0 e^{-x/\lambda}$$ (13)

It is now clear why $\lambda$ is called a space-constant, it determines the voltage attenuation with distance.

Alternatively, we could demand that the current at x=0 be specified as I₀. From (14) we see that

$$-I_0 = \frac{1}{r_i} \frac{dV}{dx}$$

and since $V(x) =V_0e^{-x/\lambda}$ we get

$$V(x) = r_i I_0 \lambda e^{-x/\lambda}$$

The input resistance is the ratio of V(0)/I₀ is thus  

$$R_\infty = r_i \lambda = (4 R_A/(\pi d^2))\sqrt{(d/4)R_M/R_A} = \frac{2\sqrt{R_MR_A}}{\pi d^{3/2}}$$ (14)

The input conductance for the semi-infinite cable is thus

$$G_\infty =\frac{\pi d^{3/2}}{2\sqrt{R_MR_A}}$$

Note: It is often convenient to introduce the dimensionless space variable, $X=x/\lambda$ and the electrotonic length of the cable, $L=l/\lambda.$ This gets rid of all the divisions by $\lambda$in the exponentials and hyperbolic functions.