Answer to Question #281682 in Differential Equations for Code_X_

Question

Let the electric equipotential lines (curves of constant potential) between two concentric cylinders with the z-axis in space be given by u(x,y) = x^2:+:y^2 = c (these are circular cylinders in xyz-space). Using the method in the text, find their orthogonal trajectories (the curves of electric force).

Answer 1

Solution:

I) Differentiate the equation $x^2+y^2=c$ with respect to x:

$$2x+2yy^{'}=0\iff y^{'}=-\frac{x}{y}$$

II) Since the equation of orthogonal trajectories is:

$$y^{'}=-\frac{1}{f(x,y)}$$

and in this case $f(x,y)=-\frac{x}{y},$ we obtain:

$$y^{'}=-\frac{1}{\frac{x}{y}}=\frac{y}{x}$$

III) Now, let`s solve the differential equation:

$$y^{'}=\frac{y}{x}\iff \frac{dy}{dx}=\frac{y}{x}\iff\frac{dy}{y}=\frac{dx}{x}$$

IV) Integrate the left side in relation to y, and the right side in relation to x:

$$\int \frac{dy}{y}= \int \frac{dx}{x} \iff ln|y|=ln|x|+c$$

V) By taking exponents, we obtain:

$$y=Cx$$