Graphical Meaning
Previously, we have been looking for a family of solution curves when concerning a (first-order) ODE. Suppose that the ODE has the general form of
\begin{equation}
f(x,y,\frac{dy}{dx}) = 0 \tag{1}
\end{equation}
which has a set of solutions:
\begin{equation}
g(x,y,c) = 0 \tag{2}
\end{equation}
\( c \) is the integration constant, acting as a parameter to control the appearance of the solution curves. A related problem is to find the orthogonal trajectories with respect to (2), which are another family of curves always orthogonal (perpendicular) to those indicated by (2).
Remember that in high-school geometry, for a curve \(y(x)\) which has a slope of \( dy/dx \) locally, another curve to cross perpendicularly needs to have a negative reciprocal slope of \( -1/(dy/dx) = -dx/dy \). For any orthogonal trajectory this needs to hold everywhere. Therefore, to derive orthogonal trajectories, we simply replace \( dy/dx \) by \( -dx/dy \) in (1) and solve this newly modified ODE:
\begin{equation}
f(x,y,-\frac{dx}{dy}) = 0 \tag{3}
\end{equation}
Sometimes we are given (2) only and we can obtain (1) by an implicit differentiation of (2) against \(x\) and eliminate \( c \) by using (2) again.
Example
Here we tackle a very simple example, consider the family of circles centered at the origin:
\begin{equation}
x^2 + y^2 = c^2 \tag{4}
\end{equation}
The corresponding ODE is, by differentiating (4) with respect to \( x \):
\begin{align}
2x + 2y \frac{dy}{dx} &= 0 \\
\frac{dy}{dx} &= -\frac{x}{y} \tag{5}
\end{align}
To obtain the orthogonal trajectories, replace \( dy/dx \) by \( -dx/dy \) as suggested above and solve the new equation:
\begin{align}
\frac{dy}{dx} &= \frac{y}{x} \tag{6} \\
\int \frac{dy}{y} &= \int \frac{dx}{x} \\
\ln y &= \ln x + C \\
y &= Ax \tag{7}
\end{align}
where \( A = e^C \). So the orthogonal trajectories are the family of straight lines through the origin, each with a slope of \( A \). (See the schematic below)
Exercise
Find the orthogonal trajectories for the family of parabolas with foci at the origin:
\begin{equation}
x^2 = 4c(y+c) \tag{8}
\end{equation}
Answer
Differentiating (8) against \( x \) gives
\begin{align}
2x &= 4c\frac{dy}{dx}
\end{align}
The trick is to write
\begin{align}
4x^2 &= 16c^2(\frac{dy}{dx})^2 \\
x^2\frac{dx}{dy} &= 4c^2\frac{dy}{dx}
\end{align}
by squaring. Then we can eliminate \( c \) from the ODE via using (8):
\begin{align}
x^2\frac{dx}{dy} + 2xy &= 4c^2\frac{dy}{dx} + 4cy\frac{dy}{dx} = 4c(c+y)\frac{dy}{dx} \\
x^2\frac{dx}{dy} + 2xy &= x^2\frac{dy}{dx}
\end{align}
Note that replacing \( dy/dx \) by \( -dx/dy \) yields
\begin{align}
x^2(-\frac{dy}{dx}) + 2xy &= x^2(-\frac{dx}{dy}) \\
x^2\frac{dx}{dy} + 2xy &= x^2\frac{dy}{dx}
\end{align}
which is unaltered compared to the old ODE above. Therefore, the orthogonal trajectories of parabolas with foci at the origin are exactly themselves.
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