I have this homework question, and while I feel pretty comfortable with the topic, the answer I got is pretty messy so I'm not sure if I made a mistake somewhere. I have the equation for which I need to find the orthogonal trajectory:
$$
y=e^{cx} \tag{a}
$$
Taking the partial derivative with respect to $x$, I get
$$\frac{dy}{dx}=ce^{cx} \tag{b}$$
From (a), we can solve for $c$ by taking the natural log of each side, yielding
$$c=\frac{\ln(y)}{x} \tag{c}$$
Substituting into (b), we get
$$\frac{dy}{dx}=\frac{\ln(y)}{x}e^{cx} \tag{d}$$
To find the family of curves that would intersect orthogonal to (d), we take the negative reciprocal of (d), yielding:
$$\frac{dy}{dx}=-\frac{x}{e^{cx}\ln(y)} \tag{e}$$$
Solving via separation of variables, we get
$$\int\,\ln(y)\; dy = \int\,-xe^{-cx}\;dx \tag{f}$$
This is where I feel I might have made a mistake. Maybe I should have solved via an integrating factor? Anyway, the way I proceeded was to use integration by parts on both sides.
Integrating $\ln(y)$ yields
$$y[\ln(y)-1]+k$$
Integrating $-xe^{-cx}$ yields
$$-\frac{xe^{-cx}}{c}-\frac{1}{c^2}e^{-cx}$$
Finally giving an integral curve
$$y[\ln(y)-1]+k=-\frac{xe^{-cx}}{c}-\frac{1}{c^2}e^{-cx}$$
Can someone let me know if/where I made a mistake? I'm not even necessarily sure it is wrong, and I feel like I generally know what I'm doing, but its just kind of a weird problem. Thanks
Best Answer
The line $\displaystyle \frac{dy}{dx} = -\frac{1}{cy}$ in nathan.j.mcdougall's answer is incorrect; a solution to this equation (in which $c$ is arbitrary but fixed) will be orthogonal to $y = e^{cx}$ for this fixed value of $c,$ but not to every member of the family $y = e^{cx},$ which is what is required of an orthogonal trajectory.
The OP is correct to argue (in the comments) that $c$ must be eliminated -- we wish to find an equation satisfied by every member of the family. In the same spirit, however, we should write equation (d) as $\displaystyle \frac{dy}{dx} = \frac{\ln{y}}{x}y = \frac{y\ln{y}}{x},$ which is true for every member of the original family, rather than substituting $y = e^{cx},$ which is only true for the one member corresponding to this fixed (albeit arbitrary) value of $c.$ Thus, the orthogonal trajectories satisfy $\displaystyle \frac{dy}{dx} = -\frac{x}{y\ln{y}},$ which we solve by separation of variables, as suggested.