Chapter 7, Problem 10RE

To determine

To find : the orthogonal trajectories of the family of curves

Answer to Problem 10RE

The orthogonal trajectories of the family of curves is $y^2\left(1-2\ln y\right)-2x^2=4C$

Explanation of Solution

Given information : The question is

  $y=e^{kx}$

Calculation :

  $y=e^{kx}$

Take natural log of both sides

  $\begin{array}{l}\ln y=\ln e^{kx}\\ \ln y=kx\\ \frac{\ln y}{x}=k\end{array}$

Differentiate both sides with respect to $x$

  $\underset{\text{using quotient rule}}{\underbrace{\frac{x\cdot \frac{1}{y}\cdot \frac{dy}{dx}-\ln y}{x^2}}}=0$

Multiply both sides by $x^2$

  $\begin{array}{l}\frac{x}{y}\cdot \frac{dy}{dx}-\ln y=0\\ \frac{x}{y}\cdot \frac{dy}{dx}=\ln y\\ \frac{dy}{dx}=\frac{y\ln y}{x}\end{array}$

Replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$

  $\begin{array}{l}-\frac{dx}{dy}=\frac{y\ln y}{x}\\ -xdx=y\ln ydy\\ {\displaystyle \int -xdx={\displaystyle \int y\ln ydy}}\\ -\frac{x^2}{2}=\ln y{\displaystyle \int ydy-{\displaystyle \int \left[\frac{d\left(\ln y\right)}{dy}{\displaystyle \int ydy}\right]}dy}\\ -\frac{x^2}{2}=\ln y{\displaystyle \int ydy-{\displaystyle \int \frac{1}{y}\cdot \frac{y^2}{2}dy}}\\ -\frac{x^2}{2}=\frac{y^2\ln y}{2}-\frac{y^2}{4}+C\\ -2x^2=2y^2\ln y-y^2+4C\\ y^2\left(1-2\ln y\right)-2x^2=4C\end{array}$