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Q.
The equation of the trajectories which is orthogonal to the family of curves $\cos y=a e^{-x}$ is
Differential Equations
Solution:
$+\sin y \frac{d y}{d x}=+a e^{-x}$
$\sin y \frac{d y}{d x}=\cos y \Rightarrow \frac{d y}{d x}=\cot y$
equation to the orthogonal trajectory will be obtained by solving the differential equation
$-\frac{d y}{d x} =\cot y $
$c-x \left.=\ln (\sin y) \Rightarrow \sin y=c^{-x}\right]$