QuestionClairaut requation $$y=4 x y^{\prime}+\ln y^{\prime}$$

Studdy Solution
Determine the singular solution by considering the derivative $\frac{d}{dc}(4xc + \ln c) = 0$:
$$4x + \frac{1}{c} = 0$$
Solve for $c$:
$$c = -\frac{1}{4x}$$
Substitute back to find $y$:
$$y = 4x(-\frac{1}{4x}) + \ln(-\frac{1}{4x})$$
$$y = -1 + \ln(-\frac{1}{4x})$$
This is the singular solution.
The general solution is $y = 4xc + \ln c$ and the singular solution is $y = -1 + \ln(-\frac{1}{4x})$.