"Why solving trigonometry equation for xy′=ycos⁡ln⁡yx" solved and explained

anadyrskia0g5

Answered question

2022-03-23

Why solving trigonometry equation for

x y^{'} = y \cos \ln \frac{y}{x}

lernarfnincln6g

Beginner2022-03-24Added 14 answers

With

u = \ln y - \ln x

u^{'} = \frac{y^{'}}{y} - \frac{1}{x} = \frac{1}{x} (\cos u - 1)

This has stationary solutions

u = c = c o n s t

y = e^{c} x

for the roots of

\cos u - 1 = 0

which are at

c = 2 k π

. For all other points you can separate the equation to get

\cot (\frac{u}{2}) = \ln x + c

Malia Booth

Beginner2022-03-25Added 16 answers

Well, we have

x \cdot y^{'} (x) = y (x) \cdot \cos (\ln (\frac{y (x)}{x}))

Substitute

y (x) = x \cdot P (x)

x \cdot (x \cdot P^{'} (x) + P (x)) = x \cdot P (x) \cdot \cos (\ln (P (x)))

So, we also get:

\frac{P^{'} (x)}{P (x) \cdot \cos (\ln (P (x))) - P (x)} = \frac{1}{x}

Integrate both sides with respect to x:

\int \frac{P^{'} (x)}{P (x) \cdot \cos (\ln (P (x))) - P (x)} d x = \int \frac{1}{x} d x

So:

\cot (\frac{\ln (P (x))}{2}) = \ln | x | + C

We also now know:

\cot (\frac{\ln (\frac{y (x)}{x})}{2}) = \ln | x | + C

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