Why solving trigonometry equation for \(\displaystyle{x}{y}'={y}{\cos{{\ln{{\frac{{{y}}}{{{x}}}}}}}}\)

anadyrskia0g5

anadyrskia0g5

Answered question

2022-03-23

Why solving trigonometry equation for xy=ycoslnyx

Answer & Explanation

lernarfnincln6g

lernarfnincln6g

Beginner2022-03-24Added 14 answers

With u=lnylnx
u=yy1x=1x(cosu1)
This has stationary solutions u=c=const , y=ecx for the roots of cosu1=0 which are at c=2kπ. For all other points you can separate the equation to get
cot(u2)=lnx+c
Malia Booth

Malia Booth

Beginner2022-03-25Added 16 answers

Well, we have
xy(x)=y(x)cos(ln(y(x)x))
Substitute y(x)=xP(x)
x(xP(x)+P(x))=xP(x)cos(ln(P(x)))
So, we also get:
P(x)P(x)cos(ln(P(x)))P(x)=1x
Integrate both sides with respect to x:
P(x)P(x)cos(ln(P(x)))P(x)dx=1xdx
So:
cot(ln(P(x))2)=ln|x|+C
We also now know:
cot(ln(y(x)x)2)=ln|x|+C

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