Question

I'm asked to solve the differential equation \[\frac{ dy }{ dt }=-yln(\frac{ y }{ 2 })\] and given the condition y(0)=1.

So far to approach this problem I have separate the variables, which leaves me with \[\frac{ dy }{ -yln(\frac{ y }{ 2 } ) }=dt\] My problem is, taking the integral of the left side. Any advice? I dont want to resort to using wolfram alpha, I would like to know to to do it by hand if possible. I feel like maybe I am missing an opportunity for substitution or simplifying?

Answer 1

Rewriting the left side term as \[\frac{ 1 }{ \ yln(\frac{ 2 }{ y } ) }\] makes it a bit simpler. And I know with properties of logs it could also be \[\ln(\frac{ 2 }{ y })^{y}\]
or \[y(\ln(2)-\ln(y))\]

Answer 2

Anything ideas? WA gives me the -ln(ln(y/2) which as far as I know does help at all in solving this.

Answer 3

@xartaan is correct, by bringing all the y values over to one side you can perform integration by parts:
\[\huge{\frac{dy}{-yln(\frac{y}{2})}=dt}\]
\[\huge{u=\ln(\frac{y}{2})}\]
\[\huge{\frac{du}{dy}=\frac{1}{(\frac{y}{2})} *\frac{1}{2}}\]
\[\huge{du=\frac{1}{y}dy}\]

Answer 4

sorry,not by parts....through direct substitution

Answer 5

Through this substitution method i got:\[\ln(\ln(\frac{y}{2}))=-t +c\]
taking exonetial of both sides:
\[\ln(\frac{y}{2})=kt^{-t}\]
subbing in for the constant k becomes:
\[k=\ln(\frac{1}{2})\]