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megannicole51
 2 years ago
solve the differential equation. assume x,y,t>0
dy/dt=yln(y/2). y(0)=1
megannicole51
 2 years ago
solve the differential equation. assume x,y,t>0 dy/dt=yln(y/2). y(0)=1

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megannicole51
 2 years ago
Best ResponseYou've already chosen the best response.0so i think im doing this right so far but I'm not really sure can someone work through the steps with me?

ddog1437
 2 years ago
Best ResponseYou've already chosen the best response.0are you taking calculus though?

megannicole51
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ ylny } = \frac{ 1 }{ 2 }dt\]

megannicole51
 2 years ago
Best ResponseYou've already chosen the best response.0ill show the steps i have so far and can u tell me if im right or not?

megannicole51
 2 years ago
Best ResponseYou've already chosen the best response.0ive never seen it done like that actually....my professor hasnt really done any examples so i am trying to figure it out from our book which isnt helping either

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{y \ln (\frac{y}{2})} dy= dt \]

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1He is doing separation of variables.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1I think he meant to divide yln(y/2) on both sides.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1and then he multiplied both sides by dt

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.1Thanks to @myininaya for pointing out my mistake: \(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles \[\frac{dy}{dt}=y\ln\left(\frac{y}{2}\right)\\ \frac{1}{y\ln\left(\frac{y}{2}\right)}~dy=dt\] Integrate both sides: \[\int\frac{1}{y\ln\left(\frac{y}{2}\right)}~dy=\int dt\] First, a substitution: \(u=\dfrac{y}{2}\), or \(2u=y\), so that \(2du=dy\): \[\int\frac{1}{2u\ln u}~(2~du)=\int dt\\ \int\frac{1}{u \ln u}~du=\int dt\] Got everything so far? \(\color{blue}{\text{End of Quote}}\)
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