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Mathematics 22 Online
OpenStudy (anonymous):

general solution to u'=u^1/2 when u(0)=1

myininaya (myininaya):

So you know to integrate both sides?

OpenStudy (anonymous):

yeah you get 2u^1/2=t+C right?

myininaya (myininaya):

\[\frac{du}{dt}=u^\frac{1}{2}\] \[\frac{du}{u^\frac{1}{2}}=1 dt\]

OpenStudy (anonymous):

this is a differential equation problem with separation of variables

myininaya (myininaya):

now you integrate both sides lol

OpenStudy (anonymous):

\[\frac{du}{dt}=\sqrt{u}\] \[\frac{du}{\sqrt{u}}=dt\] \[\int_{}{}\frac{du}{\sqrt{u}}=\int_{}{}dt\]

myininaya (myininaya):

So yes what you have is right :) so far

OpenStudy (anonymous):

\[\frac{1}{\sqrt{u}}=u^{-1/2}\]

myininaya (myininaya):

Now apply you initial condition

OpenStudy (anonymous):

so C=2u^1/2 - 1?

myininaya (myininaya):

u(0)=1 Means we want to find C when t=0 and u=1

OpenStudy (anonymous):

oh ok so C=2

myininaya (myininaya):

Yep :)

OpenStudy (anonymous):

thanks then there was one i didnt understand u'=u/4+u^2 do you multiply by the reciprocal to both sides so all your u's are on one side?

OpenStudy (anonymous):

giving you integral of (4+u^2/u)=t+C

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