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Mathematics
OpenStudy (anonymous):

Solve the differential equation dy dt = e−3y by separation of variables. You should obtain an explicit formula for y as a function of t, containing a constant c (“general solution”). For what values of t is your formula valid?

OpenStudy (anonymous):

thats e^3y, sorry

OpenStudy (anonymous):

e^-3y actually

OpenStudy (ash2326):

\[ dy/dt= e^{-3y}\] \[dy/e^{-3y}=dt\] \[e^{3y}dy=dt\] integrate both sides \[e^{3y}/3=t+c\] \[e^{3y}=3t+c1\] take log bot sides \[3y=log(3t+c1)\] \[y=(1/3)log (3t+c1)\] 3t+c1>0 or t>-c1/3 so the formula is valid for t>-c1/3

OpenStudy (anonymous):

\[\frac{dy}{dt} = e^{-3y}\]\[e^{3y}\frac{dy}{dt} = 1\]\[\int e^{3y}\frac{dy}{dt}dt = t + C_1\]\[\frac{1}{3}e^{3y} + C_2 = t + C_1\]\[e^{3y} = 3t + C_3\]\[y = \frac{1}{3}\ln (t+c)\]

OpenStudy (anonymous):

oops! didn't get it done in time. \[y = \frac{1}{3} \ln (3t+c)\]

OpenStudy (anonymous):

why are there different C's in your calculation?

OpenStudy (turingtest):

Just because the integration constant keeps changing as the formula evolves. C1+(another constant)=C2, for example. Numbering them isn't really necessary in this case, you can just call them all C. When the function is found you can apply the initial condition (if you have one) to find the final value of C.

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