Separable Differential Equation dy/dt = 6y
The question is as follows:
$$\frac{dy}{dt}=6y$$
$$y(9)=5$$
I tried rearranging the equation to $\frac{dy}{6y} = dt$ and integrating both sides to get $(1/6)\ln|y| + C = t$. After that I tried plugging in the $9$ for $y$ and $5$ for $t$ and solving but I can't quite seem to get it.
$\endgroup$ 23 Answers
$\begingroup$You should plug in 9 for $t$ and 5 for $y$.
$\endgroup$ $\begingroup$$\frac 16\ln|y| = t+C$ I would be inclined to not solve for C at this time.
You could say
$\frac 16\ln|5| = 9+C\\ C = \frac 16\ln|5| - 9$
But I think that makes for more work in the long run
$\frac 16\ln|y| = t+C\\ \ln|y| = 6t+C$
Notice that $C$ just stays $C.$ $6C, C,$ it is just an arbitrary constant.
$y = e^{6t+C}\\ y=e^Ce^{6t}\\ y = C e^{6t}$
I did it again! And once you have done these for a week, you will drop the intermediate steps and go:
$\frac 16\ln|y| = t+C\\ y = C e^{6t}$
We have to find C now.
$y(9) = 5\\ 5 = C e^{54}\\ C = 5e^{-54}\\ y = 5e^{-54}e^{6t}$
or
$y = 5e^{6t-54}$
$\endgroup$ $\begingroup$You haven't actually asked a question: what are you trying to do? Assuming it is to solve the IVP, your next step is to find $C$. From your own formula, you have: $$ C = t - \ln |y| /6. $$ If $t = 5$ and $y(5) = 9,$ then $$ C = 5 - (\ln 9)/6. $$ Now solve your formula for $y$, and you are done.
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