Separable Differential Equation dy/dt = 6y The question is as follows: (dy)/(dt)=6y y(9)=5

Separable Differential Equation dy/dt = 6y
The question is as follows:
$\frac{dy}{dt}=6y$
$y\left(9\right)=5$
I tried rearranging the equation to $\frac{dy}{6y}=dt$ and integrating both sides to get $\left(1/6\right)\mathrm{ln}|y|+C=t$. 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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Lena Gomez
$\frac{1}{6}\mathrm{ln}|y|=t+C$ I would be inclined to not solve for C at this time.
You could say
$\frac{1}{6}\mathrm{ln}|5|=9+C\phantom{\rule{0ex}{0ex}}C=\frac{1}{6}\mathrm{ln}|5|-9$
Notice that C just stays C. 6C,C, it is just an arbitrary constant.
$y={e}^{6t+C}\phantom{\rule{0ex}{0ex}}y={e}^{C}{e}^{6t}\phantom{\rule{0ex}{0ex}}y=C{e}^{6t}$
I did it again! And once you have done these for a week, you will drop the intermediate steps and go:
$\frac{1}{6}\mathrm{ln}|y|=t+C\phantom{\rule{0ex}{0ex}}y=C{e}^{6t}$
We have to find C now.
$y\left(9\right)=5\phantom{\rule{0ex}{0ex}}5=C{e}^{54}\phantom{\rule{0ex}{0ex}}C=5{e}^{-54}\phantom{\rule{0ex}{0ex}}y=5{e}^{-54}{e}^{6t}$
or
$y=5{e}^{6t-54}$