Question

solve: dy/dt = 0.12y, where y(0) = 3

Answer 1

separate the variables, and integrate

Answer 2

Separate the variables?

Answer 3

\[\frac{dy}{dt} = 0.12y \\ \\ \frac{1}{y}dy=0.12dt\]

Answer 4

Integrate both sides

Answer 5

What is the reasoning behind this? If I were looking at this on a test, how would I know to separate the variables and integrate?

Answer 6

You basically want to separate the y's with the dy and t's with dt

Answer 7

Then you will able to integrate this

Answer 8

And if I do this, I end up with ln(y)=0.12t, but I don't know what to do from there.

Answer 9

Please don't forget the constant of integration...
\[\Large \ln(y)= 0.12t + C_o\]

Answer 10

And from here, you can take it further...

Answer 11

You generally want the answer to a differential equation (that's what this kind of problem is called) as a function, generally still, a function of t.

What you can do is raise e to both sides of the equation...
\[\LARGE e^{\ln(y)}=e^{0.12t + C_o}\]\[\LARGE y = e^{C_o}e^{0.12t}\]

And to simplify things, since Co was an arbitrary constant, then \(\Large e^{C_o}\) is also just some constant, so...
\[\huge y = \color{blue}Ce^{0.12t}\]

Answer 12

Amazing, thank you very much.

Answer 13

I just polished the edges :)

Answer 14

And if you're still here, allow me to present a problem where separation of variables is a more pressing need... oh, and if any of the following is confusing, just ignore them... the last thing we want is to overcomplicate things :D

\[\Large \frac{dy}{dt}=2t^2y -2t^2+3ty-3t\]

does it look menacing? :)

Answer 15

Upon closer inspection, you'll find that "menacing" expression to the right may actually be factored...
\[\Large \frac{dy}{dt}= (y-1)(2t^2 +3t)\]

Answer 16

So, at this point, you may treat the dy and the dt as variables in their own right, which may be multiplied, divided, cancelled, etc.
In particular, multiply both sides of the equation by "dt"
\[\Large \cancel{dt} \cdot \frac{dy}{\cancel{dt}}=(y-1)(2t^2+3t)dt\]

Answer 17

And divide both sides of the equation by (y-1)
\[\Large \frac1{y-1}dy = (2t^2+3t)dt\]

Answer 18

And now, what was once a rather "menacing" looking differential equation is ripe for integration.
\[\Large \int\frac1{y-1}dy =\int (2t^2+3t)dt\]

Answer 19

Skipping all the nitty-gritty details of integration, and bringing all constants of integration to the right-hand-side, we arrive at
\[\Large \ln(y-1) = \frac23t^3 + \frac32t^2+C_o\]

Answer 20

Using a similar trick, we raise e to both sides of the equation to get rid of the "ln" at the left side, such that this expression may be rearranged as a function of t.

\[\LARGE e^{\ln(y-1)}=e^{\frac23t^3+\frac32t^2+C_o}\]

\[\LARGE y-1 = e^{C_o}\times e^{\frac23t^3 + \frac32t^2}\]

Finally
\[\LARGE y = Ce^{\frac23t^3 + \frac32 t^2}+1\]

Answer 21

In the first part, how did e^C become simply C?

Answer 22

Because Co was just some arbitrary constant, e^Co would also just be an arbitrary constant, so I just replaced e^Co with C, for simplicity