Question

Calculus Help
Fan + Medal

Answer 1

\[\large\rm y'=\frac{y}2,\qquad\qquad\qquad y(0)=80\]
You can do separation of variables,
or integrating factor,
have a preference?

Answer 2

I'm getting
ln y= 0.5t+ c

Answer 3

So I think its B

Answer 4

Its telling em to use separation of variables.

Answer 5

\[\large\rm \frac{dy}{dx}=\frac{y}{2}\]So then let's divide both sides by y,\[\large\rm \frac{dy}{y dx}=\frac{1}{2}\]And move the dx to the other side,\[\large\rm \frac{dy}{y}=\frac{1}{2}dx\]And integrate from this point, ya?

Answer 6

Yes.....
dy/y= dt/2
Now integrate both side

Answer 7

Oh, it's a t, not x, sorry :d

Answer 8

I get my work confused. Trying to work it out

Answer 9

I cant integrate it for some reason

Answer 10

No? Forgot this integral?\[\large\rm \int\limits \frac{1}{y}dy\]

Answer 11

ln|y| = x/2 +C ?

Answer 12

Mmm k looks good so far.\[\large\rm \ln|y|=0.5t+c\]

Answer 13

Recall that the log and exponential functions are inverses.
So to get rid of the log, we exponentiate.\[\large\rm e^{\ln|y|}=e^{0.5t+c}\]

Answer 14

\[\large\rm y=e^{0.5t+c}\]

Answer 15

RIght, what about the 80 portion

Answer 16

We'll apply an exponent rule to rewrite our expression like this:\[\large\rm y=e^{0.5t}\cdot e^c\]But notice that e^c is just a fancy looking constant.
Let's call it something else.\[\large\rm y=A e^{0.5t}\]
And then use your initial data to solve for A.
y(0)=80\[\large\rm 80=Ae^{0.5\cdot0}\]

Answer 17

A=80

Answer 18

So the answer would be B.

Answer 19

yes, good job