Find the solution to the differential equation, subject to the given initial condition.
\[\frac{ dy }{ dx }+\frac{ y }{ 9 }=0, y(0)=16\]
Did you try separation of variables.
yeah i got dy+y/9=dx but im not really sure what to do after that
I don't think you did the algebra there correctly.
Why don't you try writing as f(x) dx=g(y) dy
This is what I mean by separation of variables
Ah. I have no idea what I'm doing then.
Do you know how to undo the addition to isolate the dy/dx part?
\[\frac{dy}{dx}+\frac{y}{9}=0\]
You see addition you subtract.
can you do that?
so just subtract y/9 from both sides?
That is how you get dy/dx by itself now you want g(y)dy=f(x)dx
soo dy/y=-1/9 dx?
that's right
now integrate both sides
okay, i'll try it. :)
is ln(y)=-1/9x+c correct?
now use the initial condition to find C you are given when x is 0 y is 16. Use that to find C.
c=ln(16)
yep
dont I need to plug that back into the equation somewhere though?
Yep C
sooo ln(y)=-1/9x+ln(16)? i put that into the homework and it said it was wrong
It might want you to solve for y. Also I would write it as ln|y|=-x/9+ln(16) I don't know if it would count off for not having those | | instead of ( ). But yeah if wants you to solve for y then solve for y.
I don't even know how to go about doing that. ;_;
So you don't know the inverse of natural log function?
I don't remember it, no.
what about the exponential function?
uh wat.
\[e^{\ln|x|}=x\]
the exponential ( where base is e) function and natural log function are inverses
ohhh yes, okay I remeber that now.
use that to solve for y
\[\text{ if } \ln|y|=f(x) \text{ then } y=e^{f(x)}\]
steffw have you got that part?
Once you do that. I will show you how to simplify the equation you get once you show me you have gotten that far.
oops, i realized i was trying to solve for the wrong thing. xD i ended up with y=16e^-1/9x Thanks!
omg you did good :)
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