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1d
comment Integrating DiffEq (and solving for C) yields $C^2=9$ Do you use +3, -3, or both?
Sorry, but where has anyone explained why C=-3 is incorrect ?
2d
comment Integrating DiffEq (and solving for C) yields $C^2=9$ Do you use +3, -3, or both?
There is no acceptable correct reply, hence no green check.
2d
comment Integrating DiffEq (and solving for C) yields $C^2=9$ Do you use +3, -3, or both?
But why should WHEN you apply the init. condition matter? With your solution, you can get the problem wrong by substitution at the wrong time? Sounds weird.
Nov
24
comment When do you drop the absolute value from ln|x| + C when integrating $\frac{1}{u}du$
Right, but my question is: why can you assume the quantity is positive?
Nov
24
comment When do you drop the absolute value from ln|x| + C when integrating $\frac{1}{u}du$
Solve which? I can't solve for P(t) until I resolve the issue with using ln|x| vs. ln(x), so I wouldn't know the result yet. Trying to determine how I'd know to use ln(x) at this very step.
Nov
24
accepted When do you drop the absolute value from ln|x| + C when integrating $\frac{1}{u}du$
Nov
24
comment When do you drop the absolute value from ln|x| + C when integrating $\frac{1}{u}du$
Yes, when p=800, dp/dt = 0, and there can be more more cats. So, P maxes out at 800, and therefore, 800-P can never be negative.
Nov
24
comment When do you drop the absolute value from ln|x| + C when integrating $\frac{1}{u}du$
I have written the given verbatim as it was provided.
Nov
24
comment When do you drop the absolute value from ln|x| + C when integrating $\frac{1}{u}du$
Right, but why can we assume that $800-p$ can't be negative?
Nov
24
asked When do you drop the absolute value from ln|x| + C when integrating $\frac{1}{u}du$
Nov
21
comment Do we recognize higher degree asymptotes beyond Horizontal and Oblique?
Thanks! Feel free to post a formal answer for approval.
Nov
21
asked Getting hung up on notation for $\frac{d}{dx}e^u$ vs. $\int{e^u}du$
Nov
21
asked Do we recognize higher degree asymptotes beyond Horizontal and Oblique?
Nov
18
accepted DiffEq: When to sub init. cond. to determine +C
Nov
18
comment DiffEq: When to sub init. cond. to determine +C
4 regions b/c Y can not equal 0, and X can not equal (−1/9)1/3. Ok, so 2 dashed lines, if you will. The the solution (a function) can't exist on multiple regions, b/c it would have to cross the boundries
Nov
17
asked DiffEq: When to sub init. cond. to determine +C
Nov
17
comment Integrating DiffEq (and solving for C) yields $C^2=9$ Do you use +3, -3, or both?
So, in this one, once I got to y^2 = .... (Bottom left) I subbed right there, and got 1 solution. But, can't you solve for y = +-sqrt(...) and get 2 solutions? Is there an accepted method? ![enter image description here][1] [1]: i.stack.imgur.com/h3Ods.jpg
Nov
16
comment Integrating DiffEq (and solving for C) yields $C^2=9$ Do you use +3, -3, or both?
ie: Just b/c s is (+), why does that mean the square root must be positive? Is s = 100, $\sqrt{s}$ can still equal -10
Nov
16
accepted How does one integrate Newton's Cooling Law formula?
Nov
16
comment How does one integrate Newton's Cooling Law formula?
Damn, that was fast and accurate. I am rusty.