Question

Given the differential equation \[\frac{ dH }{ dt } = k(65-H)\] and the fact that at t=0 H=40, solve the equation and graph the solution.

So far, I have separated the variable,s integrated, then exponentiated. This gives: \[65-H = e^{kt+c}\]
I've used A to represent e^c, so finally we have: \[H=65-A e^{kt}\]
Solving for A is simple (A=25), but where I am stuck is finding k. Any suggestions? My problem is we are only give H at t=0 and I am not sure as to my next step. Any help or suggestions?

Answer 1

Plugging in A the equation looks like \[H=65-25e ^{kt}\] how do I get k?

Answer 2

1) Let's see that antiderivative for H.
2) You don't find two parameters with only one initial condition.

Answer 3

\[\ln(65-H)=kt+c\] is what i get from separating and integrating.

Answer 4

Same error. Please repair the H-side.

Answer 5

\[\int\limits_{}^{}\frac{ dH }{ 65-H }\] = \[\int\limits_{}^{}\frac{ 1 }{ 65-H }dH\] which is ln(65-H) right?

Answer 6

First, \(\int \frac{1}{x}\;dx\) = ln(|x|) + C

Do you KNOW that 65 - H is Positive?

Second, don't forget your chain rule. \(\int \dfrac{1}{65 - H}\;dh = -ln(|65-H|) + C\)

Answer 7

Whoops. s/b dH, not dh.

Answer 8

ahhhhhh yes I forgot the inside. Thanks

Answer 9

As for the second part, you are saying there is no way to solve for k with just those initial conditions? When I redo my work with the right integral, I will still end up being able to solve for the constant e^c (which I called A) but, I cant solve for k eh? So Im not crazy?

Answer 10

Two parameters, two conditions - unless you can demonstrate that it is only one parameter - which it isn't, in this case.