# differential equation problem

• Mar 20th 2009, 04:25 AM
swimmergirl
differential equation problem
I almost have the answer but I can't quite get there.

Differential equation is

dH/dt= (40-H)^2/100

I solved it through here:

H= 40 - 200/(2t + 5)
However, when I plug in a value t=5 I get the wrong answer. My professor has this but the following step says:

H=80t/(2t+5)

How do I get to that?
• Mar 20th 2009, 04:29 AM
mr fantastic
Quote:

Originally Posted by swimmergirl
I almost have the answer but I can't quite get there.

Differential equation is

dH/dt= (40-H)^2/100

I solved it through here:

H= 40 - 200/(2t + 5)
However, when I plug in a value t=5 I get the wrong answer. My professor has this but the following step says:

H=80t/(2t+5)

How do I get to that?

$\displaystyle \frac{dt}{dH} = \frac{100}{(40 - H)^2}$

$\displaystyle \Rightarrow t = \int \frac{100}{(40 - H)^2} \, dH = \frac{100}{40 - H} + C$

where C is a constant whose value cannot be calculated without additional data (such as H = 0 at t = 0 ....?)
• Mar 20th 2009, 04:44 AM
swimmergirl
correct...initial conditions are H(0)=0. Sorry these weren't given in the problem but its about the height of a tree over time. Therefore, I just put that initially it isn't growing which suffices H(0)=0 and I forgot to mention that.
• Mar 20th 2009, 05:10 PM
mr fantastic
Quote:

Originally Posted by swimmergirl
correct...initial conditions are H(0)=0. Sorry these weren't given in the problem but its about the height of a tree over time. Therefore, I just put that initially it isn't growing which suffices H(0)=0 and I forgot to mention that.

$\displaystyle t = \frac{100}{40 - H} + C$

Substitute $\displaystyle H = 0, ~ t = 0$: $\displaystyle 0 = \frac{100}{40} + C \Rightarrow C = - \frac{5}{2}$.

Therefore:

$\displaystyle t = \frac{100}{40 - H} - \frac{5}{2} \Rightarrow 2t + 5 = \frac{200}{40 - H}$

$\displaystyle \Rightarrow \frac{1}{2t + 5} = \frac{40 - H}{200} \Rightarrow \frac{200}{2t + 5} = 40 - H$

$\displaystyle \Rightarrow H = 40 - \frac{200}{2t + 5} = \, ....$