# Thread: Applied Maths: Diff Eqns

1. ## Applied Maths: Diff Eqns

We were given sample qs in a lecture and I have a couple of parts of questions that I really need help with so I can complete the question as a whole on my own. Please help me where possible, thanks

1. How would you go about solving an equation of the form
dNB/dt=k1NA-k2NB ... where k1,k2 are different constants. [I have an equation for NA that I solved: NA= No e^(-k1*t) if that helps]. I'm not sure how to separate everything as I tried and got into a huge mess...
I need the solution to substitute into another equation... which I can do... once I have the solution...

2. " A vase is shaped so that it's cross-sectional area at a height h from its base is:
A=f(h), O \< h \< H
where H is the total height of the vase. The vase is filled to the top with water, which then evaporate at a rate proportional to the surface area so that:
dV/dt= -k*A

where V(t) is the volume of water and A(t) is the surface area at time t. Show that the height of water in the vase obeys a simple eqn of the form
dh/dt=-k , with k as a constant, regardless of what the function f is and hence.... ....."

I don't know where to start with this q, I think you need to change dV so it is in terms of area and height but not sure how to go about it...

Thanks again! x x x

2. Originally Posted by tiger_in_the_mist
We were given sample qs in a lecture and I have a couple of parts of questions that I really need help with so I can complete the question as a whole on my own. Please help me where possible, thanks

1. How would you go about solving an equation of the form
dNB/dt=k1NA-k2NB ... where k1,k2 are different constants. [I have an equation for NA that I solved: NA= No e^(-k1*t) if that helps].

It more than helps it is necessary- you can't solve a single equation for two unknown values, whether numbers or functions!

Putting $N_A$ into that equation gives $\frac{dN_B}{dt}= kN_0e^{-k_1y}- k_2 N_B$

I'm not sure how to separate everything as I tried and got into a huge mess...
Yes, because that is not a separable equation. I hope that's not the only method you know to solve first order equations. That is a "linear equation" and there is a standard formula for an integrating factor. Do you know that.

I need the solution to substitute into another equation... which I can do... once I have the solution...

2. " A vase is shaped so that it's cross-sectional area at a height h from its base is:
A=f(h), O \< h \< H
where H is the total height of the vase. The vase is filled to the top with water, which then evaporate at a rate proportional to the surface area so that:
dV/dt= -k*A

where V(t) is the volume of water and A(t) is the surface area at time t. Show that the height of water in the vase obeys a simple eqn of the form
dh/dt=-k , with k as a constant, regardless of what the function f is and hence.... ....."

I don't know where to start with this q, I think you need to change dV so it is in terms of area and height but not sure how to go about it...

Thanks again! x x x