# Differential Equation Hemispherical Bowl: show that k=3pia^2/2T given that x=0 when t=T and x=a when t=0.

#### Skelly4444

##### New member
Trying to solve the DE below and show that k=3pia^2/2T given that x=0 when t=T and x=a when t=0.

The DE given in the question is pi(2ax-x^2)dx/dt = -kx

Every time I try to solve it, I get the same answer which is not like the one I'm supposed to be showing.

#### MarkFL

##### Super Moderator
Staff member
Hello, and welcome to FMH!

I would write the ODE as:

$$\displaystyle \frac{\pi}{k}(x-2a)\,dx=dt$$

Integrating, we then get:

$$\displaystyle \frac{\pi}{2k}x\left(x-4a\right)=t+C$$

Now, using the given point $$(t,x)=(T,0)$$ we get:

$$\displaystyle 0=T+C\implies C=-T$$

And so we have:

$$\displaystyle \frac{\pi}{2k}x\left(x-4a\right)=t-T$$

Next, using the given point $$(t,x)=(0,a)$$ we get:

$$\displaystyle \frac{\pi}{2k}a\left(a-4a\right)=-T$$

Solving this for $$k$$, we obtain:

$$\displaystyle k=\frac{3\pi a^2}{2T}\quad\checkmark$$

#### Skelly4444

##### New member
Good morning Mark,
Thank you for your swift response.

Your integration decision slightly confuses me, as I always taught to separate the constant k and integrate the right hand side with respect to t.

When I separated my variables, I had pi(2ax-x^2) / x = -kt.

How can you write the bracketed terms as (x-2a)?

A little confused here if I'm honest.

Thanks
Simon

#### MarkFL

##### Super Moderator
Staff member
It seems you've introduced the variable $$t$$ and dropped the derivative $$\displaystyle \d{x}{t}$$ from the ODE. We begin with:

$$\displaystyle \pi(2ax-x^2)\d{x}{t}=-kx$$

Divide through by $$-kx$$:

$$\displaystyle \frac{\pi}{k}(x-2a)\d{x}{t}=1$$

Now, this may be written in the differential form:

$$\displaystyle \frac{\pi}{k}(x-2a)\,dx=dt$$

Does that make sense?

#### topsquark

##### Full Member
It seems you've introduced the variable $$t$$ and dropped the derivative $$\displaystyle \d{x}{t}$$ from the ODE. We begin with:

$$\displaystyle \pi(2ax-x^2)\d{x}{t}=-kx$$

Divide through by $$-kx$$:

$$\displaystyle \frac{\pi}{k}(x-2a)\d{x}{t}=1$$

Now, this may be written in the differential form:

$$\displaystyle \frac{\pi}{k}(x-2a)\,dx=dt$$
There is a detail here and I don't know how to fix it.

The original equaiton is $$\displaystyle \pi(2ax-x^2)\d{x}{t}=-kx$$ and an initial condition is (t, x) = (T, 0). But in your solution you divided both sides of the differential equation by x, which can't be done at the initial condition, as x = 0.

Could we use the initial condition as a limit? It's not quite correct but the only thing I can think of.

-Dan

#### MarkFL

##### Super Moderator
Staff member
There is a detail here and I don't know how to fix it.

The original equaiton is $$\displaystyle \pi(2ax-x^2)\d{x}{t}=-kx$$ and an initial condition is (t, x) = (T, 0). But in your solution you divided both sides of the differential equation by x, which can't be done at the initial condition, as x = 0.

Could we use the initial condition as a limit? It's not quite correct but the only thing I can think of.

-Dan
Good catch, Dan. I realized I was losing the trivial solution $$x\equiv 0$$, but didn't notice it coincided with an initial value. I suspect the author of the problem overlooked this as well. Offhand, I don't know how to fix this either.

#### Skelly4444

##### New member
Hi Mark,
I understand what you have done but I wasn't aware we could divide through by -kx and just leave a 1 on the right hand side with respect to t.
Like I said in my previous message, I always left k on the right side when separating the variables.
Thanks again.
Simon