Question

A simple model for the shape of a tsunami is given by dW/dx = W √(4 − 2W), where W(x) > 0 is the height of the wave expressed as a function of its position relative to a point offshore.
By inspection, find all constant solutions of the DE. (Enter your answers as a comma-separated list.)

Answer 1

Answer:

a) $W=0,2$

b) $W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)$

Step-by-step explanation:

Part a

For this case we have the following differential equation:

$W \sqrt{4-2W}=0$

If we square both sides we got:

$W^2 (4-2W) =0$

And we have two possible solutions for this system $W=0, W=2$

So then that represent the constant solutions for the differential equation.

So then the solution for this case is :

$W=0,2$

Part b: Solve the differential equation in part (a)

For this case we can rewrite the differential equation like this:

$\frac{dW}{dx} =W \sqrt{4-2W}$

And reordering we have this:

$\frac{dW}{W \sqrt{4-2W}} = dx$

Integrating both sides we got:

$\int \frac{dW}{W \sqrt{4-2W}} = \int dx$

Using CAS for the left part we got:

$-tanh^{-1} (\frac{1}{2} \sqrt{4-2W})= x+c$

We can multiply both sides by -1 we got:

$tanh^{-1} (\frac{1}{2} \sqrt{4-2W})=-x-c$

And we can apply tanh in both sides and we got:

$\frac{1}{2} \sqrt{4-2W} = tanh(-x-c)$

By properties of tanh we can rewrite the last expression like this:

$\frac{1}{2} \sqrt{4-2W} = -tanh(x+c)$

We can square both sides and we got:

$\frac{1}{4} (4-2W) = tanh^2 (x+c)$

$1-\frac{1}{2}W = tanh^2 (x+c)$

And solving for W we got:

$W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)$

And that would be our solution for the differential equation