Respuesta :
Answer:
a) [tex] W=0,2[/tex]
b) [tex] W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)[/tex]
Step-by-step explanation:
Part a
For this case we have the following differential equation:
[tex] W \sqrt{4-2W}=0[/tex]
If we square both sides we got:
[tex] W^2 (4-2W) =0[/tex]
And we have two possible solutions for this system [tex] W=0, W=2[/tex]
So then that represent the constant solutions for the differential equation.
So then the solution for this case is :
[tex] W=0,2[/tex]
Part b: Solve the differential equation in part (a)
For this case we can rewrite the differential equation like this:
[tex] \frac{dW}{dx} =W \sqrt{4-2W}[/tex]
And reordering we have this:
[tex] \frac{dW}{W \sqrt{4-2W}} = dx[/tex]
Integrating both sides we got:
[tex] \int \frac{dW}{W \sqrt{4-2W}} = \int dx[/tex]
Using CAS for the left part we got:
[tex] -tanh^{-1} (\frac{1}{2} \sqrt{4-2W})= x+c[/tex]
We can multiply both sides by -1 we got:
[tex] tanh^{-1} (\frac{1}{2} \sqrt{4-2W})=-x-c[/tex]
And we can apply tanh in both sides and we got:
[tex] \frac{1}{2} \sqrt{4-2W} = tanh(-x-c)[/tex]
By properties of tanh we can rewrite the last expression like this:
[tex]\frac{1}{2} \sqrt{4-2W} = -tanh(x+c)[/tex]
We can square both sides and we got:
[tex] \frac{1}{4} (4-2W) = tanh^2 (x+c) [/tex]
[tex] 1-\frac{1}{2}W = tanh^2 (x+c)[/tex]
And solving for W we got:
[tex] W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)[/tex]
And that would be our solution for the differential equation
