isolate f(x) in d/dx(f(x)) = sqrt(1 - 4sqrt(2)f(x) + 4(f(x))^2)

[jimbimley]

Hi everyone,

Working on a project and got stuck at this step:

I need to isolate f(x) in the equation

. . .\(\displaystyle \dfrac{d}{dx}\, f(x)\, =\, \sqrt{\strut 1\, -\, 4\, \sqrt{2\,}\, f(x)\, +\, 4\, (f(x))^2\,}\)

Please let me know if you can help me!

 

[Deleted member 4993]

Hi everyone,

Working on a project and got stuck at this step:

I need to isolate f(x) in the equation

. . .\(\displaystyle \dfrac{d}{dx}\, f(x)\, =\, \sqrt{\strut 1\, -\, 4\, \sqrt{2\,}\, f(x)\, +\, 4\, (f(x))^2\,}\)

Please let me know if you can help me!

y = f(x):

. . .\(\displaystyle \displaystyle{\dfrac{dy}{\sqrt{\strut 1\,- \,y \cdot 4\sqrt{2\,}\, +\, 4y^2\,}} \, =\, dx}\)

 

[jimbimley]

Square wheel problem. Can someone help me understand the final step?

I was following through this website https://www.maplesoft.com/applications/view.aspx?SID=6322&view=html until this step:

< link removed >

d/dx f(x) = sqrt[1 - 4*sqrt(2)*f(x) + 4*f(x)^2]

I know this is an ordinary differential equation and when I solved I got this is my answer:

< link removed >

f(x) = 1/4 * [e^(-2x) + e^(2x) + 2*sqrt(2)]

rather than:

< link removed >

f : = x -> -1/4 * [(e^[-2x])^2*sqrt(2) - 2*e^(-2x)*sqrt(2) + sqrt(2) + (e^[-2x])^2 - 1]/e^(-2x)

The article's solution places the curve on the x axis, while mine has the same shape but is a translated version of it. Can someone point out how the author achieved the final step? Thanks

 

[mmm4444bot]

… d/dx f(x) = sqrt[1 - 4*sqrt(2)*f(x) + 4*f(x)^2]

I know this is an ordinary differential equation and when I solved I got this [as] my answer:

f(x) = 1/4 * [e^(-2x) + e^(2x) + 2*sqrt(2)]

rather than [the author's result]:

f : = x -> -1/4 * [(e^[-2x])^2*sqrt(2) - 2*e^(-2x)*sqrt(2) + sqrt(2) + (e^[-2x])^2 - 1]/e^(-2x)

… Can someone point out how the author achieved [their result]?

The author's annotation for that step is, "solve DE interactively".

I think that means the author asked Maple to solve the differential equation for them, but I note that the result displayed on that page does not show a Maple output. It shows the syntax for a function-mapping definition. In other words, it shows a definition for f input by the user.

(The syntax f := x -> expression is used to define a function in Maple so that function notation f(x) may be subsequently used, as opposed to assigning a simple variable name, like f:=expression.)

I don't know what version of Maple the author used, but I have access to Maple V Release 5, and it generates the following solution.

f(x) = 1/4*[1 + e^(4x + 4C₁) + 2e^(2x + 2C₁)*sqrt(2)]*e^(-2x - 2C₁)

I found C₁=0, using the initial condition that f(0)=0.

Expanding and then factoring out 1/4 yields your result. :cool:


WolframAlpha also agrees. Their output looks a bit goofy; perhaps, I entered goofy syntax for Mathematica.

It sure seems like the author omitted some subsequent steps and explanations (eg: translations and why).

I disclose that I did not carefully study that entire page. Cheers!