A very interesting differential equation

Non-linear differential equations have interesting solution developments and some require us to be creative with our solution development. This equation is one of those cases.
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[@edmundwoolliams1240]

What I find so fascinating about non-linear DEs is that one tiny change to the equation usually completely changes the solution/solution method and/or makes a doable problem into a completely hopeless one. For example, just changing any one of those x or y terms to x^2 or y^2 would just suddenly make it impossible

[@Monokumaaaaaaaaa]

We have a fraction of 2 linear functions, so we can solve the system:
y + 2 = 0 => y = -2
x + y + 1 = 0=> x = 1
Then we can substitute x = x* + 1, y = y* -2; (x*)' = x', (y*)' = y'
So we actually have:
dy* = (y*)² • dx* / (y* + x*)²
Here goes another substitution: y* = ux*, (y*)' = u'x* + u
Now we can rewrite it as:
du•x* + dx*•u = u² • dx* / (u+1)²
Then simplify:
du•x* = dx* • (u²/(u+1)² - u)
And now we can just separate the variables and finish the job

[@neg2sode]

It seems like an extraordinarily neat solution to the equation to me! May I ask the exact name for the first step where a system of the two linear function is solved? I am interested in the idea behind this solution.

[@raghavkumar4914]

​@@neg2sode idea behind the soln is that the rate of chnge does not chnge when we chnge the origin.. 6

[@raghavkumar4914]

​@@neg2sode idea behind the soln is that the rate of chnge does not chnge when we chnge the origin.. 6

[@edmundwoolliams1240]

Maybe you could rewrite the inverse tan as a complex logarithm, then combine the logs and solve for y?

[@stevenkrantz6580]

exactly what i thought of. arctan has an explicit complex logarithmic form derived by integrating 1/(x^2+1) with partial fraction decomp.

[@thomasstokes9412]

Here is an alternative form for the answer using the complex logarithm definition of arctanx.
tanx = sinx/cosx = (e^{ix}-e^{-ix})/(i[e^{ix}+e^{-ix}])
Re-arranging everything the following formulae for arctanx: arctanx = 1/(2i) ln([1+ix]/[1-ix]).
Plugging this into the answer to the differential equation in the video. The answer 'simplifies' to:
ln(y+2)+iln([1+i(y+2/1-x)]/[1-i(y+2/1-x)])=c
Let e^c = a
(y+2)[(1-x+i(y+2))/(1-x-i(y+2))]^i = a

[@potterhead0720]

When considering messy x and y thing as u, instead of taking the log on both sides, i would have preferred taking the denominator to the other side and multiply by u, then differentiate both sides wrt x, this i believe would have been less lengthy in comparison to the method you demonstrated

[@TheArtOfBeingANerd]

From a lowly calc 2 student, it seems like you can get the answer in a form x = f(y)? Or does that not matter as much as getting y in terms of x?

[@maths_505]

Oh it matters just as much it's just that math people are too stubborn to admit that 😂

[@moeberry8226]

At 714 it’s (1+x)*dz/dx you have to factor out the negative 1 for substitution.

[@jkid1134]

You did an algebra step you said to check and then a partial fraction decomp that you also said to check, but one just undoes the other and you could've skipped them both :P

[@ericthegreat7805]

Can't we express arctan as a log approximation?

[@Grundini91]

10:11 This is why you don't skip steps

[@insouciantFox]

(y+2)(y+2-i(x-1))^i(y+2+i(x-1))^i=K

[@Mephisto707]

2:17 Can you log both sides at will without verifying if they are always positive?

[@alphazero339]

Yes if you assume complex logarithm

[@Mephisto707]

@@alphazero339 but it's not the case here, right? I was expecting him to verify that applying log is possible, but he didn't.

[@alphazero339]

@@Mephisto707 I mean you can always log both sides no?

[@Mephisto707]

@@alphazero339 only if the argument is positive.

[@alphazero339]

@@Mephisto707 I think you can take log of negative number too
Suppose 𝐱∈ℝ⁺. Then if we want log of negative number -𝐱 we can compute like this:
𝐥𝐧(-𝐱) = 𝐥𝐧(𝐱 × 𝐞𝐱𝐩(𝐢π)) = 𝐥𝐧(𝐱) + 𝐥𝐧(𝐞𝐱𝐩(𝐢π)) = 𝐥𝐧(𝐱) + 𝐢π
So as long as argument is not 0 its okay and can be negative

[@shivanshnigam4015]

Hey try this one
Sum from n=0 to infinity of 1/(3n)!
Ans is (e+e^w+e^w²)/3
Where w is the complex cube root of unity

[@DKAIN_404]

Bro did you watched jee simplified

[@shivanshnigam4015]

@@DKAIN_404 hell yeah bro 👌

[@holyshit922]

Quite easy , substitution u(x) = (y(x)+2)/(x+y(x)+1)
will reduce this equation to separable differential equation

[@bandishrupnath3721]

At this point ,its saddening that im not worthy enough to watch these awesome videos and fully enjoy and understand them:),one day ill be worthy.
btw love ur vida

[@giuseppemalaguti435]

Come per gli integrali,il problema è sempre trovare il cambio variabile opportuno

[@azmath2059]

How long did this take you to develop a solution?

[@lakshay3745]

I think thats the best implicit form you can possibly get yeah

[@nightmareintegral5593]

Is this is not complete equation?

[@MrWael1970]

Thank you for your featured solution

[@abdoshaat3304]

Where's isolated Y

[@yoav613]

Nice!! And oh no i do not want to try anything to get something better than what you got..😂

[@maths_505]

😂😂😂

[@RTX-tv6cq]

I really love your vids.

[@MohamedachrafKadim-jm5yr]

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