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
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
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.
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
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
@@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
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
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
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
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
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.
@@neg2sode idea behind the soln is that the rate of chnge does not chnge when we chnge the origin.. 6
@@neg2sode idea behind the soln is that the rate of chnge does not chnge when we chnge the origin.. 6
Maybe you could rewrite the inverse tan as a complex logarithm, then combine the logs and solve for y?
exactly what i thought of. arctan has an explicit complex logarithmic form derived by integrating 1/(x^2+1) with partial fraction decomp.
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
At 714 it’s (1+x)*dz/dx you have to factor out the negative 1 for substitution.
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
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
Bro did you watched jee simplified
@@DKAIN_404 hell yeah bro 👌
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?
Oh it matters just as much it's just that math people are too stubborn to admit that 😂
Can't we express arctan as a log approximation?
Thank you for your featured solution
I really love your vids.
10:11 This is why you don't skip steps
2:17 Can you log both sides at will without verifying if they are always positive?
Yes if you assume complex logarithm
@@alphazero339 but it's not the case here, right? I was expecting him to verify that applying log is possible, but he didn't.
@@Mephisto707 I mean you can always log both sides no?
@@alphazero339 only if the argument is positive.
@@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
(y+2)(y+2-i(x-1))^i(y+2+i(x-1))^i=K
Quite easy , substitution u(x) = (y(x)+2)/(x+y(x)+1)
will reduce this equation to separable differential equation
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
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
How long did this take you to develop a solution?
Is this is not complete equation?
Where's isolated Y
I think thats the best implicit form you can possibly get yeah
Nice!! And oh no i do not want to try anything to get something better than what you got..😂
😂😂😂
Come per gli integrali,il problema è sempre trovare il cambio variabile opportuno
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