For real solutions you want to make sure the radicand is >= 0: 2x lnx - 2x + 2 > 0 => 2x (lnx - 1) > -2. x has to be > 0 or ln x blows up, so the only thing which can be negative is the (lnx - 1). ln x - 1 < 0 => lnx < 1 => x < e. For x >= e it's always going to be non negative,. The range 0 < x < e has to be investigated further.
The plot of both branches of y(x) and y'(x) is unexpected, at least to me. I used 2C in the plots rather than C as shown here. There is a discontinuity at x=1 which "splits" the solutions. Calling them ym(x) for - branch and yp(x) for + branch. Combining ym(x1) with ym(1)=yp(1) yields a continuous curve as does yp(x1) with yp(1)=ym(1). A similar combination of the derivatives yp'(x) and ym'(x) greater and less than 1 can be put together though the point x=1 has to be treated carefully as a limit. A quick check of y''(x) and y'''(x) shows similar behavior.
Re: dividing by y. I take it we're assuming y cannot be identically zero, else y^y' makes no sense (or y(0) = 1 > 0 => y > 0 on some open interval containing 0)
Reminds me of my old maths teacher. Double 1st class honours in maths and physics at Cambridge, so what he was doing teaching all his life in a boy's school I don't know. However, one day he filled two whole blackboards with a particularly complicated algebra problem only to come to the end with the wrong answer. At which point, I (being the maths nerd of the class) pointed out he'd made a mistake near the top of board 1. He was not amused, but the guy had beaten the entire chess club in a simultaneous display, taught my father when he was at the same school, so who was I to question his working?
y' ln y = y ln x y'/y ln y = ln x (ln y)' ln y = ln x let ln y = u ∫ u du = ∫ ln x dx u^2 = 2x ln x - 2x + C u = ±√(2x ln x - 2x + C) ln y = ±√(2x ln x - 2x + C) y = exp(±√(2x ln x - 2x + C)) y(1) = 1 => C = 2 y = exp(±√2√(x ln x - x + 1)) check y(1) = exp(±√2√(1 ln 1 - 1 + 1)) = exp(±√2√(1*0 - 1 + 1)) = 1
Interesting differential equation: y^y'=x^y y(1)=1 log(y)y'/y=log(x) log(y)dy/y=log(x)dx z=log(y) dz=dy/y Integral zdz = Integral log(x)dx (z^2)/2 = xlog(x)-x+c c with 2c z = +-sqrt(2xlog(x)-2x+c) y=e^+-sqrt(2xlog(x)-2x+c) y(1)=1 1=e^+-sqrt(-2+c) c=2 y=e^+-sqrt(2xlog(x)-2x+2) Calculating y^y' was a nightmare I had to check with wolframalfa that in fact c=2.
Well, actually solving this differential equation is not very difficult. Follow my approach to solve; y^(y')=x^(y) and y(1)=1 Rewriting the equation in this way; y^(dy/y)=x^(dx) We get the integral from both sides according to the initial condition ∫[y,1]y^(dy/y)=∫[x,1]x^(dx) These types of integrals are known and we know how to convert them to regular integrals In fact, we use this relationship ∫f(x)^(dx)=e^(∫ln(f(x))dx) Considering that two integrals are of the same type, it is not necessary to write the exponential base on both sides ∫[y,1]ln(y)d(ln(y))=∫[x,1]ln(x)dx ln²(y)/2=xln(x)-x+1
7:52 +2 (not +1)
I used to love spotting the teacher (or textbook) mistake. :)
Yep - I was about to point out that the function as written doesn't have a real value at the point x = 1.
Will there be a fact check disclaimer for this video?
True ❤
he even sort of gives a rationale for the 1, i don't get it
The first time I tried one of these on my own and got it before watching! Super proud of myself here
As a jee aspirant, can do this shit mentally. But good job anyway
Same, did it mostly in my head but it's the first time I did it before watching the video.
@@scienc-ification2539 You all act the same
+2
At 8:00, you also need to multiply c by 2...
this non-linear DE is wild enough that i would have liked to see a check of the solution just to have a look at all the moving parts
i agree
For real solutions you want to make sure the radicand is >= 0: 2x lnx - 2x + 2 > 0 => 2x (lnx - 1) > -2. x has to be > 0 or ln x blows up, so the only thing which can be negative is the (lnx - 1). ln x - 1 < 0 => lnx < 1 => x < e. For x >= e it's always going to be non negative,. The range 0 < x < e has to be investigated further.
Graphing this in Desmos gives a pretty cool graph.
Does it gives a plus(+) centered at x=1 ?
The plot of both branches of y(x) and y'(x) is unexpected, at least to me. I used 2C in the plots rather than C as shown here. There is a discontinuity at x=1 which "splits" the solutions. Calling them ym(x) for - branch and yp(x) for + branch. Combining ym(x1) with ym(1)=yp(1) yields a continuous curve as does yp(x1) with yp(1)=ym(1). A similar combination of the derivatives yp'(x) and ym'(x) greater and less than 1 can be put together though the point x=1 has to be treated carefully as a limit. A quick check of y''(x) and y'''(x) shows similar behavior.
Re: dividing by y. I take it we're assuming y cannot be identically zero, else y^y' makes no sense (or y(0) = 1 > 0 => y > 0 on some open interval containing 0)
Given the IC of y(1)=1 we know that y (as a function) is not y=0.
Micheal please teach Abstract Algebra division algorithm.
Mistake stated for the first time at 7:47 when writing the expression for z(x).
yes, the final 1 should be a 2
+2 not +1
Reminds me of my old maths teacher. Double 1st class honours in maths and physics at Cambridge, so what he was doing teaching all his life in a boy's school I don't know. However, one day he filled two whole blackboards with a particularly complicated algebra problem only to come to the end with the wrong answer. At which point, I (being the maths nerd of the class) pointed out he'd made a mistake near the top of board 1. He was not amused, but the guy had beaten the entire chess club in a simultaneous display, taught my father when he was at the same school, so who was I to question his working?
Brilliant - as always!
if C=1, does not the double equal to 2 ? 😇
just a detail, but great sub... Again !
If you graph the functions, you also see you have to take the plus before x=1 and the minus after or vice versa
What? Care to expand? I don't understand.
@@SeeWeeee 2xlnx-2x+2 ~(x-1)^2 at x=1, meaning that sqrt(2xlnx-2x+2)~|x-1| at x=1 if you take alternate signs than it becomes a smooth function
Thank you dr pen im msc math algebra
I could tell that would be the answer just by looking at the differential equation 😂
typo, c=2
Nope, c=1 is correct, but z^2=2xlnx-2x+2c, but Michael forgot to multiply c by two too.
y^(y'/y)=x
y^(d/dx(ln(y)))=x
ln(y)•d/dx(ln(y))=ln(x)
u=ln(y)
du/dx=d/dx(ln(y))
u(1)=0
u•du=ln(x)•dx
u^2/2=xln(x)-x+1
u^2=2xln(x)-2x+2
u=+-sqrt(2)sqrt(xln(x)-x+1)
y=e^u=e^+-sqrt(2xln(x)-2x+2)
9:35 Identity = x ^ identity 🗿
If with "identity" you mean the function y = x, then no, that's not a solution here.
Cool !!
y' ln y = y ln x
y'/y ln y = ln x
(ln y)' ln y = ln x
let ln y = u
∫ u du = ∫ ln x dx
u^2 = 2x ln x - 2x + C
u = ±√(2x ln x - 2x + C)
ln y = ±√(2x ln x - 2x + C)
y = exp(±√(2x ln x - 2x + C))
y(1) = 1 => C = 2
y = exp(±√2√(x ln x - x + 1))
check y(1) = exp(±√2√(1 ln 1 - 1 + 1)) = exp(±√2√(1*0 - 1 + 1)) = 1
Soo cool. Any real life example of this diff. eq. occurring?
More lie algebras on the second channel?
Maybe even other courses? Multivariable calc?
Interesting differential equation:
y^y'=x^y
y(1)=1
log(y)y'/y=log(x)
log(y)dy/y=log(x)dx
z=log(y)
dz=dy/y
Integral zdz = Integral log(x)dx
(z^2)/2 = xlog(x)-x+c
c with 2c
z = +-sqrt(2xlog(x)-2x+c)
y=e^+-sqrt(2xlog(x)-2x+c)
y(1)=1
1=e^+-sqrt(-2+c)
c=2
y=e^+-sqrt(2xlog(x)-2x+2)
Calculating y^y' was a nightmare I had to check with wolframalfa that in fact c=2.
While these problems are of theoretical interest, showing how they can occur in a practical application makes the solution more fascinating.
Well, actually solving this differential equation is not very difficult.
Follow my approach to solve;
y^(y')=x^(y) and y(1)=1
Rewriting the equation in this way;
y^(dy/y)=x^(dx)
We get the integral from both sides according to the initial condition
∫[y,1]y^(dy/y)=∫[x,1]x^(dx)
These types of integrals are known and we know how to convert them to regular integrals
In fact, we use this relationship
∫f(x)^(dx)=e^(∫ln(f(x))dx)
Considering that two integrals are of the same type, it is not necessary to write the exponential base on both sides
∫[y,1]ln(y)d(ln(y))=∫[x,1]ln(x)dx
ln²(y)/2=xln(x)-x+1
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