Can be done without substitution as well by multiplying both sides by (y-x) and rearranging to y'y - y'x - y = 2 - x. Now you recognize that the left hand side is the derivative of 0.5y^2 - yx, so you get d/dx(0.5y^2 - yx) = 2 -x. Integrate both sides and you get the same solution as in the video.
More or less what I did, except I rearranged it to y,dy + x.dx = x.dy+y.dx + 2.dx, and recognizing that the 1st 2 terms on the right are just d(xy). Now every term is integrable, leaving (y^2 + x^2)/2 = xy + 2x + C. I stopped there, too lazy to peek around the next corner to find a neat y(x) to wrap things up. I didn't see the value of the u-sub until I watched the video - very nice!
@@TedHoppwell, I didn't see the u-substitution coming until he set it up - it was unexpected & beautiful, and it tickled. My approach was pretty typical plug & chug - nothing wrong with plug & chug, but there's something fun about the surprise endings of some "method 2s".
You can rearrange your equation to get y = x + 2/(y' − 1) which is actually a nonlinear differential equation of a special type known as D'Alembert's equation y(x) = x·g(y'(x)) + f(y'(x)) with g(t) = 1 and f(t) = 2/(t − 1) and these are solvable, see the German Wikipedia for an explanation how to do this. However, your equation is even easier to solve. When at 0:44 you have arrived at y'(y − x) = y − x + 2 the easiest way to proceed is to write this as (y' − 1)(y − x) = 2 Now if we set v = y − x where v is a function of x then we have v' = y' − 1 and so our equation becomes v'·v = 2 which of course gives v² = 4x + c₁ and therefore (y − x)² = 4x + c₁ as per your second method.
Whether or not you take the plus or minus square root depends on your initial condition. For example, if the initial condition is set at x=0, the sign of the square root will be equal to the sign of the initial value of y.
Can be done without substitution as well by multiplying both sides by (y-x) and rearranging to y'y - y'x - y = 2 - x. Now you recognize that the left hand side is the derivative of 0.5y^2 - yx, so you get d/dx(0.5y^2 - yx) = 2 -x. Integrate both sides and you get the same solution as in the video.
More or less what I did, except I rearranged it to y,dy + x.dx = x.dy+y.dx + 2.dx, and recognizing that the 1st 2 terms on the right are just d(xy). Now every term is integrable, leaving (y^2 + x^2)/2 = xy + 2x + C. I stopped there, too lazy to peek around the next corner to find a neat y(x) to wrap things up. I didn't see the value of the u-sub until I watched the video - very nice!
Neat
That's where I thought he was headed with his first method. Don't know why he bailed on that approach.
@@TedHoppwell, I didn't see the u-substitution coming until he set it up - it was unexpected & beautiful, and it tickled. My approach was pretty typical plug & chug - nothing wrong with plug & chug, but there's something fun about the surprise endings of some "method 2s".
@@TedHopp He definitely has a bias towards substitution. Doesnt he? Hehe
Nice! BTW I think it'd be cool if you could start exploring problems involving tetration?
You can rearrange your equation to get
y = x + 2/(y' − 1)
which is actually a nonlinear differential equation of a special type known as D'Alembert's equation
y(x) = x·g(y'(x)) + f(y'(x))
with g(t) = 1 and f(t) = 2/(t − 1) and these are solvable, see the German Wikipedia for an explanation how to do this.
However, your equation is even easier to solve. When at 0:44 you have arrived at
y'(y − x) = y − x + 2
the easiest way to proceed is to write this as
(y' − 1)(y − x) = 2
Now if we set
v = y − x
where v is a function of x then we have v' = y' − 1 and so our equation becomes
v'·v = 2
which of course gives
v² = 4x + c₁
and therefore
(y − x)² = 4x + c₁
as per your second method.
Nice
Whether or not you take the plus or minus square root depends on your initial condition. For example, if the initial condition is set at x=0, the sign of the square root will be equal to the sign of the initial value of y.
Great video as always😊
Thank you
@@SyberMathWhy didn't you make u equal to y prime instead just curious? Thanks for sharing.
A possible variant : write u' = 2/u as 1/2(u^2 ) ' = 2 to get u^2 = 4x + c . Of course one gets the same solution as in the video.
5:46 is normal that I laughed so hard at this moment? xd
I love integrating
Was that uh-oh deliberately timed? Because the ads came on right away.
y = x + √(4*x+ C)
Noice
y' - 1 = 2/(y - x). z = y - x so z' = y' - 1. Thus z' = 2/z, hence zz' = 2. (z^2)' = 2zz', so zz' = (1/2)(z^2)' = 2, thus (z^)' = 4, therefore z^2 = 4x + C. Hence z = sqrt(4x + C), and therefore
y = sqrt(4x + C) + x.
easy
(y - x)' = 2/(y -x)
let u = y - x
u' = 2/u
udu = 2dx
u^2/2 = 2x + const
u^2 = 4x + C
(y - x)^2 = 4x + C
y - x = ±√(4x + C)
y = x ± √(4x + C)
y-x=t....y=x+√(4x+c).…
You sound unwell. Take care.