Which means piecewise solutions exist for the equation😃. We can find the tengency point which is b=1/a^2 then we can choose for xb the parabola solution or vice versa. Those would be all the piecwise solutions. I wouldn't have noticed this without stambling across your comment!😄
I just was wondering about the tangency condition and thought "what if the line solution intersects the parabola solution at another point?" which is obviously impossible; there is only 1 tangent point, and then I realized: I am not obliged to have 1 line! I could choose 2 different line, this will give us 2 bifurication points and thus more piecewise solutions. So, here is the full set of solutions: for x in D , y = ax + 1/a or y = ±2√x for xb , y = sgn(a)2√x (b=1/a^2) for xb , y = ax + 1/a for x
You can obtain further (continuous and differentiable) solutions by appropriate piecewise combinations of the ones you found. E.g. f(x) = 2x+1/2 for x = 1/4 is one such solution.
You can also add another bifurication point (is that its name??), such that for the left interval, the solution is of the form y = a1x + 1/a1 and for the middle interval, the solution is of the form y = ±2√x (the sign is the same as that of a1) and for the right interval, the solution goes back to the form a2x + 1/a2 with a different constant but should have the same sign as a1 and be less than a1.
Differential equations of the form y=xy'+f(y') are called Clairaut equations. This is the case f(y')=1/y'. They all have a solution method like this one.
@@primenumberbuster404we can also generalize it more y=xf(y')+g(y') This equation is called Lagrange equation od d'Alembert equation but solution method is also similar y' = f(y')+xf'(y')y''+g'(y')y'' Now we can introduce parameter p = y' p = f(p)+xf'(p)p'+g'(p)p' p = f(p)+(xf'(p)+g'(p))p' p-f(p)=(xf'(p)+g'(p))p' p' = (p-f(p))/(xf'(p)+g'(p)) dp/dx = (p-f(p))/(xf'(p)+g'(p)) dx/dp = (xf'(p)+g'(p))/(p-f(p)) dx/dp = f'(p)/(p-f(p)) * x + g'(p))/(p-f(p) And we have linear equathon for x x = F(C,p) y = f(p)x+g(p) is parametric solution We can also try to get rid of parameter p
I'm assuming there are a couple of different ways to solve this, but I think I have a good one: xy'+1/y'=y' differentiating with respect to x (d/dx), we get: y'+xy''-y''/(y')^2=y' xy''-y''/(y')^2=0 which factors into: y''(x-1/(y')^2)=0 solving for y + checking solutions + having in mind the function is probably continuous so y is one distinct solution out of the 2 equations received earlier, we get: 1) y = cx+1/c 2) y = plus/minus 2sqrt(x)
Second solution is ±√x just a small correction. It's very satisfying differentiating when everything cancels and you are left with a simple line equation and a first-order ode
Piecewise solutions exist. Divide the domain into x≤b and x≥b where b=1/a^2 . Choose one type of the solutions for half of the domain and the other type for the other half. Those would be 2 piecewise solutions. [I was about to post the reply until I realized there could be 2 bifurication points] So, here is the full set of solutions: for x in D , y = ax + 1/a or y = ±2√x for xb , y = sgn(a)2√x (b=1/a^2) for xb , y = ax + 1/a for x
I used Maple to solve this DE and the result was identical to Professor Penn's solution. I'm wondering how computer programs are able to understand such a wonderful trick that Prof. Penn used. Well done!
The general solution to this first order differential equation has no continuous degrees of freedom. Is that common? I was expecting a constant of integration to stick around in the final answer, but the original DE constrained the value of a.
As others have pointed out in other comments, you can get two degrees of freedom because piecewise-defined solutions also solve this equation (and are continuously differentiable because the line equations exactly give you tangent lines to one of the +/-2sqrt(x) solutions). So you can have a "middle" part where the function is +2sqrt(x), switching to a tangent line to that function at some point a in (0,infty) to the left, and the same at some point b > a to the right. Nonlinear ODEs in general need not give you a family parametrised by some constant that you add or multiply to your "baseline" solution, that is very much a linear algebra thing (spanning an affine subspace via a particular solution plus spanning vectors), even if y' = y and similar separation-of-variables equations also happen to work out this way (with a multiplicative constant instead of an additive one). Nonlinear ODEs may have much weirder solution sets, just like polynomial equations can have n-point solution sets (intersecting a line and a circle will generally give you 2 points, for example).
@@droid-droidsson This doesn't go against any of what you said If ODEs are nice enough, their solutions could be parametrized by n constants where n is the order of the equation. By nice enough I don't mean linear necessarily but some more general property that I can't remember.
@@droid-droidsson I think it has something to do with Lipschitz-continuity. Consider the ODE y'(x) = f(x,y(x)). If f is Lipschitz-continuous on the y variable then the initial value problem would have a unique solution and we can later prove that a general solution can be parametrize by the initial value, that is the parameters is y(0). And for an ODE with order n and with a similar Lipschitz-continuity condition, the parameters are going to be y(0) , y'(0) , y^(2)(0) and so on until y^(n-1)(0) . At least that's what I can remember. Hope you find this interesting to read!
for y"=0 there is a solution : y=ax+b but this isn't the only solution for example y=|x | the absolute value function and the family of the parabolic trig function which Michael penn explained in the last vedio
It all depends on the domain of y, which should have been stated in the problem itself. If 0 wasn't in the domain, then the absolute value function would be a solution. The general solution would be f(x) = ax+1/a for x0 , different constants on each interval . He also got the last solution incomplete. it is also defined piecewisely, f(x) = ±2√−x for x0 , and this will give us 4 cases. Now, tell me, what are the parabolic trig function you talk about? Edit: I checked the degenerate solution and it turned out I was wrong, f can't be ±2√−x for x
OK, I watched the parabolic trig functions video and I didn't like it. He claim that they should parametrize a unit parabola. I find that arbitrary. However, the functions he derived don't solve the differential equation and many people said so in the comments and gave an alternative. The proposed functions in comments are trivial but nevertheless more natural and in accordance to the spherical and hyperbolic functions.
I said that the absolute value function and the parabolic trig function are a solutions to this equation : y"=0 not the main equation in this vedio and for the main equation in this vedio I have not checked if those functions are a solutions or not . but i liked the parabolic trig function because I haven't seen it before that vedio and I want to know and discover new amazing functions. but what is the spherical function you say ? I know the hyperbolic functions but I haven't seen the spherical function before?
@@conanedojawa4538 >> "I said that the absolute value function and the parabolic trig function are a solutions to this equation : y"=0 not the main equation in this vedio" Yes I know, I never said otherwise. I am talking about the equation y"=0 and what I was saying is this; the parabolic functions he proposed in that video do not solve the equation y"=0 .
@@mnokeee It negates the need for a modulus for y, but not for x, I'd say? The end result should be y = +- 2 sqrt(|x|), not only y = +- 2 sqrt(x). Edit: Ok, I checked... y = +- 2 sqrt(|x|) solves the equation 2 x y' = y, but not the original equation. So you (and Michael) are right, only y = +- 2 sqrt(x) works.
His work is lacking in rigor. There should be one or several intervals introduced, for y'', and he integrates in case 2 improperly ! Why would we be in the case x>0 ? Then, you combine the 2 cases in order to get a solution with y' continuous, someone already gave an example.
@@digxx Absolutely not. Suppose y''(1)=0 so 2xy'=y can be false in x=1 so you have to solve 2xy'=y on (-oo,1) and on (1,+oo). I remember you, that solving a differential équation means to solve it, on each interval ! If y'' has 10 roots there are 11 generic intervals without roots to check plus 10 generic intervals with only one root, plus 9 generic intervals with two consecutives roots and so on and so forth and if y'' has an infinity (like cosine), you have an infinity intervals to check !
@@mathunt1130 Yeah he messed up a bit there, but this is how it should be: ln |y| = ln |a*sqrt(x)| |y| = |a*sqrt(x)| = a*sqrt(x) y = A*sqrt(x); where A = ±a. Remember, the anti-derivative of 1/x is ln |x|; ln x doesn't cover all cases.
This differential equation is called Clairaut's equation.
Interesting. The family of lines y = ax + 1/a are the tangents to the parabola y = ±2√x
Which means piecewise solutions exist for the equation😃. We can find the tengency point which is b=1/a^2 then we can choose for xb the parabola solution or vice versa. Those would be all the piecwise solutions. I wouldn't have noticed this without stambling across your comment!😄
I just was wondering about the tangency condition and thought "what if the line solution intersects the parabola solution at another point?" which is obviously impossible; there is only 1 tangent point, and then I realized: I am not obliged to have 1 line! I could choose 2 different line, this will give us 2 bifurication points and thus more piecewise solutions.
So, here is the full set of solutions:
for x in D , y = ax + 1/a or y = ±2√x
for xb , y = sgn(a)2√x (b=1/a^2)
for xb , y = ax + 1/a
for x
@@RUclips_username_not_found Very interesting observations!
You can obtain further (continuous and differentiable) solutions by appropriate piecewise combinations of the ones you found. E.g. f(x) = 2x+1/2 for x = 1/4 is one such solution.
You can also add another bifurication point (is that its name??), such that for the left interval, the solution is of the form y = a1x + 1/a1 and for the middle interval, the solution is of the form y = ±2√x (the sign is the same as that of a1) and for the right interval, the solution goes back to the form a2x + 1/a2 with a different constant but should have the same sign as a1 and be less than a1.
Is there a wide class of differential equations where this trick is useful, or is it just for very specific examples like this?
You can solve any equation of the form :-
y=xy'+f(y')
=>y'=y'+xy"+y"f'(y')
=>0=y"[x+f'(y')]
Now, you could do the two cases:
y"=0 or x+f'(y')=0
Differential equations of the form y=xy'+f(y') are called Clairaut equations. This is the case f(y')=1/y'. They all have a solution method like this one.
@@primenumberbuster404 Thanks!
@@joshuatilley1887 I looked up Clairaut's equations and they have some interesting properties. Thank you for mentioning it!
@@primenumberbuster404we can also generalize it more y=xf(y')+g(y')
This equation is called Lagrange equation od d'Alembert equation but
solution method is also similar
y' = f(y')+xf'(y')y''+g'(y')y''
Now we can introduce parameter p = y'
p = f(p)+xf'(p)p'+g'(p)p'
p = f(p)+(xf'(p)+g'(p))p'
p-f(p)=(xf'(p)+g'(p))p'
p' = (p-f(p))/(xf'(p)+g'(p))
dp/dx = (p-f(p))/(xf'(p)+g'(p))
dx/dp = (xf'(p)+g'(p))/(p-f(p))
dx/dp = f'(p)/(p-f(p)) * x + g'(p))/(p-f(p)
And we have linear equathon for x
x = F(C,p)
y = f(p)x+g(p)
is parametric solution
We can also try to get rid of parameter p
I'm assuming there are a couple of different ways to solve this, but I think I have a good one:
xy'+1/y'=y' differentiating with respect to x (d/dx), we get:
y'+xy''-y''/(y')^2=y'
xy''-y''/(y')^2=0 which factors into:
y''(x-1/(y')^2)=0
solving for y + checking solutions + having in mind the function is probably continuous so y is one distinct solution out of the 2 equations received earlier, we get:
1) y = cx+1/c
2) y = plus/minus 2sqrt(x)
That's the same solution I applied 🙂
Second solution is ±√x just a small correction. It's very satisfying differentiating when everything cancels and you are left with a simple line equation and a first-order ode
Piecewise solutions exist. Divide the domain into x≤b and x≥b where b=1/a^2 .
Choose one type of the solutions for half of the domain and the other type for the other half. Those would be 2 piecewise solutions.
[I was about to post the reply until I realized there could be 2 bifurication points]
So, here is the full set of solutions:
for x in D , y = ax + 1/a or y = ±2√x
for xb , y = sgn(a)2√x (b=1/a^2)
for xb , y = ax + 1/a
for x
I used Maple to solve this DE and the result was identical to Professor Penn's solution.
I'm wondering how computer programs are able to understand such a wonderful trick that Prof. Penn used.
Well done!
Very nice ODE. Thanks👌
Very nice. It reminds me of Hooke's Law: y ' ' = -ky. Multiply both sides by 2y '.
You can solve that one with a Laplace transform though - you can’t do that here
@@NathanSimonGottemer True.
Pedant: should point out that x^(1/2) = +/- sqrt(x). However the negative root must be rejected as it's inside a logarithm.
Love it. Two different solutions.
I figured the negative x values would be a trivial extension with a solution like a*sqrt(-x) but it appears to not work.
You can really stoke the boilers on this one, if you can keep your head above still waters.
The general solution to this first order differential equation has no continuous degrees of freedom. Is that common? I was expecting a constant of integration to stick around in the final answer, but the original DE constrained the value of a.
As others have pointed out in other comments, you can get two degrees of freedom because piecewise-defined solutions also solve this equation (and are continuously differentiable because the line equations exactly give you tangent lines to one of the +/-2sqrt(x) solutions). So you can have a "middle" part where the function is +2sqrt(x), switching to a tangent line to that function at some point a in (0,infty) to the left, and the same at some point b > a to the right.
Nonlinear ODEs in general need not give you a family parametrised by some constant that you add or multiply to your "baseline" solution, that is very much a linear algebra thing (spanning an affine subspace via a particular solution plus spanning vectors), even if y' = y and similar separation-of-variables equations also happen to work out this way (with a multiplicative constant instead of an additive one). Nonlinear ODEs may have much weirder solution sets, just like polynomial equations can have n-point solution sets (intersecting a line and a circle will generally give you 2 points, for example).
@@droid-droidsson This doesn't go against any of what you said
If ODEs are nice enough, their solutions could be parametrized by n constants where n is the order of the equation. By nice enough I don't mean linear necessarily but some more general property that I can't remember.
@@droid-droidsson
I think it has something to do with Lipschitz-continuity. Consider the ODE y'(x) = f(x,y(x)). If f is Lipschitz-continuous on the y variable then the initial value problem would have a unique solution and we can later prove that a general solution can be parametrize by the initial value, that is the parameters is y(0). And for an ODE with order n and with a similar Lipschitz-continuity condition, the parameters are going to be y(0) , y'(0) , y^(2)(0) and so on until y^(n-1)(0) . At least that's what I can remember.
Hope you find this interesting to read!
Wait, shouldn't the derivate of a*sqrt(x) have a minus sign in front? I did it that way, got a complex coefficient and it worked.
Isnt this like claurouts equation someway like we do similar thing in that
for y"=0 there is a solution : y=ax+b but this isn't the only solution for example y=|x | the absolute value function and the family of the parabolic trig function which Michael penn explained in the last vedio
It all depends on the domain of y, which should have been stated in the problem itself.
If 0 wasn't in the domain, then the absolute value function would be a solution. The general solution would be f(x) = ax+1/a for x0 , different constants on each interval .
He also got the last solution incomplete. it is also defined piecewisely, f(x) = ±2√−x for x0 , and this will give us 4 cases.
Now, tell me, what are the parabolic trig function you talk about?
Edit: I checked the degenerate solution and it turned out I was wrong, f can't be ±2√−x for x
OK, I watched the parabolic trig functions video and I didn't like it.
He claim that they should parametrize a unit parabola. I find that arbitrary.
However, the functions he derived don't solve the differential equation and many people said so in the comments and gave an alternative. The proposed functions in comments are trivial but nevertheless more natural and in accordance to the spherical and hyperbolic functions.
I said that the absolute value function and the parabolic trig function are a solutions to this equation : y"=0 not the main equation in this vedio and for the main equation in this vedio I have not checked if those functions are a solutions or not .
but i liked the parabolic trig function because I haven't seen it before that vedio and I want to know and discover new amazing functions.
but what is the spherical function you say ? I know the hyperbolic functions but I haven't seen the spherical function before?
@@conanedojawa4538 Did you receive my reply? I think it got shadowbanned and I don't know why.
@@conanedojawa4538
>> "I said that the absolute value function and the parabolic trig function are a solutions to this equation : y"=0 not the main equation in this vedio"
Yes I know, I never said otherwise. I am talking about the equation y"=0 and what I was saying is this; the parabolic functions he proposed in that video do not solve the equation y"=0 .
9:05 Shouldn't we use ln|y| = 0.5 ln|x| ?
it doesnt really matter since the next step negates the need for a modulus
@@mnokeee It negates the need for a modulus for y, but not for x, I'd say? The end result should be y = +- 2 sqrt(|x|), not only y = +- 2 sqrt(x).
Edit: Ok, I checked... y = +- 2 sqrt(|x|) solves the equation 2 x y' = y, but not the original equation. So you (and Michael) are right, only y = +- 2 sqrt(x) works.
That's a good place to stop differentiating until. . .
How does lnx being onto effects the constant of integration????
Since it is onto, we are guaranteed that the constant can be written as ln(something), no matter what the 1st constant was.
thanks :)
I got a y=i(x-1) as a solution and it works
I did not know that constant can be complex
This is the solution y = ax+1/a, with a=i
His work is lacking in rigor. There should be one or several intervals introduced, for y'', and he integrates in case 2 improperly ! Why would we be in the case x>0 ? Then, you combine the 2 cases in order to get a solution with y' continuous, someone already gave an example.
Nice!
Alexis Clairaut , yes Alexis is a male name
Great
I don't agree.
2xy'y''=yy'' doesn't imply that
Y''=0 or 2xy'=y because y is a function !
Then how would you solve it if this is not the case?
@@Iron_uksus
I don't know yet ...
@@Iron_uksus Need to consider piecewise functions.
@NoahRebeiWhat roots? The equation holds for all x, so y'' would have to have uncountably many roots, which is equivalent to y''=0 ?!?
@@digxx
Absolutely not.
Suppose y''(1)=0 so 2xy'=y can be false in x=1 so you have to solve 2xy'=y on (-oo,1) and on (1,+oo).
I remember you, that solving a differential équation means to solve it, on each interval !
If y'' has 10 roots there are 11 generic intervals without roots to check plus 10 generic intervals with only one root, plus 9 generic intervals with two consecutives roots and so on and so forth and if y'' has an infinity (like cosine), you have an infinity intervals to check !
You can get rid of the solution with the minus sign, as it leads to non-real solutions.
I'm pretty sure it actually does work, though.
Why does it lead to non-real solutions?
@@bjornfeuerbacher5514 if a
@@mathunt1130 Yeah he messed up a bit there, but this is how it should be:
ln |y| = ln |a*sqrt(x)|
|y| = |a*sqrt(x)| = a*sqrt(x)
y = A*sqrt(x); where A = ±a.
Remember, the anti-derivative of 1/x is ln |x|; ln x doesn't cover all cases.
@@mathunt1130 Yes, but that only means that ln(y) is complex. If you exponentiate that to get y, the end result will be a real number again.
lost solution y = 0
if y=0 then y'=0 and so it doesn't work with the initial equation as there's a division by y' (1/y') and you can't divide by 0