To find the solution without series expansion, observe that xy' + y = (xy)', and from here we can simply integrate to get y' = xy + c. This is a linear ODE, which can be integrated using normal techniques.
y'' = xy' + y = (xy)', and thus y' = xy + C. At x=0, we have 0 = 0 + C, so C = 0. Now to solve y'=xy, we have (log(y))' = x, so log(y) = 0.5*x^2 + B. Therefore, y = e^B * e^(x^2 / 2). Plugging in initial condition y(0) = 1 implies that B = 0.
You could have just written down xy'+ y as (xy)'. The diff. equation then reduces to y''= (xy)' which is easily solved. I'd say that's a rather long way to solve this equation. But it still feels nice to work out the solution using power series.
This one occurs all of the time in physics of biological systems, specifically cellular structures and membranes. I would guess it also appears in a lot of other areas involving statistical mechanics. Because cells are generally overdamped systems, i.e. the inertial terms vanish (there is no acceleration, as soon as things stop being pushed, they stop moving), the velocity of an object living inside the cell or on the membrane will be proportional to the force exerted on it. This is because all of the force exerted on an object is "used up" to overcome friction, which is, to a good approximation, proportional to velocity. If we choose p(x,t) to be the probability distribution of the particle (or the local density of a particle ensemble) and J to be the current, the Fokker-Planck equation (that's just the more general version of the diffusion equation which also includes a directed current), omitting constants everywhere, is a partial differential equation of the type dp/dt = d/dx (dp/dx - J) Now the current is velocity*density, but velocity is proportional to the force, so we can write J=F*p. If the system is subject to an elastic force F=-x, and you are interested in the stationary state of the system in which the density doesn't change in time, i.e. dp/dt=0, you get the equation (p' + x*p)' = 0, which can be expanded into an equation that is very similar to the one discussed here (the sign is different, and there would be a constant multiplying the second term for it to be meaningful in terms of a real physical system). It also seems much easier to solve in this form. Note that, if C is some constant, the equation can be written as p' + x*p = C In order for p(x) to be a stationary probability distribution, it has to (a) be positive and (b) vanish at both infinities. That means there has to be some finite point x0 for which the probability is maximal (otherwise, we would just have a trivial solution p=0). We can then do a coordinate shift of the type y=x-x0 and for this shifted equation, p'(y=0)=0, which gives us C=0. Now we just need to solve a much simpler equation p' = -y*p This is fairly easy to do just by guessing. We know that the solution has to be an exponential of some function of y, p=exp(f(y)), and plugging this back into the equation gives f(y)=-(y^2)/2, i.e. p(y) is a Gaussian. Finally, we have to shift the solution back into x and obtain p(x) = A*exp[-0.5*(x-x0)^2], where the constant A is determined using the normalisation condition and x0 is a property of the system.
@@murdockZS For understanding how to use statistical physics to model cells, organelles, membranes etc., look at either "Physical Biology of the Cell" by R. Phillips et al. or "Mechanics of Motor Proteins and the Cytoskeleton" by J. Howard. If you are more interested in the math, the absolute best introductory book on stochastic analysis, which is the most important field of math when it comes to modelling in cell biophysics, is "Stochastic processes in physics and chemistry" by N.G. Van Kampen, it's absolutely amazing. If you want something more advanced, there is "Stochastic Methods - A Handbook for the Natural and Social Sciences" by Gardiner and Crispin or "The Fokker-Planck Equation - Methods of Solution and Applications" by H. Risken.
You can use Ladder Operators. Note it's an eigenvalue problem if you move y to the other side of the equal sign and treat what remains on the left as a differential operator. Note applying this operator to y' yields twice y'. In other words, y' is a solution with a different eigenvalue, ie increased by 1. Differentiation is a raising operator. So assume the initial eigenvalue is zero, the equation is amenable to reduction of order. TheN the derivative of that solution solves the original problem.
Maxima derived the full solution of this differential Equation, it also had the error function: y = (sqrt(%pi)*%k1*%e^(x^2/2)*erf(x/sqrt(2)))/sqrt(2)+%k2*%e^(x^2/2)
Defining y as a power series with a(n) the coefficient of x^n, we use the ode and initial conditions to find that a(0) = 1 a(1) = 0 a(n+2) = a(n)/(n+2) for each n. This comes from plugging the power series for y into the ode and combining like terms (I can’t go through the computations in text). We see immediately from the recursive formula that a(2k+1) = 0 for each k. That is, odd indexed terms of a(n) are 0. We focus on the even index terms. We have a(2k) = a(2(k-1))/2k = a(2(k-2))/(2k*2(k-1)) = ... Using that a(0)=1, we eventually find a(2k) = 1/((2^k)k!). Again I’m skipping the straightforward computations. Thus y = sum_{k=0}^inf (1/2^k*k!)x^2k =sum (x^2/2)/k! = exp(x^2/2). One can easily check that this function satisfies the ode and boundary conditions, so this is our solution.
Hi, There is a shorter differentiel equation for that function, that is : y'- xy = 0 For fun: 5:29 : "and so on and so forth", 6:17 : "and then, we'll finish it off".
HOMEWORK : Let f(0) = f(1) = 0 and f(n + 2) = 4^(n+2) · f(n + 1) - 16^(n+1) · f(n) + n · 2^(n²) for n = 0, 1, 2, 3… Show that the numbers f(1989), f(1990), f(1991) are divisible by 13. SOURCE : Shortlisted problem from IMO 1990, submitted by Greece.
SOLUTION Let f(n) = g(n) · 2^(n²) for all n. The recursion then transforms into g(n+2) - 2g(n+1) + g(n) = n · 16^(-n-1) for n integer including 0. By summing this equation from 0 to n-1, we get g(n+1) - g(n) = 1/(15²) · (1 - (15n+1)·16^(-n)). By summing up again from 0 to n-1, we get g(n) = 1/(15³) · (15n - 32 + (15n+2) · 16^(-n+1)). Hence, f(n) = 1/(15³) · (15n + 2 + (15n-32) · 16^(n-1)) · 2^((n-2)²). Now let's look at the values of f(n) modulo 13 : f(n) ≡ 15n + 2 + (15n-32) · 16^(n-1) ≡ 2n + 2 + (2n-6) · 3^(n-1). We have 3³ ≡ 1 (mod 13). Plugging in n ≡ 1 (mod 13) and n ≡ 1 (mod 3) for n = 1990 gives us f(1990) ≡ 0 (mod 13). We similarly calculate f(1989) ≡ 0 and f(1991) ≡ 0 (mod 13).
Nice , an other problem Let M (1,2,3.......,n) and A=(a1,a2.a3..........,am ) in M such that m is a natural number #0 And for any b in M there exist ai , aj in group A such that 1
Interesting problem, but to me saying that its "solution is a well known non-elementary function" suggested that the solution would be a power series not expressible in closed form as an elementary function. But here, exp(x^2/2) is an elementary function, it just doesn't have an elementary antiderivative. In other words, you cannot solve y' - exp(x^2/2) = 0 in terms of elementary functions, but that's a different problem from this one which doesn't seem all that unusual.
I agree completely. In general a real elementary function is one that can be constructed from exp, trig functions, ln and inverse trig functions using sum, difference, product, quotient, powers (including roots) and compositions.
Actually, the general solution of this particular equation involves the Error function, which is the integral of exp(x^2/2) and as you note, it is non-elementary, so Professor's statement is actually correct although the solution demonstrated is an elementary function.
thanks for your comment but I disagree. multiplication by x is equivalent to differentiation in Fourier space. you are correct that multiplication in general is convolution but in the special case of x it simplifies to differentiation. check the wiki page on Fourier transform.
y’’ -xy’-y = 0 iff d/dx(y’) -d/dx(xy) = d/dx(y’-xy) = 0 iff y’-xy = c1 for some constant c1. The integral of -x is -x^2/2, exponentiating gives the integrating factor exp(-x^2/2). exp(-x^2/2)y’-x*exp(-x^2/2)y = c1*exp(-x^2/2) d/dx(y*exp(-x^2/2)) = c1*exp(-x^2/2). Integrating gives y*exp(-x^2/2) = c1*integral (0 to x) exp(-t^2/2) dt +c2, which then further rearranges to give y(x) = c1*exp(x^2/2)*int(0 to x) exp(-t^2/2) dt + c2*exp(x^2/2)
It goes deeper than this - you can treat all recurrence relations as formal DE's/functional eq's on power series and get a fuckton of combinatorics simply for free. Algebra is more powerful than you would think.
The Cauchy Kowalevski Theorem implies the existence of analytic solutions to ODEs locally near a point assuming that the coefficients of the differential equation are sufficiently nice. See en.m.wikipedia.org/wiki/Cauchy%E2%80%93Kowalevski_theorem In particular, for this example the theorem can be used to imply there exists an analytic solution in a neighborhood of x=0. Now, we also know that all analytic functions can be expressed as a power series locally as well. Once we've found such a power series, it is then easy to analyze where it converges and so forth. This problem, the solution converges for every x and so is a global solution.
i did this. y''-xy'-y=0 note that -xy'-y=(-xy)' to get y''-(xy)'=0 now integrate to cancel one derivative and get. y'-xy=c y'=xy+c . From one inital condition we have y'(0)=0*y+c=0 => c=0 Now y'=xy is seperable and we get ln(y)=x^2/2 +C y=e^(x^2)/2*e^c From last initial condition y(0)=1 we get e^c=1 and we have y=e^(x^2)/2
@@angelmendez-rivera351 ... Thank you, but why do you say that no need to evaluate 0^(-1)? At around 1:30 he has y'(x) = sum [n=0 to inf] (n * An * x^(n-1)) and he says "We can start at zero or one, given that we multiply by n so the zero term is zero times some stuff". Well, that "some stuff" contains x^(n-1) which for n=0 is 0^(-1) when the derivative is evaluated in x=0, what you HAVE to do since one f the givens is that y'(0)=1.
Two reasons: First is linear ODE, second it has not constant coefficients. Therefore, space of solutions is restricted, for practical purposes, to exponential or polynomical forms.
It's an "ansatz", an assumption to solve it. Infinite polynomials are very powerful, they will allow finding any function that is infinitely differentiable. Personally I was more floored by the assumption that since the sum is zero, then all the elements of the sum must be zero. Sure, that's one way, but the elements might be non-zero and balance out.
I know you’re showing a method of solving integrals, I did it a bit of a different way and got the same answer and it was much faster but I don’t know if you’d call it a proof. It was more intuition. If you start with the general y” + fy’ + gy = 0 and you plug in y = e^Int[Y]dz, then it reduces to Y’ + Y^2 + fY + g = 0. In the problem above, f = -x and g = -1 so we have Y’ + Y^2 - xY - 1 = 0. From there it’s pretty easy to see that Y = x is a solution. Then you just integrate it and exponetiate that to get back to your original y value. Y = x Int[x]dz = x^2/2 + c y = e^Int[Y]dz = Ce^(x^2/2) where C = e^c, both are constants.
There is only one solution for that ODE given the initial conditions y(0)=1, y'(0)=0. You can get another, linearly independent solution by solving for y(0)=0, y'(0)=1. I believe you will then find that the even coefficients are zero and the odd coefficients have a simple recursion formula. Try to work this out, and let us know if you need more help. (Note that going from the coefficients to the expression is more difficult than in the video, since the erf function is involved.)
why did we start with assumption that the function y was of the form of an infinite polynomial? can every solution of such a differential equation be represented by an infinite polynomial? i surely dont have the background for this, just curious
Basically, we assume that y is a sufficiently nice function that it can be expressed as a Taylor series. Then we insert that series with the unknown coefficients into the differential equation and solve for those coefficients. Most of the time though, there is not anyway to write the final solution in a form that isn't a series.
Some functions (they all need to be infinitly differentiable) can be expressed this way. Here, since it is a differential equation with fairly simple coefficients, the Cauchy-Lipschitz theorem can be used and shows that there is an only solution to this differential equation with those initial conditions. To find this unique solution, Michael is looking for a solution that can be expressed this way. We don't know for sure that there is a solution that can be expressed this way, but if we find a function solution to the problem with this method, we know by construction that this function will be solution of the differential equation. Hope that helps !
@@nathane2079 helped indeed, ty so much for your answer :D to see if i understood correctly: if i were to solve a differential equation and found it to have a unique, yet undetermined solution under the theorem you mentioned, then i would TRY to assume it is a "well-behaved" function, aka expressable in the way shown. Then i would pray (and worked hard on the problem :P ) and hope i come up with a solution. otherwise i would have to come up with other stratergies to look for it (if possible). Is this correct? ( Note: i just did a quick wikipedia lookup to write this answer, i dont understand maths this deep, nor do i know the relevant literature, im simply curious :D)
in my opinion "non-elementary" isn't the right term to use in this context, but what Michael means is that the solution to the differential equation does not have an antiderivative (indefinite integral) that is expressible in terms of elementary functions - that is, the solution has a non-elementary antiderivative. (personally I would only use the term "non-elementary differential equation" if the solution itself is not expressible in terms of elementary functions, i.e. if the solution is non-elementary)
I tried using Laplace transforms. Perhaps I could keep going although knowing the final result as exp(x^2/2), looking at tables I can't see this appearing. I did get to Y(s) = (s+(1/s^2))/(s^2-(1/s)-1) just quickly. I believe it's correct algebraically. Next I'd want to fiddle with denominators, factor, use impartial fractions. You can multiply out some s terms but just looking it it's not going to work or I'm starting to get complex roots etc and just don't trust my penmanship
We're allowed to assume the function we're looking for is continuous (otherwise, knowing the values at 0 won't help constrain answers). So instead of working with y(0), we can work with lim x->0 y(x), and disregard any terms with zero coefficients and negative exponents and treat x^0 as 1 everywhere. This ends up just being treated as conventions for working with power series without justifying it each time.
"Double prime"? "Triple prime"? why not "second" and "third" like in music notation? lagrangian notation is not different then other notations. if it was written d^2y/dx^2 you don't read it "double derivative".
Either I don't understand your point or you are just arbitrarily adversarial. Every notation, nomenclature, definition and convention are by definition prescriptive. that's literally the point of it.
Laplace Transform do not work with non-linear differential Equations. However some non-linear differential Equations work with the Fourier Transform, and this particular One should work, since it Will obtain derivatives of the Fourier Transform itself.
After some fidling, I found the Laplace Transform of this Equation, which contains the derivative of the Laplace Transform. laplace(diff(y(x),x,2)-x*diff(y(x),x)-y(x),x,t) t*'diff('laplace(y(x),x,t),t,1)-%at('diff(y(x),x,1),x = 0)+t^2*'laplace(y(x),x,t)-y(0)*t
Nice video. I did the problem somewhat differently though. First, its not too hard to notice that y"-xy'-y=0 is already a total derivative. Namely, (y'-xy')'=0 which can be integrated up to y'-xy=C for some C. Now we can use our initial conditions to see that at x=0 this equation is 0, and thus C=0 giving us y'-xy=0. And this can be solved using separarion of variables to and the initial condition again to get the answer in the video.
That was easy. How about finding the other linearly independent solution to this equation? The Abel's identity gives that in the form of an integral of e^(-3x²/2).
@@angelmendez-rivera351 I often try to solve before I open the video, as do many other people here. The problem as seen on the thumbnail picture is *not* an initial condition problem.
I got a recurrence relation of A sub (n+2) = A sub (n) divided by (n+2) and a power series of y = 1+(1/2)x^2+(1/8)x^4+(1/48)x^6+···. I'd by excstatic if my old self did this right.
More or less. Since it's a 2nd order linear in y, It can be transformed to SL form with the integrating factor exp(-x^2/2). It's also lacking an eigenvalue.
To find the solution without series expansion, observe that xy' + y = (xy)', and from here we can simply integrate to get y' = xy + c. This is a linear ODE, which can be integrated using normal techniques.
great approach!
@@Chill---- why is it wrong ?
the general solution does require the error function tho
@@shivaramkratos4018 ıt ıs not wrong. It ıs fıne sımple and correct. But I think Michael uses the problem to show us a generel method.
@@klausg1843 Yes, I agree with you, I posted the other method, to illustrate why the final solution turns simpler than it is expected to be.
y'' = xy' + y = (xy)', and thus y' = xy + C. At x=0, we have 0 = 0 + C, so C = 0. Now to solve y'=xy, we have (log(y))' = x, so log(y) = 0.5*x^2 + B. Therefore, y = e^B * e^(x^2 / 2). Plugging in initial condition y(0) = 1 implies that B = 0.
Slick...
Nice!
General solution : y(x)=exp(x^2/2)*integral(t to x of exp(-k^2/2) dk), t 1. If you choose t =x you find the null solution y=0, etc,etc.
You could have just written down xy'+ y as (xy)'. The diff. equation then reduces to y''= (xy)' which is easily solved. I'd say that's a rather long way to solve this equation. But it still feels nice to work out the solution using power series.
This one occurs all of the time in physics of biological systems, specifically cellular structures and membranes. I would guess it also appears in a lot of other areas involving statistical mechanics. Because cells are generally overdamped systems, i.e. the inertial terms vanish (there is no acceleration, as soon as things stop being pushed, they stop moving), the velocity of an object living inside the cell or on the membrane will be proportional to the force exerted on it. This is because all of the force exerted on an object is "used up" to overcome friction, which is, to a good approximation, proportional to velocity. If we choose p(x,t) to be the probability distribution of the particle (or the local density of a particle ensemble) and J to be the current, the Fokker-Planck equation (that's just the more general version of the diffusion equation which also includes a directed current), omitting constants everywhere, is a partial differential equation of the type
dp/dt = d/dx (dp/dx - J)
Now the current is velocity*density, but velocity is proportional to the force, so we can write J=F*p. If the system is subject to an elastic force F=-x, and you are interested in the stationary state of the system in which the density doesn't change in time, i.e. dp/dt=0, you get the equation
(p' + x*p)' = 0,
which can be expanded into an equation that is very similar to the one discussed here (the sign is different, and there would be a constant multiplying the second term for it to be meaningful in terms of a real physical system).
It also seems much easier to solve in this form. Note that, if C is some constant, the equation can be written as
p' + x*p = C
In order for p(x) to be a stationary probability distribution, it has to (a) be positive and (b) vanish at both infinities. That means there has to be some finite point x0 for which the probability is maximal (otherwise, we would just have a trivial solution p=0). We can then do a coordinate shift of the type y=x-x0 and for this shifted equation, p'(y=0)=0, which gives us C=0. Now we just need to solve a much simpler equation
p' = -y*p
This is fairly easy to do just by guessing. We know that the solution has to be an exponential of some function of y, p=exp(f(y)), and plugging this back into the equation gives f(y)=-(y^2)/2, i.e. p(y) is a Gaussian. Finally, we have to shift the solution back into x and obtain
p(x) = A*exp[-0.5*(x-x0)^2],
where the constant A is determined using the normalisation condition and x0 is a property of the system.
now back up and explain it to me like i'm 10.
Please, can you provide some bibliography for biophysics and statistical mechanics? It would be great
@@murdockZS
For understanding how to use statistical physics to model cells, organelles, membranes etc., look at either "Physical Biology of the Cell" by R. Phillips et al. or "Mechanics of Motor Proteins and the Cytoskeleton" by J. Howard.
If you are more interested in the math, the absolute best introductory book on stochastic analysis, which is the most important field of math when it comes to modelling in cell biophysics, is "Stochastic processes in physics and chemistry" by N.G. Van Kampen, it's absolutely amazing. If you want something more advanced, there is "Stochastic Methods
- A Handbook for the Natural and Social Sciences" by Gardiner and Crispin or "The Fokker-Planck Equation
- Methods of Solution and Applications" by H. Risken.
what the-
You can use Ladder Operators. Note it's an eigenvalue problem if you move y to the other side of the equal sign and treat what remains on the left as a differential operator. Note applying this operator to y' yields twice y'. In other words, y' is a solution with a different eigenvalue, ie increased by 1. Differentiation is a raising operator. So assume the initial eigenvalue is zero, the equation is amenable to reduction of order. TheN the derivative of that solution solves the original problem.
Maxima derived the full solution of this differential Equation, it also had the error function:
y = (sqrt(%pi)*%k1*%e^(x^2/2)*erf(x/sqrt(2)))/sqrt(2)+%k2*%e^(x^2/2)
Defining y as a power series with a(n) the coefficient of x^n, we use the ode and initial conditions to find that
a(0) = 1
a(1) = 0
a(n+2) = a(n)/(n+2) for each n.
This comes from plugging the power series for y into the ode and combining like terms (I can’t go through the computations in text).
We see immediately from the recursive formula that a(2k+1) = 0 for each k. That is, odd indexed terms of a(n) are 0. We focus on the even index terms.
We have a(2k) = a(2(k-1))/2k = a(2(k-2))/(2k*2(k-1)) = ...
Using that a(0)=1, we eventually find a(2k) = 1/((2^k)k!). Again I’m skipping the straightforward computations.
Thus y = sum_{k=0}^inf (1/2^k*k!)x^2k =sum (x^2/2)/k! = exp(x^2/2). One can easily check that this function satisfies the ode and boundary conditions, so this is our solution.
Hi,
There is a shorter differentiel equation for that function, that is : y'- xy = 0
For fun:
5:29 : "and so on and so forth",
6:17 : "and then, we'll finish it off".
One can get it by integrating original equation and using initial conditions. Michael liked it the hard way.
HOMEWORK : Let f(0) = f(1) = 0 and f(n + 2) = 4^(n+2) · f(n + 1) - 16^(n+1) · f(n) + n · 2^(n²) for n = 0, 1, 2, 3… Show that the numbers f(1989), f(1990), f(1991) are divisible by 13.
SOURCE : Shortlisted problem from IMO 1990, submitted by Greece.
SOLUTION
Let f(n) = g(n) · 2^(n²) for all n. The recursion then transforms into g(n+2) - 2g(n+1) + g(n) = n · 16^(-n-1) for n integer including 0. By summing this equation from 0 to n-1, we get g(n+1) - g(n) = 1/(15²) · (1 - (15n+1)·16^(-n)). By summing up again from 0 to n-1, we get g(n) = 1/(15³) · (15n - 32 + (15n+2) · 16^(-n+1)).
Hence, f(n) = 1/(15³) · (15n + 2 + (15n-32) · 16^(n-1)) · 2^((n-2)²).
Now let's look at the values of f(n) modulo 13 : f(n) ≡ 15n + 2 + (15n-32) · 16^(n-1) ≡ 2n + 2 + (2n-6) · 3^(n-1).
We have 3³ ≡ 1 (mod 13). Plugging in n ≡ 1 (mod 13) and n ≡ 1 (mod 3) for n = 1990 gives us f(1990) ≡ 0 (mod 13). We similarly calculate f(1989) ≡ 0 and f(1991) ≡ 0 (mod 13).
Nice , an other problem
Let M (1,2,3.......,n) and A=(a1,a2.a3..........,am ) in M such that m is a natural number #0
And for any b in M there exist ai , aj in group A such that 1
Hey you've to id's Good place to stop
And Good place to start
Interesting problem, but to me saying that its "solution is a well known non-elementary function" suggested that the solution would be a power series not expressible in closed form as an elementary function. But here, exp(x^2/2) is an elementary function, it just doesn't have an elementary antiderivative. In other words, you cannot solve y' - exp(x^2/2) = 0 in terms of elementary functions, but that's a different problem from this one which doesn't seem all that unusual.
I agree completely. In general a real elementary function is one that can be constructed from exp, trig functions, ln and inverse trig functions using sum, difference, product, quotient, powers (including roots) and compositions.
Actually, the general solution of this particular equation involves the Error function, which is the integral of exp(x^2/2) and as you note, it is non-elementary, so Professor's statement is actually correct although the solution demonstrated is an elementary function.
Still another, linearly independent solution to this equation *is* non-elementary function.
Very nice solution🙌 I love your channel!!!
This method is awesome, I'm seeing it for the 1st time😲, the An finding part was pretty hard though
Nice! Power series are really nice, totally agree.
This looks working well. Hope to see more solutions using this method. Final step converting back to function is maybe difficult...
Ah! I loved this problem. Thanks for all your amazing content
alternative approach: fourier transform, simplify, inverse fourier transform, done.
Sorry man, but the xy' forced your transformed equation to one that contain convolution, which is harder to deal with compare to the video's method
thanks for your comment but I disagree. multiplication by x is equivalent to differentiation in Fourier space. you are correct that multiplication in general is convolution but in the special case of x it simplifies to differentiation. check the wiki page on Fourier transform.
y’’ -xy’-y = 0 iff d/dx(y’) -d/dx(xy)
= d/dx(y’-xy) = 0 iff y’-xy = c1 for some constant c1. The integral of -x is -x^2/2, exponentiating gives the integrating factor exp(-x^2/2).
exp(-x^2/2)y’-x*exp(-x^2/2)y
= c1*exp(-x^2/2)
d/dx(y*exp(-x^2/2)) = c1*exp(-x^2/2). Integrating gives y*exp(-x^2/2)
= c1*integral (0 to x) exp(-t^2/2) dt +c2, which then further rearranges to give
y(x) = c1*exp(x^2/2)*int(0 to x) exp(-t^2/2) dt + c2*exp(x^2/2)
idk why but i love it when you can see the chalk break in his hand
10:01
How are you so quick? Only 56 seconds
Oo
hello my friends, is there any good book recommendation for studying diff. equations?
Finally it make sense now (the relation between recurrence relation and differential equation)
It goes deeper than this - you can treat all recurrence relations as formal DE's/functional eq's on power series and get a fuckton of combinatorics simply for free.
Algebra is more powerful than you would think.
I need channels like this, but with physics problems
1. Andrew Dotson 2. Physics explained
@@gavasiarobinssson5108 It doesn't look like they are focused on solving physics books-like exercises.
The other solution - y(0) =0, y'(0) = 1, I find more interesting, involving, in addition the error function.
Amazing solution
Thank you, professor!
Great explanation
Why can we assume that y can be represented as a sum of powers of x? Do we not have to show it converges?
The Cauchy Kowalevski Theorem implies the existence of analytic solutions to ODEs locally near a point assuming that the coefficients of the differential equation are sufficiently nice. See en.m.wikipedia.org/wiki/Cauchy%E2%80%93Kowalevski_theorem
In particular, for this example the theorem can be used to imply there exists an analytic solution in a neighborhood of x=0. Now, we also know that all analytic functions can be expressed as a power series locally as well. Once we've found such a power series, it is then easy to analyze where it converges and so forth. This problem, the solution converges for every x and so is a global solution.
Knowing the initial conditions, could we just use the Laplace transform?
If we modify DE with all + signs we get Gauss distribution function.
i randomly stumbled upon the bessel function and it has a definition similar to this
I like v.m. your demonstration!
Nice... what an unexpected result!
i did this. y''-xy'-y=0 note that -xy'-y=(-xy)' to get y''-(xy)'=0 now integrate to cancel one derivative and get. y'-xy=c
y'=xy+c . From one inital condition we have y'(0)=0*y+c=0 => c=0
Now y'=xy is seperable and we get ln(y)=x^2/2 +C y=e^(x^2)/2*e^c
From last initial condition y(0)=1 we get e^c=1 and we have y=e^(x^2)/2
I started with the series a_n*x^(n+
ho), and worked it out like that. I forget the name of the method...
wat is a differential equation?
Professor penn please consider completing the roger ramunujan identities playlist
How is it known that y must be a polynomial?
Are you taking 0^0=1? And 0^(-1)=1/0 as something that exists and has a finite value?
@@angelmendez-rivera351 ... Thank you, but why do you say that no need to evaluate 0^(-1)?
At around 1:30 he has y'(x) = sum [n=0 to inf] (n * An * x^(n-1)) and he says "We can start at zero or one, given that we multiply by n so the zero term is zero times some stuff". Well, that "some stuff" contains x^(n-1) which for n=0 is 0^(-1) when the derivative is evaluated in x=0, what you HAVE to do since one f the givens is that y'(0)=1.
so how do we know that y is a polynomial? That seems to be an initial assumption that is not given in the question.
Two reasons: First is linear ODE, second it has not constant coefficients. Therefore, space of solutions is restricted, for practical purposes, to exponential or polynomical forms.
It's an "ansatz", an assumption to solve it. Infinite polynomials are very powerful, they will allow finding any function that is infinitely differentiable. Personally I was more floored by the assumption that since the sum is zero, then all the elements of the sum must be zero. Sure, that's one way, but the elements might be non-zero and balance out.
I know you’re showing a method of solving integrals, I did it a bit of a different way and got the same answer and it was much faster but I don’t know if you’d call it a proof. It was more intuition.
If you start with the general y” + fy’ + gy = 0 and you plug in y = e^Int[Y]dz, then it reduces to Y’ + Y^2 + fY + g = 0. In the problem above, f = -x and g = -1 so we have
Y’ + Y^2 - xY - 1 = 0.
From there it’s pretty easy to see that Y = x is a solution. Then you just integrate it and exponetiate that to get back to your original y value.
Y = x
Int[x]dz = x^2/2 + c
y = e^Int[Y]dz = Ce^(x^2/2) where C = e^c, both are constants.
then apply initial conditions to get C = 1
Since it is a second order ODE, there should be another solution. We could not find it this way, but I still want to know what it is.
There is only one solution for that ODE given the initial conditions y(0)=1, y'(0)=0. You can get another, linearly independent solution by solving for y(0)=0, y'(0)=1. I believe you will then find that the even coefficients are zero and the odd coefficients have a simple recursion formula. Try to work this out, and let us know if you need more help. (Note that going from the coefficients to the expression is more difficult than in the video, since the erf function is involved.)
why did we start with assumption that the function y was of the form of an infinite polynomial? can every solution of such a differential equation be represented by an infinite polynomial? i surely dont have the background for this, just curious
Also curious
Basically, we assume that y is a sufficiently nice function that it can be expressed as a Taylor series. Then we insert that series with the unknown coefficients into the differential equation and solve for those coefficients. Most of the time though, there is not anyway to write the final solution in a form that isn't a series.
Some functions (they all need to be infinitly differentiable) can be expressed this way. Here, since it is a differential equation with fairly simple coefficients, the Cauchy-Lipschitz theorem can be used and shows that there is an only solution to this differential equation with those initial conditions.
To find this unique solution, Michael is looking for a solution that can be expressed this way. We don't know for sure that there is a solution that can be expressed this way, but if we find a function solution to the problem with this method, we know by construction that this function will be solution of the differential equation. Hope that helps !
He just assume is an analytic function
@@nathane2079 helped indeed, ty so much for your answer :D to see if i understood correctly: if i were to solve a differential equation and found it to have a unique, yet undetermined solution under the theorem you mentioned, then i would TRY to assume it is a "well-behaved" function, aka expressable in the way shown. Then i would pray (and worked hard on the problem :P ) and hope i come up with a solution. otherwise i would have to come up with other stratergies to look for it (if possible). Is this correct?
( Note: i just did a quick wikipedia lookup to write this answer, i dont understand maths this deep, nor do i know the relevant literature, im simply curious :D)
Power series are Powerful .
Good Place To Start At 0:25
Please anyone explain what does it mean by non elementary
in my opinion "non-elementary" isn't the right term to use in this context, but what Michael means is that the solution to the differential equation does not have an antiderivative (indefinite integral) that is expressible in terms of elementary functions - that is, the solution has a non-elementary antiderivative. (personally I would only use the term "non-elementary differential equation" if the solution itself is not expressible in terms of elementary functions, i.e. if the solution is non-elementary)
I tried using Laplace transforms. Perhaps I could keep going although knowing the final result as exp(x^2/2), looking at tables I can't see this appearing.
I did get to Y(s) = (s+(1/s^2))/(s^2-(1/s)-1) just quickly. I believe it's correct algebraically. Next I'd want to fiddle with denominators, factor, use impartial fractions. You can multiply out some s terms but just looking it it's not going to work or I'm starting to get complex roots etc and just don't trust my penmanship
Simplify, and try to use the Residue Theorem.
Isn't it true that representing the function as a sum starting at 0 makes y(0) undefined, since y(0)=a_0*0^0?
0^0=1 by convention : x^0 is always 1 as it is the product of no element and it should equal to the neutral element of mutliplication
0^0 appears just because of the notation use. There is no such weird stuff going on here.
We're allowed to assume the function we're looking for is continuous (otherwise, knowing the values at 0 won't help constrain answers). So instead of working with y(0), we can work with lim x->0 y(x), and disregard any terms with zero coefficients and negative exponents and treat x^0 as 1 everywhere. This ends up just being treated as conventions for working with power series without justifying it each time.
"Double prime"? "Triple prime"? why not "second" and "third" like in music notation? lagrangian notation is not different then other notations. if it was written d^2y/dx^2 you don't read it "double derivative".
Found the prescriptivist!
Either I don't understand your point or you are just arbitrarily adversarial. Every notation, nomenclature, definition and convention are by definition prescriptive. that's literally the point of it.
Couldn't we take Laplace transformation ?
Laplace Transform do not work with non-linear differential Equations. However some non-linear differential Equations work with the Fourier Transform, and this particular One should work, since it Will obtain derivatives of the Fourier Transform itself.
After some fidling, I found the Laplace Transform of this Equation, which contains the derivative of the Laplace Transform.
laplace(diff(y(x),x,2)-x*diff(y(x),x)-y(x),x,t)
t*'diff('laplace(y(x),x,t),t,1)-%at('diff(y(x),x,1),x = 0)+t^2*'laplace(y(x),x,t)-y(0)*t
What if the function is not analytic and has only 1st and 2nd derivatives?
@@angelmendez-rivera351 I haven't taken differential equation course, could you tell the full name of the theorem please?
Nice video. I did the problem somewhat differently though. First, its not too hard to notice that y"-xy'-y=0 is already a total derivative. Namely, (y'-xy')'=0 which can be integrated up to y'-xy=C for some C. Now we can use our initial conditions to see that at x=0 this equation is 0, and thus C=0 giving us y'-xy=0. And this can be solved using separarion of variables to and the initial condition again to get the answer in the video.
Small note:
It's (y'-xy)'=0, in 4th line.
That was easy. How about finding the other linearly independent solution to this equation? The Abel's identity gives that in the form of an integral of e^(-3x²/2).
@@angelmendez-rivera351 I often try to solve before I open the video, as do many other people here. The problem as seen on the thumbnail picture is *not* an initial condition problem.
@@angelmendez-rivera351 go teach somebody else, please. I’m fine without your advices.
Thank for premath
He's done the impossible! He solved a differential equation of y and x where ∫y dx was a non-elementary function!
I got a recurrence relation of A sub (n+2) = A sub (n) divided by (n+2) and a power series of y = 1+(1/2)x^2+(1/8)x^4+(1/48)x^6+···. I'd by excstatic if my old self did this right.
So I watched the end of the video and I got it right. Yay!
I’m not sure if this is right, but can’t you do y’’=(xy)’ so y’=xy?
Then you would get ln(y)=x^2/2+C so y=Ce^(x^2/2), and then you plug in the initial conditions
strictly y'=xy+c but since y'(0)=0 c=0.
@@reneszeywerth8352 and the general case where c is not 0 would have an additional term - if my calcs are correct - c * erf(x /sqrt(2)) * exp(x^2/2)
Substitute y=-vx
Your mic is only consistent when your facing directly forward and not standing on the left side. The lapel mic was working much better.
your =/= you're = you are
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y"+xy'+y=0 try this
I've waited so long, I didn't know we supposed to solve for y, I got confused on why you're leaving out y'' and xy' and then it clicks
:(
Laplace transform solves this equation much easier
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it's a Strum-Liouville equation.
More or less. Since it's a 2nd order linear in y, It can be transformed to SL form with the integrating factor exp(-x^2/2). It's also lacking an eigenvalue.
totally lost- not a good teacher