A nice suggested differential equation
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- čas přidán 21. 08. 2024
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instead of overloading the notation by introducing u, you could instead write the equation as t''=t*t' and rewrite both sides as derivatives. that would looks like dt'/dx=d(t²/2)/dx. integrating both sides gives t'=t²/2+c.
Thanks. I had the same thought watching the video. I am glad I was not hallucinating
Same 💬💭
An interesting solution
Same thing I thought
How do we determine the constants A & B?
The classification of this ordinary differential equation is "does not contain the variable x explicitly".
Then, according to the theory, it allows lowering the order by one by replacing dy(x)/dx =p(y(x)).
Then y"(x)= d(dy(x)/dx)/dx =dp(y)/dx =[dp(y)/dy]*dy(x)/dx =p*dp/dy.
The original equation takes the form dp/dy - p/y =lny.
This linear differential equation can be solved either by the Bernoulli method or by the constant variation method. But in this case it is easier to write
d(p/y)/dy= lny /y=>
d(p/y)=lny*dy /y => p/y = (1/2)*(lny)^2+c1 =>dy/dx=y*[(1/2)*(lny)^2+c1 ]=>
dy/{y*[(1/2)*(lny)^2+c1 ]} =dx =>
dlny/[(lny)^2+2*c1] =(1/2)*dx. (*)
It is necessary to consider three cases.
1. c1=0 => -1/lny=x/2+C/2.
Answer:y=e^[-2/(x+C)], C - any.
2. c1>0. Then from (*) we get
[1/sqrt(2*c1)]*(atan(lny/sqrt(2*c1))=x/2 +c2 .
Re-assign (1/2)*sqrt (2*c1)=>c1>0, c2*sqrt(2*c1)] =>c2- any.
Answer: y(x)= exp[2*c1*tan(c1*x+c2)]. c1>0, c2- any.
3. c1
@@Vladimir_Pavlov Excellent proof. There's a 4th case: in this case y(x) = 1 for all x, which is indeed the only constant solution of the original equation. This solution can't be deduced from the first case; it could be inferred from the second or the third case, posing c_1 = 0, but this would be a contradiction because those cases arise only for c_1 different from 0.
@@MarcoMate87 If y(x)=1, then y'(x)=0. and this does not satisfy the original equation, as it was written.
If we write the equation in the form y"- (y')^2/y= y'*lny, then there will be another solution
y(x)=C>0.
You're right. I only considered the differential equation involving p(y) = y'. That equation has the special solution y = 1.
I am always curious whether problems like this are derived from an application in physics, engineering, or other field, even a branch of pure or applied mathematics.
We can take two others options: -1/2A^2 and 0. So we will infer three others types of solutions: exp(-A coth((Ax+B)/2)) and exp(-A tanh((Ax+B)/2)) where A is not equal to zero, and exp(2/(B-x)).
why c = contant > 0 in 8:36?
if c = 0, then y = e^(-2/(t + c_1)) - it is also solution,
if c < 0, then y = e^(c * tanh(c_1 - c^2/2 * t)) - one more solution
yeah I am thinking the same
Yes professor didn't mention this solution ;)
he said he wont go on all the details. He just wants to solve for 1 solution because it's too long
@@thangnguyen-iw8tb ok, my english is bad
It's also doable via the quotient rule. It can be rewritten as (y'' y - (y')^2)/y^2 = (ln y / y) y', or equivalently (y'/y)' = (ln y/ y) y'. Integrating, this becomes y'/y = 1/2 (ln y)^2 + C. Setting u = ln y we have u' = 1/2 u^2 + C, which is now a separable first order equation which can be done using standard methods.
x2
That's actually the first differential equation I managed to solve by myself and I'm quite proud of it.
I have just practiced solving it after watching this video. How did you do that on your own? Are you a math major? Just asking.
I feel like introducing u ( at 6:30 ) made everything more complicated than necessary, why not just write t'' = t't = 1/2 (t²)' and then integrate both sides? That gets us the equation from 9:00 immediately.
Yeah it was unnecessary to introduce the u
whenever you see equations with independent variable missing,
do subsitution f(y) = u = y',
u' = dy'/dt = dy'/dy dy/dt = uů
where ů = dy'/dy = du/dy = f'(y)
(so the idea is simply to turn the eq into function of y only)
uů/u - u/y = lny
ů - (1/y)u = lny
notice that this is none other than classic 1st ODE
ů/y - u/y² = lny/y
(u/y)' = lny/y
integrate both sides
u/y = ½ln²y + C
dy/(y(½ln²y+C)) = dx
the rest is left as an exercise for readers
The first half would have been simpler by directly assuming y=exp(t). Therefore, y'=exp(t)*t', y"=exp(t)*((t')²+t"). Plugging this into the original equation directly gives t"/t' = t.
I had the same thought, but still I prefer the way Michael did it, because it shows how we arrive at this particular substitution. Getting to know the techniques we can use to solve these problems is as important as the solution itself, perhaps more so.
I concur 100%
@@Nikolas_Davis ln(y) = t is same as y = e^t since both functions are inverse functions.
That would nicely introduce the concept of using an "Ansatz".
(But I'm sure Michael can do that in a next video with an even trickier problem!)
@@JosBergervoet german words everywhere today. What does it mean in english?
Fun fact: This differential equation belongs to a miniscule class of differential equations that can be analytically solved 💀
with these i always wish you went and plugged the answer in to check, i like watching that when people do it
This is clearly an equation like F(y, y', y'') = 0 -> make substituion y' = p(y) -> y'' = p(y) * p(y)' and it becomes usual ordinary first linear order equation
Does this method have a name? So that I can study it. Thanks!
@@nahblue I think this is a special case of a Poincaré map (correct me if I'm wrong, please), but that's likely not too helpful for study. I would call it something like order reduction of an autonomous ODE.
Thanks for opening up by explaining what the jargon means.
It seems one more solution is here. When the first constant A is below zero then integral is arcth and logarithms will disappear.
The constant added didn't need to be positive. There are three scenarios. The other two are one with natural log of a ratio of linear functions if the constant was negative and the other one being a simple hyperbolic if the constant was 0.
I think we should still check the case when our constant in first integration is negative. Also you said it's not negative but later checked only positive case, because the integration you made doesn't work for zero constant. Then there is different formula. Still nice work.
According to WolframAlpha, and writing the constant as -A^2, we actually just get -2/A*arctanh(x/A). Note the h: This is the inverse hyperbolic tangent function. It is only real-valued for -A < x < A. And this makes fairly intuitive sense to me, because if we rewrite our original constant as (A*i)^2, we get -2/(A*i)*arctan(x/(A*i)). Without going into details, hyperbolic trig functions are closely related to regular trig functions evaluated at imaginary values.
Another idea I considered is if we take the limit as A goes to zero, we get lim[A->0] (2/A*arctan(x/A)) = sgn(x)*inf. More or less the same thing happens with arctanh. It would be really neat if these limits were exactly -2/x, but alas.
All the first 5:40 can be simplified if we put z=lny e^z=y
We will reach the same result z"=z'z
Means 2z'=z^2+£a^2 where £=-1,0,1
£=-1 => z= -a + 2a/(1-b.exp(ax))
£=0 => z= 2/(b-x)
£=1 => z= -a + a.tan((ax+b)/2)
11:38
The text book to checkout for problems like this is this textbook "Advanced Mathematical Methods for Scientists and Engineers " by Steven Orszag and some other author.
"and that's a pretty good place to"
*gets nervous af*
"be then"
phew. he has us trained like pavlovian dogs to perk up and respond to a key phrase
So, we got t'' / t' = t. I'm going to say that there exists a solution of the form A x^N there. So, t' = AN x^(N-1) and t'' = AN(N-1) x^(N-2). Smashing all of that back into the equation, we get AN(N-1) x^(N-2) / AN x^(N-1) = A x^N. There is some obvious cancellation, which leaves us with (N-1)x^-1 = A x^N. So, N = -1, and A = N - 1 = -1 - 1 = -2. Which means that t = -2/x is a solution. Plugging that in, we see that it works. But all solutions are of the form t = A tan ((Ax + B)/2). So, there exists some values A and B such that A tan ((Ax + B)/2) = -2/x?
I feel like I've done something obviously wrong, but I can't see it.
Yeah, negative constant and zero constant give pretty different stuff.
I followed up until u got involved. I didn't quit watching though, and was able to ... follow isn't quite right, but after substituting back for u things made enough sense that I wasn't going to argue.
Thank you.
If I re-learn differential equations, what are some domains where they are applied?
further simplification will result in exp( 2Atan(Ax + B) )
Why the constant must be positive? If it's equal to 0 you get t=2/(k-x), wich is a solution of t*t'=t''
Just *GREAT* .
Thank you so much Professor
Thoroughly enjoyed this. Thank you!
please dont use the final solution
I ended up with a different solution. I started by getting rid of lny.
Z=lny. The the differential equation becomes Z''/Z' = Z. So just like yours.
I then did a trial solution where Z = C/X. it solves the equation and Z = -2/x.
So y = e^(-2/x).
more generally y=e^(-2/(x+B)) is a solution if you assume A is 0 vs A>0
@@ThAlEdison you mean B = 0 and B>0?
@@ChargeOfGlory No, I mean the A like in the video.
You get to a point where t''=tt' the constant picked when you integrate this changes the form of the final answer, if you pick the constant (A) to be 0, you get t'=t^2/2, and integrating gets you to the y=e^(-2/(x+B)) equation. Your particular solution is if both A and B are 0. As opposed to the solution in the video, which takes A>0.
@@ThAlEdison oh now I know what you mean. Thanks.
I tried to rewrite LHS as one fraction, which yields something that looks like quotient rule, the derivative of (y'/y), but it didn't go so good
It's all good, after that there will be a riccati equation for z=ln(y) as following:
z'=z^2/2+c1,
where c1 - constant of integration.
This equation you can solve using a quadrature really quick.
@@Lamiranta Ahh I see it, very nice
Can someone else confirm that the "positive constant +A² " can actually be negative, which by integration would give a -artanh or -arcoth (both works), and finally in the expression of y, a -tanh or -coth instead of tan. (with an extra negative sign)
i think he’s implicitly assuming the original ODE is for a real valued function only, and due to the log y term the image of y under x is restricted to the positive real numbers at most and as the derivative of log is always positive that’s why he assumed the constant is positive so that there are no contradictions there
Great video. Thank you
Our favourite problem suggester. Not the integral suggester anymore
Did someone got also y=exp(int(W(ax+b)dx))?
int is the indefinite integral and W is the lambert function.
How do we know u is a function of t instead of a relation? Also, how do we deal with the fact that ln is multivalued for negative inputs?
For each value of t there is at most one possible value of t' assuming we're dealing with nice functions. For the second question, you can see at the end that if you stay with real values of x then you never take a log of a negative input.
@@frankjohnson123 But what about for example t=x^2? Then, for t=9, t' could be 6 or -6.
Great work! I love it!
Today I learned that a pretty good place to be in is not a good place to stop.
Nice solution, I like it. But clearly y = exp(-2/x) is also a solution. Is this one somehow included in the solution derived by Michael by choosing proper constants?
I don’t think so. It seems like your solution can be derived if we solve for a case when the constant at 8:35 is equal to 0. Check out a Natrium OH’s comment in this comment section. According to it, we get different solutions when solving for cases when the constant is 0 or negative.
See Владимир Павлов comment above. He gave the three cases whether the constant is positive, negative or zero.
"If you're the smartest in the room, you're in the wrong room." Well, I think I've found my room - I was able to follow along only with much rewinding. No pain no gain, right? Good stuff, keep 'em coming!
When my nephews and nieces are old enough, I'll tutor algebra and trig all day, but > calc 1 might be a good place for me to stop. 😀
It does look nice! How neat!!
A^2/2 is always non negative. Why can't we use negative constant?
at 8:40 how can we assume the constant of integration is nonnegative?
Thanks!!!
Michael ,do more sum's resolution.
Isn't y = exp (-2/x) one solution to this ?
There is not x, so we can replace dy(x)/dx =p(y(x))...
Why can we assume the constant is non negative
What is the point of keeping the 1/2 ? The 1/2 can be absorbed into the A and B, no?
t = A tan((Ax+B)/2)
You could do that but watch out for the A outside of the tan function.
All those 七 variables are going to be confusing for chinese/japanese/korean/vietnamese veiwers... They're going to wonder why u depends on 7 and what does the derivative of 7 have to do with anything. XD
Started writing a post on measuring non-linearity, but got stuck on defining what operations keep the degree of non-linearity of an expression intact 😟😟
I think that would be a good place to start another video of yours.
It really depends on what you mean by non-linearity, but if it involves polynomial approximations of functions I'm going to assume that only linear combination of functions with real (or complex) coefficients will preserve the degree of nonlinearity.
@@hydraslair4723 Ok, sure. That includes de-rationalizing, i.e. multiplying an expression by something's denominator, right?
@@perappelgren948 it doesn't. By linear combination I mean that if you have functions f, g, h, a linear combination of them is af + bg+ ch where a, b, c are real or complex numbers.
Derationalising seems to me that it would change the degree of linearity of a function: consider (x²-9)/(x-3). This is a linear function defined everywhere except at 3. If you rationalise it, you get x²-9 which is a parabola (decidedly not a linear function).
@@hydraslair4723 Ah, you are right! Interesting. So polynomization is not the path to exhibiting true non-linearity. Got to examine this further. Mathematicians must have pondered upon degrees of non-linearity before.
Who suggested this problem?
are there any field of physics where this differential equation must be solved?
Hi Dr.!
Why is he making the easy things difficult? I would immediately substitute u=lny to get rid of the logarithm and the result is u''=uu' etc.
y"/y'-y'/y=lny
(y"y'-y'y')/y^2=(lny/y)*y'
d(y'/y)=(lny/y)*y'
y'/y=(lny)^2/2
The rest is left as an exercise for the reader(also I'm lazy)
nice
Nice vid!!! Am I the only person that just assumed the exponential answer to solve the problem?
fantastic
The only way to solve this sort of problem is with an exorcist. eerrggggh!!
Can anybody please clarify this for me: while doing calculus, especially while finding indefinite integrals, or solving DEs, we don't really care about any constraints on the functions we are dealing with. For example when we (while solving an integral) are doing a trig substitution cos(t)^2 = x, we don't care that 0≤cos(t)^2≤1, but x is not anyhow constrained.
My question is how can we assume something like cos(t)^2=x, when for most of the values of x there are no values of t to make it work?
I think this is more of a formality of its kind when solving this kind of equations, of course, if we want to plot a solution surface/planar graph, then, due to the limitations of the argument, the function will be limited. Our task is simply to find a function that would generally satisfy the equation.
Very nice😀
You're missing other solutions, though.
The only reason this works as a trick
is because students are taught the logarithm as a shorthand function.
With the understanding that a logarithm is just a description of arithmetic
the initial expression is tautological.
That s beautiful Michael ;)
35seconds late
asnwer= y isit hmm tobe honest isit 🤐