A nice suggested differential equation 

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.

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.

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.

That's actually the first differential equation I managed to solve by myself and I'm quite proud of it.

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

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

with these i always wish you went and plugged the answer in to check, i like watching that when people do it

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)).

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.

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.

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.

Thanks for opening up by explaining what the jargon means.

Thoroughly enjoyed this. Thank you!

Just *GREAT* . Thank you so much Professor

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.

Fun fact: This differential equation belongs to a miniscule class of differential equations that can be analytically solved 💀

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

Great work! I love it!

"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

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?

It seems one more solution is here. When the first constant A is below zero then integral is arcth and logarithms will disappear.

11:38

Great video. Thank you

It does look nice! How neat!!

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.

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)

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.

further simplification will result in exp( 2Atan(Ax + B) )

Yeah, negative constant and zero constant give pretty different stuff.

Thanks!!!

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''

please dont use the final solution

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.

Michael ,do more sum's resolution.

Did someone got also y=exp(int(W(ax+b)dx))? int is the indefinite integral and W is the lambert function.

A^2/2 is always non negative. Why can't we use negative constant?

Today I learned that a pretty good place to be in is not a good place to stop.

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)

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?

at 8:40 how can we assume the constant of integration is nonnegative?

are there any field of physics where this differential equation must be solved?

Our favourite problem suggester. Not the integral suggester anymore

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

Why can we assume the constant is non negative

Isn't y = exp (-2/x) one solution to this ?

fantastic

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).

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?

Hi Dr.!

nice

Very nice😀

Who suggested this problem?

There is not x, so we can replace dy(x)/dx =p(y(x))...

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.

What is the point of keeping the 1/2 ? The 1/2 can be absorbed into the A and B, no?

"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. 😀

Nice vid!!! Am I the only person that just assumed the exponential answer to solve the problem?

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

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)

You're missing other solutions, though.

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.

That s beautiful Michael ;)

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?

35seconds late

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.

asnwer= y isit hmm tobe honest isit 🤐