if you don't make the replacement B=exp(b), you can directly write the final result as y=cosh(ax+b)

@elengul

Жыл бұрын

At the point where you have y' = a*sqrt(y^2 - 1), one can also make the substitution y = cosh(u(x)), and that simplifies the equation to u' = a, so u = ax + b, so y = cosh(ax + b)

@paulkohl9267

Жыл бұрын

I thought the same exact thing about cosh (ax + b).

@minamagdy4126

Жыл бұрын

Technically ±, since the logarithm really should've left behind an absolute value.

@leif1075

Жыл бұрын

@@elengul But Why On Earth would.amyone EVER rhinknofncldine there is just no reason to at all..or sine..so why even bring it up?? Surely there is a way to solve with just algebra,

@leif1075

Жыл бұрын

Whybdidnt he start by taking the swuare rootnto get rid of rid of swuare in y prime..surely that's how most ppl wpuld.start? And then maybe taking the derivative of the whole thing to look for patterns?

There is indeed something sketchy, but it happens at 2:48 : When integrating y y' / (y^2 - 1) you will get 1/2 ln |y^2 - 1| with absolute value inside the logarithm. This is needed because y^2 - 1 can be negative, which happens for your solution y = sin x.

@pageboysam

Жыл бұрын

Would this also be the case when integrating y” / y’ ? Should there be 4 branches?

@jgray2718

Жыл бұрын

He also dropped the absolute values on the integration of sec(x).

@NathanSimonGottemer

Жыл бұрын

Does this mean y=sin(x) is extraneous then? I did imagine non-linear ODEs would be a bit unpredictable though

On an intuitive level: though the equation is second order, it is nonlinear, therefore we should expect 'weirdness' in the solutions. More analytically, the solutions might be expected to be unique and linearly independent, but only when the function (or the equation?) is continuously defined. Since we y= -1,0,1 are 'discontinuity ' values in the equation, each of our linearly independent solutions may be applied to one of the regions defined by these crossover values. hyperbolic cosine's image is y>=1 so it should only hold there, and I'm guessing cosine and sine take up -1

@user-en5vj6vr2u

Жыл бұрын

Good point. But generally cos and sin are interchangeable as are cosh and sinh, so i think |y|1 could be either cosh or sinh depending on initial conditions

@pageboysam

Жыл бұрын

How is hyperbolic sine achievable? The coefficients of the exponential functions need to be opposite parity, but B and 1/B are the same parity, when restricted to real numbers. Normal sine is achievable because 1/i = -i, but that’s not the case for hyperbolic sine when B is restricted to real numbers.

@OdedSpectralDrori

Жыл бұрын

@@user-en5vj6vr2u Sorry in advance if this is a bit tedious I see your point... we need to be a bit more precise in terminology for some of my claims to make sense then :) cos and sine are only interchangable, but only because we usually neglect to complete the sentence "cosine and sine are linearly independent and therefore are two viable separate solution to an ODE" . What we should mention and often do not, is this only holds for real constants and with no difference in phase. i.e. cos(x) and sin(x) are linearly independent over the real numbers -> there are no real constants A,B such that Acos(x) + Bsin(x) = 0, for a real x this is important because 1. it explicitly states we only need two constants (if we wanted to include a phase difference 'phi' in this, we will need another boundary/intiial value) 2. it seemingly breaks if A,B are allowed to be complex. but even in the case where we would allow the constants to be complex (say z1 and z2), and the sin/cos would become linearly dependent, and therefore can be combined into a single function z1Cos(x)+z2Cos(x) = z3Sin(x), we are still left with determining two free real unknowns, Re(z3) and Im(z3) so this really is all just a big soup of notations and terminology so cos and sin are interchangable , but are not ;) for cosh and sinh, I'm not actually sure they are as interchangable as cos and sin, but that's very plausible provided the arguments are allowed to be complex. interesting point. finally (and probably the only important point I make), the discontinuity in 0 means that even if after all of this, we come to the conclusion that a single function, (for example) Asin(x+phi) is THE solution, it may be a diffent A,phi for -1

@OdedSpectralDrori

Жыл бұрын

@@pageboysam good shout! I've fiddled with this for a little while and I think you're right. even though we could let B,a be complex, so long as the eventual y we find is real valued, there are no such values that would give us a real valued constant times a real valued hyperbolic sine. We can probably also figure that by the fact that no real valued C,D such that y=C * Sinh(x)+D will work if we plug them into the original equation. I'm tempted to say this means that negative cosh probably holds for the negative end of the domain, and positive cosh holds for the positive. This is not a trivial observation as it might've been for a linear ODE where any multiplication by a constant still solves the equation, we have to actually check this. given y is a solution, let u=-y . then the original equation transforms into -u'' / (u')^2 = -u/(u^2-1) which simplifies into u'' / (u')^2 = u/(u^2-1) which means if y is a solution, then -y is also a solution, and the choice between the two may actually just be a matter of boundary values

@khoozu7802

Жыл бұрын

How do u know y=0 is discontinuous? If imaginary numbers are allowed, we can let x=0, B=i to get y=0

if you put B = exp(b) you can write final solution as y = cosh(a*x+b). More beauty. You have multitude of linearly independent functions because initial equation is not linear. Still you have 2-parametric set of solutions. If it were linear one, only two linearly independent solutions are possible. Upd: Misprints.

There is no contradiction, a second order DE yields two constants of integration in a solution, with no further guarantees. A second order LINEAR DE guarantee having two LINEARLY independent solutions (that can be spanned by taking a linear combination of them, which also uses two constants to describe). So the complaint here is kinda backwards: with linear DE we would find those two solutions and put them together as a general solution = A * first function + B * second function. With nonlinear DE there's no such thing, but the general solution is still spanned by two independent constants, just mixed in a way that's also not linear, so general solution = some function (x, A, B), in this case, cosh(Ax + B).

@md2perpe

Жыл бұрын

Correct. The solution space of an n:th order differential equation is an n-dimensional complex manifold. Only for linear differential equations that manifold is a linear space.

Around 2:35, if you rewrite the RHS as d[(1/2)*ln(1-y^2)]/dx, you will eventually get a solution y=sin(ax+b), which also covers cos(x).

I see "y²-1", I try the cosh function Note : I believe that there are no problems in the end, we get a function with 2 degrees of freedome (the function being cosh(ax+b)) and this function is the same as the cosine or sine if you choose the right a and b

Hi Michael. I’m a big fan of your channel. Thanks for scratching my math itch every day ! I had an idea that I think could improve your channel : I think it might be cool if you would show the graph of the solutions in order to have a visual representation of it. Thanks again for the awesome videos.

the solution has to be separated for the range abs(y)>1 and abs(y)

13:34

@DepozidoX

Жыл бұрын

I thought I came early, but you are already here :Oo Marvelous dedication

I loved this problem. Thank you, professor.

Trig functions and their hyperbolic equivalents are not independent when you allow complex parameters?

@j.d.kurtzman7333

Жыл бұрын

This is what I was thinking as well

@bjornfeuerbacher5514

Жыл бұрын

They are not independent in general, right. But they are still _linearly_ independent. And it's the latter which Michael was specifically talking about.

@QuantumHistorian

Жыл бұрын

@@bjornfeuerbacher5514 Right. But here you have the solutions cosh(a_1 x) and cos(a_2 x), where the a's are free parameters - until fixed by boundary conditions. But whatever solution you pick from the cos(a_2 x) family, you can map that as a cosh(a_1 x) solution by choosing a_1 = i a_2. So the two families of solutions are equivalent, they're just written in (marginally different) functional forms. The same thing with sin and sinh. In fact, using @Daniel Milyutin's solution y = cosh(a*x+b), it's easy to see that all of those solutions can be written in a single functional form with two free parameters.

Part of me wants to say that everything is sketchy as soon as you rewrote y''/y' = yy'/(y²-1) as d/dx(ln y') = d/dx(ln(y²-1)) without absolute values within the logarithms, thus restricting the validity of your solution to the case where y' > 0 and y² > 1. On the other hand, I want to say the solution is not at all sketchy because sinh and cosh are not independent from sin and cos once i is introduced. (I.e. sinh(x) = i sin(h) and cosh(x) = cos (ix).) This has to be somehow related to the fact that y'' = ay + b might, depending on the exact values of a and b, oscillate, grow, or shrink.

I believe the reason there are 3 solutions is because there are only actually 2, sin and cos can be linearly transformed between by applying a phaseshift to x, which will have no affect on the solution process, as D(x) = D(x+a)

@bjornfeuerbacher5514

Жыл бұрын

That sin and cos can be linearly transformed into each other by a phase shift doesn't change the fact that these are 2 linear independent solutions.

@vladimir10

Жыл бұрын

sin(x) and cos(x) are linearly independent functions in the sense that none of their linear combinations gives zero except for the trivial combination. Phaseshift or translation is not a linear operator generally.

Great video. Keep up the good work!

I think part of the sketchiness might arise when the anti-derivative is a logarithm which technically/rigorously requires absolute values. Furthermore, whenever performing trig substitutions, one has to be careful of the regions of integration. Therefore, I am guessing (haven't verified) that the boundary conditions may determine which solutions apply. So, there is probably am uber-general solution that includes specific regions for x and y.

I have a very picky correction… technically it’s natural log of absolute value not just natural log

Interesting & easy DE ! After first step where SQR(ABS(y^2-1)) appears, one have to split : for y>1 substitution is y=cosh and for y

exp(a) could be absorbed into a constant too, giving solution in form y=(Ba^x+1/B*a^(-x))/2. But I guess the exponential form is more instructive to see the form of familiar expressions for sin, cos, etc...

Great video. Thank you

this can be re-written as cosh(ax+b). So i guess the two-ness of the differential equation is that all solution can be written as cosh( b(a(x)), where b is a translation and a is scaling function. In that sense of the two solutions being multiplied and added, our two solutions are actually the generators of scaling and translations, and they are being raised to a power and composed (then composed by cosh(x))

Need to revise the old hyperbolic functions there Michael. Catenary curves. Well known in Calculus of Variations

Consider a scaled solution y = f(ax+b) . The differential equation in terms of f becomes: a^2 f''(ax + b) / a^2 f'(ax + b)^2 = f(ax + b) / f(ax + b)^2 - 1 As you can see, if f(x) is a solution, so is f(ax+b) for any a, b. So for example, f(x) = (1/2)(exp(ax+b) + exp(-ax-b)) = cosh(ax+b) a = i, b = 0 cos a = i, b = i pi/2 sin etc.

You have discarded the possibility of √1-y^2 in beginning as given function can be √1-y^2 also if negative sign is taken common initially otherwise we would have obtained sin cosine also

sin x and cos x are the same function if you ignore the actual x value and only look at y and its derivatives, which is what the original was asking.

It's nice to see a Nonlinear Differential Equation spanning the trig and hyperbolic functions as solution. At first glance it seems sketchy how we get the sines, cosines, and hyperbolic cosines as solution. But here each and every time when you get a function you are constraining a and B, which means you are finding the solution for a particular condition. For example even in linear Differential Equations, say y" = 0, which spans y = a + bx as the complete solution set and the 2 functions are 1 and x here. When you need your function to satisfy certain condition say must pass through (0,0) and (1,1) you solve accordingly and get (a,b) = (1,1) A similar condition is being imposed here as you are explicitly substituting some values for a and B here. I don't think every 2nd order Differentia equation must have 2 functions as solution, it must be framed as we get 2 constants of integration in the process of solving and they span the complete set of solution of the Differentia equation and substituting values for these constants narrows down this set. Linear combinations of this solution need not be a solution of the differential equation incase of Non linear Differential equation,

The equation belongs to the class of ordinary differential equations that do not contain x explicitly. Lowering the order by substituting dy(x)/dx = p(y(x)). Then d2y(x)/dx2 = p*dp/dy and the original equation takes the form dp/p =y*dy/ (y^2 -1) . Answer: If |y(x)|>1, then y(x)= ± cosh(c1*x+c2), where c1≠0 , c2 is any. If |y(x)|

@Vladimir_Pavlov

Жыл бұрын

More details. dp/p =y*dy/ (y^2 -1) => ln |p|= (1/2)*ln|y^2 -1|+ln|c1| (it is convenient to write the integration constant as follows, ln|c1| ,c1≠0 ) => ln |p| = ln [|c1| *sqrt|y^2 -1|] => p≡dy/dx = c1*sqrt(|y^2 -1|) => dy/sqrt(|y^2 -1| )=c1*dx. If |y(x)|>1, then ln|y+sqrt(y^2 -1)| =c1*x +c2=> |y+sqrt(y^2 -1)| = exp(c1*x +c2). From where we get y(x)=cosh(c1*x+c2) if y(x)>1 and y(x)= - cosh(c1*x+c2) if y(x)

As a nonlinear second order ode it’s like a system of two nonlinear algebraic equations. You can get multiple solutions (x,y) because the equations are nonlinear-for example, quadratic.

It is an interesting result, since the presence of a parameter in the exponent means the solution space changes depending on the parameter a (the solution space is spanned by e^ax and e^-ax). I assume this is a result of the problem being non-linear, and thus there are more than 2 linearly independent solutions? It seems there are infinitely many.

For more generality you should include the absolute values in the logarithms

Here, you assumed that |y|>1 and the solution in this case can be written as y=cosh(C_1x+C_2), which the same as y=(Be^{ax}+ e^{-ax}/B)/2. Thus, we cannot say that y=cos(x) and y=sin(x) are solutions because they are less than 1 in magnitude. However, if |y|

The weird solutions probably come from domain weirdness. If you start with certain initial conditions then the sin and cos will give the right function, and with others sinh and cosh are the right ones. If I had to guess it's to do with whether y is initially between -1 and 1 or not.

Note that y ± √(y^2-1) are reciprocal of each other. So, 2y = Be^(ax) + 1/Be^(-ax)

To do ∫ dy/sqrt(y^2-1), just substitute y=cosh(u) if y>1, to get ∫ sinh(u)/sinh(u) du = u+C, we get y=cosh(ax+C); for y

i tried to solve this DE and got a general solution in the form of sin(ax+b) ,yours was more general actually. it's pretty easy to understand if sinx was a solution then cosx also could a solution via trigonometric relationships, but using euler's formula you'll relationships between regular and hyperbolic trig functions, as an example cos(ix)=coshx and sin(ix)=isinhx

Who said the solutions need to be linearly independent? As others started, the complete family of solutions is y = cosh(ax+b) with free coefficients a,b. Any real value of a would give you a different linearly independent solution, and they don't even have to be real... cosx and ±i*sinx can be achieved with a=±i, b = 0 or iπ/2

Michael, I really like your videos. They are great fun. A substitution with y=cosh(t) would have been better and leads to the solution y=cosh(ax+b) of the ode, Michael Baum

In the space of complex functions, {cos, sin, cosh, sinh} aren't four independant solutions, only two, since cosine and hyperbolic cosine are essentially the same functions, rescaled. cos(z) = cosh(i z), and similarly sin(z) = -i sinh(i z). That being said, those are simply particular solutions, not a base for a linear combination , since the differential equation is not linear. As a side note, the constant "a" cannot be zero, which would yield to a constant y, and an indefinite value in the given equation.

I would have cancelled out this d/dx and ln straight away without thinking (and yes, missed the constant). But the trick turning it into derivatives just "by sight" - yes either luck or experience.

You have sin, cos, sinh, and cosh as independent solutions. But cos(i*x) = cosh(x) and i*sin(i*x) = sinh (x) That means you have a specific set of two independent solutions for a specific subset of numbers. You have another set of two independent solutions in another subset of the numbers.

I believe because the equation degree is higher than one (is two), then uniqueness and existence theorem doesn't longer apply.

Why y=sec(theta) when y= Cosh(u) gives the solution directly ? a x+b=argcosh[+/- y] and y=+/- Cosh[ax+b]

My guess, its b/c of the squaring and un-squaring vis square roots along the way. But, at first glance my head went to Cosh(x) bc i recognized that (y’)^2 and (y^2-1)^(1/2) are the same thing for cosh(x) and similarly y’’ and y are the same. I think the trig functions fail bc with the squaring of the derivative you lose the sign change that I think is required to make the (y^2-1) identity work out.

I'm a little late to the game, but it comes from the step when you write ln(y^2 - 1).. it's actually ln|y^2 - 1|. So if |y| > 1 you have y'/sqrt(y^2 - 1) = C, leading to arccosh y = Cx + D or y = cosh(Cx + D), which is equivalent to your solution. When |y| < 1 you have y'/sqrt(1 - y^2) = C, leading to arcsin y = Cx + D or y = sin(Cx + D), leading to the alternate solutions. Through any (x_0,y_0) with y not equal to -1, 0 or 1 there is a unique solution as always.

Интеграл в решении - табличный, в России его называют «длинный логарифм». Можно было не делать постановку, а сразу написать ответ :) ~~~ The integral in the solution is tabular, in Russia it is called the "long logarithm". It was possible not to make a statement, but to write an answer right away :)

We can substitute x = 0 in the general solution, yielding B = y₀ ± sqrt(y₀² - 1). We can also derive the general solution, and substitute x = 0, which lets us find a = 2By₀' / ( B² - 1 ). Therefore, for -1 1, B is real, which means that a is real, which means that y has a purely exponential behaviour.

Very nice equation 💯

The same "sketchiness" (if you think it is sketchy...) arises from something a lot simpler like (y")² = y² Same types of families of sin(h)/cos(h) solutions.

You could have used y=cosh(u), which I think is a nicer thing to do.

+/-isinh(ax+b) also works

6:18 "abusing notation" 😂😂😂😂

That' awesome solution

I think it would be better if you could explain how the thumbnail is related to the equation. I'm also intrigued by the thumbnail in the AMGM episode.

If one makes the intermediate substitution z=Be^(ax), the solution takes the form y=(z+z^-1)/2, which contains a pleasing symmetry that generalizes both cos and cosh. The fact that our subject is a second-order differential equation only means that there are two degrees of freedom in the solution parameters, but these parameters need not (although they may) correspond to vectors in a vector space. The fact that it is a nonlinear differential equation means that there is no expectation that the solution set should form *any* vector space, or, if it does, that the dimension of that vector space should be 2. In "playing things fast and loose" have also neglected to address the singularities at y^2 = 1 and y' = 0, as well as the problematic cases where the logarithmic differentiations operate on values that are complex or non-positive real numbers. We have also failed to examine in detail the various cases where square roots and absolute values were taken either explicitly or implicitly.

I *guess* the reason we have 3 solutions is that the equation satisfies the conditions for the Peano existence theorem, but not the conditions for the existence and uniqueness theorem (a.k.a. the Picard-Lindelöf theorem)?

If |y|>1, then solution is y=+or-cosh(ax+b), where “a” and “b” in R and “a” is not 0. If |y|

can someone help me with my homework; we were asked to find a general term for the sequence defined by U(n+1)=U(n)-U(n)^2 with U(0)=a while a is a real non-zero number

I think that's because that exp(x)=cosx+isinx, that tells us the imaginary numbers connect with the sin, cos, etc.and e^x. The problem just because you added the imaginary numbers at last, but at first we may just thought of real numbers.

The solution we're left with can be written as cosh(ax+b) and I did some working to find every value of a & b in the complex plane that gave a real function in x, and here they are: 1. If a & b are real we have cosh(ax+b) 2. If a is real and b=v+i*pi for real v, we have -cosh(ax+v) (v-i*pi give the same result and v+2i*pi gives solution 1.) 3. If a=ui and b=vi for real u,v, we have cos(ux+v) which is the sin and cos functions for (u,v)=(1,-pi/2) & (1,0) respectively, but one can see that those functions are horizontal translations of each other, as are their negatives. This is because cosh(c+di)=cosh(c)cos(d)+i*sinh(c)sin(d) and so for real c,d, sinh(c)=0 or sin(d)=0 in order for cosh(c+di) to be real, so c=re(ax+b)=0 or d=im(ax+b)=a multiple of pi So the 3 solutions are cosh(ax+b), cos(ax+b), -cosh(ax+b), and notice that they have (almost) separate domains, and you must be careful when the function approaches 1 or -1, or its derivative approaches 0 (which happens at y=1,-1).

Cool video. I would say it's because on some level sinh, sine and cosine are the same. Though that doesn't explain why cosh isn't a solution either.

@landsgevaer

8 ай бұрын

But cosh is a solution. And so is sinh if you take B imaginary. They're all different windows on the same thing, in a way.

But can B and a really complex values? We derived those numbers by integrating a constant function, does it make sense for the independent constants to be complex?

@landsgevaer

8 ай бұрын

Why would it not? The complex number are designed to generalize the reals while keeping the same "rules" as much as possible.

Could it be that they are linearly independent over the reals but not over the complex plane? Just an intuition

The "fact" that a second order equation cannot have 3 linearly independent solutions is only true for linear equations, where the solutions form a vector space of the same dimension as its order (here, 2). As the equation is not linear, it is not restricted in this sense. Indeed, away from discontinuities, we have that the solution space for this equation constitutes a 2-dimensional manifold. The fact that a 2-dimensional manifold can have 3 "linearly independent points" (i.e. points whose position vectors are linearly independent) is evident by considering the unit sphere in R^3, which contains points identified with each of the 3 standard basis vectors.

I just can't stop rubbing my head about possible constants signs

The general solution to this equation is y=cosh(c1*x+c2), where c1 and c2 are constants. Nothing unusual here.

So the term ln y' - ln sqrt(y^2-1) is a conserved quantity. I wonder if this form has a corresponding physical mechanism.

@williamperez-hernandez3968

Жыл бұрын

A hanging chain (both ends tied at equal height) has the catenary equation y(x) = a cosh(x/a) where the minimum point of the chain is taken as (0,a). So taking a=1 gives a physical interpretation for the result of this vid.

My thought/guess on sin(x), cos(x) sketchiness: since we have an autonomous 2nd order diff. equation and sin(x) and cos(x) are actually the same function just phase shifted in respect to x?

@alxjones

Жыл бұрын

That's a really nice insight! It's not exactly the reason here, but I really like where your head is at.

y=cosh（a x+b）

Damn, those forearms are huge

Obtaining 3 solutions for 2nd order non-linear DE happens. Consider the following 2nd order nonlinear DE: y'' = yy' The reader can verify that the following 3 functions satisfy the DE: y1(x) = c, where c is a constant, y2(x) = -2/x, y3(x) = (sqrt2)tan(x/(sqrt2)). One can compute W(y1, y2, y3), the Wronskian & verify that W is nonzero for c being nonzero. Three independent solution functions for a 2nd order DE.

Equation is not continuously defined? May be that is why.

Hi Dr.!

Maybe your 3rd solutions came from the square term in the first derivative? Or your cosine and sine are in fact the same solution expressed differently in Complex?

@TheLethalDomain

Жыл бұрын

That plays a part because the square term in the first derivative is what makes this non-linear to begin with. If the solution contained a linear set of 2 solutions, then the original equation would have to be linear.

Is there some problem with y"/y'² = y/(y²-1) y dy'/dy /y'² = y/(y²-1) dy'/y'² = dy/(y²-1) 1/y' = arcoth y + C dx = (arcoth y +C) dy x + D = y arcothy - ½ln(y²-1) + Cy {|y| >1} or x + D = y artanhy - ½ln(1-y²) +Cy {|y| < 1}

Multiple solutions are allowed for nonlinear equations.

Amazing .

my proposed explanation is that it makes sense if you don't think about it lol

To be honest, my thought process when seeing the problem was something like this: "You almost have a function being equal to some ratio of its derivatives, except for the (y^2)-1 part, that just screams sin(x) and cos(x). Done. Why is this video almost 15 min?" If you couldn't tell, I'm a physicist. No clue how you got cosh(x), you probably used values of constants that are not allowed because they appeared in logs and under squares, which always fucks with things. Just my intuition, don't have a proper argument.

In elementary algebra, can't you introduce extraneous solutions by squaring both sides as we did? I'm too lazy to plug in all 3 to see which ones work but I suspect one of them won't. If they do all work then it's probably some trig or hyperbolic function relationship secretly consolidating two of the solutions somehow

@bjornfeuerbacher5514

Жыл бұрын

Both your guesses are incorrect. The right answer was given already by several people: The differential equation is not linear, that's why it can have more than two independent solutions.

It could be that the domain restrictions of y that we previously neglected will eliminate 2 of those 3 solutions. If we say that y' cannot equal 0 and y cannot equal 1 or -1, then some simple calculations will show that y = cosh x and y = cos x in fact cannot be solutions as B cannot equal 1 from those restrictions.

A different start of the calculation would be to note that y'' /(y')^2 = - (1/y ')'. Then the next steps are also not difficult.

@renesperb

Жыл бұрын

I just realized that this does not get you far . Michael Penn's procedure is better.

100% of this weirdness is because we have different names for f(x) and f(ix), as evidenced by let B = 1. y = cosh(ax). for a=i, as you said, this is cos(x). All other connections follow similarly

@landsgevaer

8 ай бұрын

Yep, I don't see what is "sketchy" about it either (beyond it being a nonlinear and 2nd-order diff.eq.). Sin/cos satisfy y"=-y and sinh/cosh satisfy y"=+y, so this is equally sketchy as the solutions to (y")²=y². (Not that these have the same solutions, but a similar "sketchiness" arises.)

It seems intuitive that your solution set would be given by two constants because it's second-order, but they wouldn't be linear combinations of two functions because it's non-linear.

SInce y should never equal 0 and 1, I think we cannot retain sin(x) and cos(x) as possible solutions

I'm so happy i solved this easily

Because you're allowing complex numbers to be used as constants, cos(x) & cosh(x) are no longer linearly independent, because cosh(ix) = cos(x) and cosh(x) = cos(ix).

@synaestheziac

Жыл бұрын

Wait, are you sure? Wouldn’t they only be dependent if you could pull the i out of the cos? Setting complex numbers aside, aren’t cosx and cos2x independent (for example)?

@buhlaigah

Жыл бұрын

@@synaestheziac I believe you're right. Thanks for correcting me

6:24 is that abusing notation or something that actually can be justified rigourously? The only real analysis I've done has been with dx's

@Noam_.Menashe

Жыл бұрын

Integrate both sides with respect to X and then in the left substitute u=y. It's really just the chain rule.

@juliang8676

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Its an abuse of notation that works, you can be more rigours but I can't remember the details

(y')^2 = (-y')^2, so d/dx ln(-y') = d/dx ln(sqrt(1-y^2)) also.

This argument about the number of independent solutions only works in the case when these solutions form a vector space.

Lineary independent functions? Sure, but the space of solutions isn't a vector space, so it doesn't matter.

others have said it, but here's a summary For all solutions, a=/=0 y=cosh(ax+b) covers y>1, x=/=-b/a y=cos(ax+b) and y=sin(ax+b) are not linearly independent because of the +b, they cover 0

@ThAlEdison

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Also, those are the functions that I use when solving algebraic questions where you're given x+1/x=a find x^4+1/x^4 if |a|

unfortunately, earlier in history we have had difficulty noticing that there is something sketchy about the final solution.

Strange solutions? More like “Superb explanations and investigations!”

very nice

The LHS is the derivative of -(y')^-1. Why didn't you start with that?

I don't see what is so sketchy here. The initial condition y(0)=1,y'(0)=0 clearly is a point of non-uniqueness as you would guess immediately by seeing both (y')^2 and y^2-1 in the denominators.

Cross multiple and integrate? 👉👈

The “sketchiness” here is actually your misinterpretation of the solution. Allow me to explain. The solution you got is dependent on two integration constants, and then you picked a few random ones and found some specific solutions and then assumed that the solution should be able to be written as a linear combination of these specific solutions. This is the case for linear differential equations, but is NOT the case here. Allow me to demonstrate this. Choose the specific solution y=sin(x) and plug that into the original differential equation and you should find that it proves correct. However, try plugging in y=c*sin(x) and you will find that it does not satisfy the differential equation. This shows that the solution cannot be expressed in any such way. Also a fun solution to this is that if you let y=sin(c*x) you will find that this solves the differential equation for all c. Same with the cos() and cosh() solutions. This is all because of what expressions in terms of sin(), cos(), and cosh() can be expressed by choosing values of A and B in your solution. So yeah, because this differential equation is nonlinear, the solution is nonlinear in terms of its integration constants so you can’t express it as y = c1*y1 + c2*y2.

Hi, As the equation is not linear, you can apply the theorem : if y_1 and y_2 are solution, then any linear combination a y_1 + b y_2 is solution as well.

It’s a NON linear differential equations, so it might have more than 2 solutions