Beyond Linear DEs
Given that y = f ( x ) , d x d y = x − 4 + y 2 , and f ( π 1 ) = − π , find f ( 3 π 4 ) .
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3 solutions
u, z, y are functions, while x and t are variables. Could you precise that in the formulas. Somehow I don't know how you account for that in the change of variable t = x − 1 . Is it necessary? Shouldn't z ( x ) become z ( t 1 ) ? Thanks for the clarification
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You are missing the point of a change of variable. If we write \x=t^{-1}) then we can write a function of x as a function of t - but it will be a different function. For example the function z = sin x − 1 , which would have a complicated DE with respect to x , would become z = sin t , and so would have a simpler DE with respect to t . The effect of first introducing the new function z = x 2 y , and then regarding z as a function of \t=x^{-1}), is to obtain a sufficiently simple DE for z with respect to t that it can be solved. We can then backtrack to find y as a function of x .
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ok thanks for the clarification. I haven't looked at DE for a loooong time :)
how do you know where to substitute what? did you just use hits and trials to get the correct substitution?
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Tricks for solving DEs are never that easy to explain. I tried the first substitution to make the RHS a separable function of x and z . The rest followed from experience with other DEs...
I found y = − x − 1 + i x − 2 to satify the dy/dx requirement by simple trial and error. No idea if that goes in the right direction, any thoughts?
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I landed there as well. Didn’t know what to do with it.. that solution satisfies the diff eqn but not the boundary condition.. concluded that uniqueness theorem has some asterisks around it.. 😁
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The uniqueness theorem states that there is a unique solution to a DE (satisfying certain general conditions) in a neighbourhood about a particular initial value pair. It does not say that the solution has to be globally unique.
It is a solution, but a limit of other solutions. If we considered general solutions of the form y = − x − 1 + x − 2 tan ( x − 1 + i c ) then your special solution comes from letting c → ∞ .
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Thank you. I don't see any way to get from my special solution to this general one, so I'd say I had a dead end.
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@Paul Blohn – Yes. Your solution is very much a special case, and won't easily yield the general solution.
Could you elaborate on this "trial and error" process? This does not resemble any form of a standard DE, and only an introductory course in DE is clearly not enough to solve the problem. I'm struggling to figure out where to even begin with something like this... (I got the answer correct by guessing.)
This DE is a particular case of Riccati equation, precisely the standard way to solve it, if you don't have Mark Hennings intuition, is to first have a particular solution y 1, then the change of varable y=y 1+1/z will lead a linear equation on z wich you have to know how to solve for example by the intagrating factor method. You just did the first step, and probaby the most dificuld one, without knowing this method!!!
For more details about it see: https://en.m.wikipedia.org/wiki/Riccati_equation#/search
The question here is if this complex particular solution will also bring you to the general solution. I tried to check it but I got to a strange expression and I was tired. Would be great if someone could elaborate better this idea.
I thought about Laplace or Fourier transform see wiki https://brilliant.org/wiki/laplace-transform/. But I was too lazy to try with them. Do you think they will work? I think with some avility, probably a previous change of variable, they will work.
Doesn't it look like a Riccati Equation ?
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Yes, it is, and that would give you another solution method.
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Yeah, I solved it that way ⌣ ¨
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@Sravanth C. – I used the Ricatti equation transformation but got stuck on y" + (1/x^4)y=0 -- embarrassing!
I went through a very "improper" way to find the solution, by elimination :
First note that the equation d x d y = x − 4 + y 2 implies that y ′ > 0 on R + . I deducted that f ( 3 π 4 ) > f ( π 1 ) . Therefore solutions 2 and 3 are excluded.
I couldn't find a proper deduction between 1 and 4, so used the numerical values provided (in the question and solutions) to see that with x ∈ [ π 1 ; 3 π 4 ] , we have roughly y ′ ∈ [ 3 0 ; 1 1 0 ] . I deducted that solution 4 : 1 6 3 π ( 3 π − 4 ) ∼ 3 . 2 was far more likely than solution 1 : 1 6 3 π ( 3 π + 4 ) ∼ 7 . 9 . I know this is very "wrong" solution, but i would like to know if womeone can find any reassoning in order to exclude solution 1 ? for a "quizz" solution rather than pure mathematic solution.
Same blood 😁
I just wrote a short Euler's Method program for approximating solutions and found the closest answer. :Prompt M :(1/pi)>X :-pi>Y :For (N,1,3M) :Y+1/(9Mpi)(x^-4+Y^2)>Y :X+1(9Mpi)>X :Disp Y :End
Simple and efective althought not very nice.
With the substitution y = x − 2 z the differential equation becomes x 2 d x d z − 2 x z = 1 + z 2 The change of variable t = x − 1 gives − d t d z − 2 t − 1 z = 1 + z 2 Then the substitution z = u − t − 1 gives − d t d u = 1 + u 2 and hence tan − 1 u = − t − c for some constant c , so that u z y = − tan ( t + c ) = − t − 1 − tan ( t + c ) = − x − tan ( x 1 + c ) = − x 1 − x 2 1 tan ( x 1 + c ) for some constant c . Since y ( π 1 ) = − π we deduce that tan ( π + c ) = 0 , so that c = n π for some integer n , and hence y = − x 1 − x 2 1 tan ( x 1 + n π ) = − x 1 − x 2 1 tan ( x 1 ) This solution is valid for the domain 3 π 2 < x < π 2 . Thus we deduce that y ( 3 π 4 ) = − 4 3 π − 1 6 9 π 2 tan ( 4 3 π ) = 1 6 9 π 2 − 4 3 π = 1 6 3 π ( 3 π − 4 )