Solving A Differential Equation
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i did it exactly like your second method. and for those saying that dy/dx is not a fraction: obviously that's technically true, BUT the reciprocal of the slope of y(x) happens to be the slope of the inverse function x(y) wherever the slope is not 0. so even though it is not a fraction, the math is correct
does this work for ln(x) and e^x?
@@HeckYeahRyan yes, if y=e^x, then dy/dx=y and dx/dy=1/y
@@demenion3521 isn't it more like the reciprocal function, not the inverse function?
@@harrabi15 the reciprocal dx/dy of dy/dx, the inverse x(y) of y(x)
It’s not a fraction, but you can just flip it because it acts like a fraction
There ought to be one free parameter. The solution of method 2 is
x = ke^y - y - 1 k constant
Which is the equation Sybermath got at 5:41 using his first method.
People love saying that dy/dx isn’t a fraction, but if you think about what it literally is, it's the limit of a ratio of a change in y over a change in x. That is a fraction. If you prefer, you can think of it as a proportionality factor or a ratio or a rate of change, but those are all different ways of thinking about a fraction. Where you need to be careful is that dy and dx on their own aren't useful, they would both =0, which you can't do with an ordinary fraction.
Ok, before watching, I'll try it myself.
First instinct was to try and separate variables, but clearly that isn't possible. Then, I realized you can flip the fractions and just try to work out x as a function of y, with the simple linear first-order inhomogeneous x' - x = y, with solution x = Ae^y - (y+1)
No idea how to invert that to get a function of y, though, nor if there even is a way to express such a function in terms of elementary functions.
Alright, time to watch.
Oh *wow*, I almost actually went ahead and did the first method but quickly shy-ed away from it not expecting it to go anywhere, but damn, that was cool, and effective. And, once again, I've been foiled by the pesky Lambert W function - I never seem to know how to make use of it.
Also, you're a brilliant explainer, and have earned a sub.
In my opinion substiitution u = x+y solves the problem
but if we want linear equation we can flip both sides
same
dx/dy = x + y - a linear equation
dx/dy = x -> x=C e^y
dx/dy = x + y, x=C(y) e^y, C' e^y + C e^y = C e^y + y, C' = y e^(-y), C(y) = -y e^(-y) - e^(-y) + D, we can take D = 0.
x = C e^y - y - 1, C is a real number.
do you mean dy/dx
10:11 i think you meant (D-1)x
Even if it's not a fraction, you can still flip it. Just use the identity dy/dx × dx/dy = 1.
Did this problem roughly a month ago. Was looking for an interesting logarithm-like function. I sort of used the second method, but I “substituted” y as x, and x as y. The equation became dy/dx = x + y. I had already figured out x in terms of y through the last problem I looked into. So all I had to do from there was find the inverse function of y = ke^x - x - 1.
Never thought about using the substitution u = x + y, it was very interesting approach to see!
A bit of a correction with the second method, you forgot to add a constant onto the homogeneous solution. It should be y = Ae^x (not a massive issue though, I can see that you have it correct when using the first method).
Second method:- just reciprocal both sides. dx/dy = x + y. Now I think everyone can do
At 6:20 Sybermath begins with the W function approach, which I've always detested. However, one can now easily solve for x in terms terms of y and exp(y), so why not call this a solution and be done with it?
We usually like to have y=f(x) if possible but there's nothing wrong with saying the solution to a differential equation is x=f(y) or even c=f(x,y)
It's nice to be able to succinctly describe the solution in terms of a "known" function. This isn't as arbitrary as it feels, we do it all the time. Lambert W just doesn't have quite the same ring to it as all the famous functions like sin, cos, and e^x etc, but it's still a perfectly valid function.
@10:26 - you forgot a constant of integration. Your "general" solution using method 2 is actually itself only a particular solution, and should be x = A*e^y - y - 1.
Further, going back to your first solution, if you substitute a different constant for -1/k, you get y = -W(A*e^x) - x - 1. Unfortunately, I am not able to get its derivative to look anything like - 1/(W(A*e^x)+1), so I can't actually verify this solution.
I got y = - x - 1 using basic arithmetics
With luck and with regards
easy solve but after integration the answer seems messy, WA expressed the solution with product logs not sure how though.
The idea of the method here is called obvious substitution, set x+y=t and take the derivative with respect to x. dx/dx +dy/dx = dt/dx
so dy/dx = dt/dx - 1
Afterwards we have a very simple separable diff equation dt/dx = 1 + 1/t
Look at his second solution. I liked his conversion to a separable equation in x and y vs your parametric equation solution. Yes the Lambert - W way was too difficult to understand. By doing homogeneous differential equations and particular solutions equations is the more versatile and widely used trick engineering students and Physicists use for "State Equations" of solutions of multiple variables in matrix forms. I would recommend the less used parametric approach of yours to the highly versatile homogeneous and particular solutions approach that by Senior collegiate years students will be equally solving not only for single variables of y in terms of x but also for a set of solutions like Y matrix = A matrix matrix multiply with set of solutions X matrix as systems of linear separable first order differential equations engineers and scientists will routinely do in their senior level courses in universities. I know because I'm a graduate student solving Electrical Systems of equations modelled in amplifiers and feedback loops and D.C. motors high level electrical engineering projects! 👍 Your parametric approach is a good middle ground to do what matrix solutions of linear first order systems of equations described in matrix forms does.
@@lawrencejelsma8118 thank you for the detailed response, though I'm afraid to say the majority of the topics you talked about aren't familiar to me. I recently finished my differential equations course that covered most of the basic solving strategies and some applications of diff equations, though I feel like i should know more. Can you refer me to a course on working with, for example, higher order differential equations? Another thing is due to a very very bad linear algebra protestor, i feel lost when it comes to work with matrices or really most linear algebra topics, can you recommend a free online book or course to help me out, thank you!
@@barberickarc3460 ... Just future education knowledge and not to be overly confusing about his method 2 advanced mathematics topics:
I am stating this in case you want to solve, like for electrical engineering all currents and voltages in a circuit system where a system of all di/dt of currents = 1/L times systems of system voltages. ... Or by dual parallel system dv/dt = 1/C times systems of currents electrical circuits solutions. By Junior University coursework You will understand it more in sciences or engineering coursework using MATLAB or PSPICE matrix inputs for solving by conputers.
I said things quick and wrong in my comments writing. There are sets of differential equations called the differential matrix Y' and that equals an A matrix matrix multiplied by an X matrix of elements linearly related to Y in two dimensions (non differential system output matrix X) then it is added to a B matrix times a system input matrix U for the first formed system of equations expressed in linear matrix form Y' = AX + BU. That was for finding the homogeneous solutions for determining what X axis set of elements matrix X system of solutions are.
As he stated in the single dimensional first order differential equations y = ax + bu non matrix case in his method 2 (when studying introduction to solving by differential equations a first order differential equation his video is solving). Then another set of matrix parameters of finding the Y matrix as the system output matrix by another linear relationship Y = CX + DU type thing to be able to solve for the particular system set of solutions he did in the non-matrix needed general first order differential equation of y = ax + b he did ending the video for the y total = y homogeneous added to y particular system set of equations for first order differential solutions of a scientific or engineering configured system of differential equations.
Prerequisites to understanding first order systems differential equations mathematics using Matrices for a system set of differential equations don't use anything advanced in Linear Algebra coursework vs matrix multiplication and matrix determinants and matrix inverse Linear Algebra concepts. The Differential Equations theory applied to matrices representative of a set of system of equations needing parallel solutions in matrix forms is Differential Equations Calculus of solving First Order Differential equations he did in the most basic y'= ax + bu and y = cx + du form of homogeneous adding to particular y solution results (study and copy his method 2 in the video to get the parallel you will need to study in engineering and sciences mathematics).
10:14 you factored D instead of factoring x
🫣
But why isn't y = -x-1 in your set of solutions ??
I hate the Lambert W function.
then u didnt discover the beauty of it yet
@@RayMyName The only times I have ever seen LambertW ever used is at the end of a problem like this, basically just to rewrite f(y)=x into f(x)=y. It can't be that important if I don't see it in any other context.
@@bsmith6276 solving of non standard equations as well
I love the idea of it, but I hate the absence of a formula for it. At least pi has a Taylor series.
Lambert hates you as well my dear
Nice!
Thanks!
Your methods?
Where is the arbitrary constant in the second method?
Honestly, it's a good exercise to follow his work, and see where he neglected it.
damn
y=-W(-exp(-(x+c+1)))-1-x
Tbh not much point having c + 1, have c as inclusive of the 1
hi
What an arse wipe solution
d ( x + y) / dx = ( x + y + 1) /( x + y)
d ( x + y) - d ( x + y) /( x + y + 1) = d x
d y = d ( x + y) /( x + y + 1)
x + y + 1 = A exp ( y)
x = A exp ( y) - ( y + 1)