Solving A Differential Equation

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❤️ dy/dx = 1/(x + y)

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[demenion]

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

[pwmiles]

There ought to be one free parameter. The solution of method 2 is

x = ke^y - y - 1 k constant

[someknave]

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.

[fahrenheit]

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.

[eirdeix]

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.

[holyshit]

In my opinion substiitution u = x+y solves the problem
but if we want linear equation we can flip both sides

[iamadooddood]

Even if it's not a fraction, you can still flip it. Just use the identity dy/dx × dx/dy = 1.

[gregstunts]

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).

[YuvrajDandotiya]

Second method:- just reciprocal both sides. dx/dy = x + y. Now I think everyone can do

[jimschneider]

@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.

[weylguy]

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?

[user]

I got y = - x - 1 using basic arithmetics

[Acssiohm]

But why isn't y = -x-1 in your set of solutions ??

[alejrandom]

10:14 you factored D instead of factoring x

[barberickarc]

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

[alexandermorozov]

Where is the arbitrary constant in the second method?

[honestadministrator]

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)