How to solve adifferential equation with u=y/x substitution

In this video, I showed how to solve a separable differential equation using the u = y/x substitution. Enjoy!

[duckyoutube]

You are like the bob ross of math.

The voice. The atmosphere. Its perfect.

[manucitomx]

What an excellent, clear, to the point explanation
Kudos

[michaelstahl]

Great video . Thanks Sire ! Very elegant separating y from x using a substitution .

[AbouTaim-Lille]

،whenever you see the expression xy' + y you should directly put it in the forme (xy)'. And when you see the expression xy' - y you should directly anticipate it would be of the form: (y/x)'. x². So in this example we could easily notice that xy'- y = (y/x)'x² = e^y/x so rewriting the equation after putting z= y/x we get:
z' x² = e^z. Thus, Z'e^-z = 1/x²
So e^z = 1/x +c. And finally:
Z= ln |1/x +c|

[iivfjxj]

king of mathematics, i didn't have any idea how to solve that

[michaelbaum]

I would leave out the absolut value in the logarithm function. The solution is y=-x*ln(1/x+K) with x<-1/K, so the argument is positive. That‘s also the solution which Wolfram Alpha delivers.

[kragiharp]

Thank you so much for reviewing!

In school we would rewrite the y' immediately into dy/dx. So we would not have any confusion about the derivative.

[vladimir]

Excellent video and nice development. However, it is clearly NOT the general solution for the problem since the step of taking ln() of both sides is not always valid, and introducing the absolute value does not change much.
The general solution should be more subtle than presented.

[boburturimov]

First isolate exponential, then divide both side by exp(y/x) and try to check derivative from exp(-y/x).

[firstolasto]

Thank you, I enjoy your videos!! 🙏👏👏👏

[surendrakverma]

Excellent explanation Sir. Thanks and Regards

[serae]

I think you should have a duel who can solve more integrals in the same time with "Black Pen Red Pen"😊

[sfundomsezane]

Do you have a video explaining bernoullis equations

[nathanbarnes]

I left the final answer as
x[e^(- y/x) - k] = 1 as I don’t like having messy fractions within logarithms haha

[kingbeauregard]

This one tricked me because I did the quick test of homogeneous equations, and this didn't pass the test. EVEN SO, the homogeneous technique worked. I think the lesson is, try the y/x substitution, or whatever other substitution looks like it might help, and see what happens.

The quick test is, replace every instance of x with ax, and every instance of y with ay. Do all the a's cancel each other out? If so, it's a homogeneous equation, and for sure the y/x substitution will work.

[skwbusaidi]

No absolute value as we are ln both side not integerating