Solving the Differential Equation y''=(y')^2

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[mcwulf]

I divided by y' to get y"/y' = y' which when integrated becomes ln(y') = y + c.

y' = exp(y+c) = k exp(y)
Integrate again after grouping the terms to the left and we have
-k.exp(-y) = X+q
y = -k ln(1/(X+q)) = k ln(X+q)

[johngreen]

When you divided both sides by u^2, you assumed that u was not zero. If u is zero you get y'=0 which gives y =c a constant. It is a trivial solution. If f(0)=1 then the constant c is 1

[NuTahn]

I believe that the lost solution at the end should be u=0 (i.e. y’=0), thus it would be y=C rather than y=0.
Thanks for the video! It’s an interesting problem to solve.

[Mathiqnice]

Giải phương trình vi phân cấp hai bằng đặt ẩn phụ. Phương pháp hay. Cảm ơn.

[JohnRandomness]

Let z = y'. Then dz/dx = z² or dx/dz = 1/z². That means x = (-1)/z + C, or z = (-1)/(x - C). Another solution is z = 0, leading to y = D -- another arbitrary constant. Zero is the limiting case as x --> ∞, so it stands to reason that z = 0 is a solution.

dy/dx = (-1)/(x-C) so y = (-1) ln|x - C| + D. Let's not forget y = D as another solution.

I think that dz/dx = z² is of a certain form taught in DE classes, but I've forgotten the solution.

[JohnRandomness]

Suppose we tried y(0) = 1 and y'(0) = 0 as initial conditions? 1 = (-1) ln |C| + D, or D = ln |C| + 1.
0 = (-1)/(-C) -- doesn't work. But y = 1 is a solution.

[alextang]

y=constant is also a solution. But, why we cannot find in any equation in your calculation showing this result. Could you tell me why? 🤔🤔🤔

[kinshuksinghania]

Not y = 0, but y = constant is the general solution in that case

[Vladimir_Pavlov]

The initial conditions appeared uninvited in the middle of solving the problem.))
If we take the initial conditions y(0) =1 and dy/dx (0)=0, then what will be the particular solution satisfying these conditions?

[damiennortier]

Let d = degree of the function y. So the degree of y' is d-1 and the degree of y'' is d-2.
Since when we multiply 2 equation, we add the power, the degree of (y') ^2 is 2d-2 right?
So we get 2d-2 = = d - 2 and then d = 0
So y = a where a is a constant. Y' = 0 and y'' = 0. So this is a valable solution

[anshumanagrawal]

What does the "dx" even mean, du/dx is just shorthand for d/dx(u) or u' you can't just treat it as a fraction

[CadeiraGritante]

Só passando para falar que seu canal é incrível, creio que o "Silvio Duarte" seja brasileiro, assim como eu, diante de tantas línguas, a matemática nos une S2

Just passing on to say that your channel is amazing, I believe that "Silvio Duarte" is Brazilian, as I am, in the face of so many languages, mathematics unites us S2

[franciscoj.f.]

Funny enough that for any y = constant, it satisfies also the equation (since y' = y'' = y''' = ... = 0)

[frentz]

Lemme try .. #1 substitute u = y', so u' = u^2. #2 (1 / u^2) du = 1 dT. #3 -1/u = T + C, or (solution) u = -1 / (T + C). Then #4 from y' = -1 / (T + C), so y = - ln | T + C | + C2. .. Hmmm. Does this actually do y'' = (y')^2 ?? Seems weird and probably wrong, but I guess there's one way to find out. #5 plug it in and check it

[francospano]

Les recuerdo que en la República Argentina hablamos castellano

[GregWeidman]

Checked the solution, and then added initial conditions. Beautiful!

[theimmux]

What do you mean the integral of 1/u^2 is -1/u when u is a function of x? Chain rule???

[fonzi]

i think the antiderivate of 1/u^2 is not -1/u ; if you derivate -1/u you get u'/u^2 not 1/u^2. Am i wrong?

[victorchoripapa]

I was happy enjoying it👌🏽☺️
Your solution was great. Didn't see coming that. Trying to solve it I got y=C for solution but... I don't know it was correct...

[ziedabdellaoui]

🤔 initial condition (the end)🤔 f(0)=1 => f'(0)= -1/e 🤔 why ? how ?