Ylnydx + (x-lny) dy = 0 to find its general solution

Ylnydx + (x-lny) dy = 0 to find its general solution

∵ ylnydx + (x-lny) dy = 0 = = > lnydx + XDY / y-lnydy / y = 0 (both ends of the equation are divided by Y) = = > lnydx + XD (LNY) - lnyd (LNY) = 0 = = > d (xlny) - D ((LNY) ^ 2 / 2) = 0 = = > xlny - (LNY) ^ 2 / 2 = C / 2

Finding the general solution of the differential equation x * dy / DX = y * LNY / X The original question is as follows: X * dy / DX = y * ln (Y / x)

x*dy/dx=y*lny/x
dy/(ylny)=dx/x^2
Integral on both sides
lnlny=-1/x+C1
lny=C2e^(-1/x)
y=Ce^[e^(-1/x)]

Find the extreme value of function f (x, y, z) = XYZ under the condition x ^ 2 + y ^ 2 + Z ^ 2 = 16

Lagrange derivative method is used to establish Lagrange function L = XYZ- λ (x ^ 2 + y ^ 2 + Z ^ 2-16), l can obtain YZ-2 by deriving x, y and Z respectively λ x=0,xz-2 λ y=0,xy-2 λ Z = 0, represented by X, y and Z respectively λ, We can get x ^ 2 = y ^ 2 = Z ^ 2, so x ^ 2 = y ^ 2 = Z ^ 2 = 16 / 3, x = y = z = 4 / 3 ^ (1 / 2)

When x > 0, Y > 0, z > 0 and a > 0, find the lowest cost of a product's cost function u = XYZ under constraints of 1 / x + 1 / y + 1 / z = 1 / A

It belongs to conditional extreme value
Lagrange least square method
Constructor:
F(x,y,z)=xyz+ λ (1/x+1/y+1/z-1/A)
Derivative x, y, Z, respectively
Fx'(x,y,z)=yz- λ/ x^2
Fy'(x,y,z)=xz- λ/ y^2
Fz'(x,y,z)=xy- λ/ z^2
And make it 0
Then yzx ^ 2 = xzy ^ 2 = XYZ ^ 2= λ
And x > 0, Y > 0, z > 0
1/x+1/y+1/z=1/A
Then x = y = z = 3A
be
xyz=27A^3

If there is a function y = x * SiNx, how can we prove that this function is unbounded,

Let x = 2K π + π / 2 K ∈ Z, then y = 2K π + π / 2 sin (2k π + π / 2) = 2K π + π / 2 because 2K π + π / 2 is unbounded
Y = x * SiNx unbounded

lim（x→0) sinx-x(x+1)/xsinx

Using the quadratic robbida's law
lim（x→0) sinx-x(x+1)/xsinx =lim（x→0) （cosx-2x-1)/(sinx +xcosx)
=lim（x→0) （-sinx-2)/(2cosx -xsinx)=(-2)/2=-1.