Find the general solution or special solution of Y '= (Y / x) (1 + LNY LNX) differential equation

Find the general solution or special solution of Y '= (Y / x) (1 + LNY LNX) differential equation

y’=(y/x)(1+lny-lnx)
Because: LNY LNX = ln (Y / x), let: y = UX, because: (UX) '= u' * x + U
It can be changed into:
xdu/dx+u=u+uln(u)
That is: Du / DX = (U / x) ln (U)
The results were as follows
du/[uln(u）]=(1/x)dx
The integral of both sides is as follows:
(1/lnu)d(lnu)=lnx+lnC
(Note: LNC is written for formal beauty, but it's a constant anyway)
ln(lnu)=ln(Cx)
So:
u=e^(Cx)
And y = UX
So: y = x * e ^ (Cx)

What is the integral of (LNX) Cube / x square

Integral: (LNX) ^ 3 / x ^ 2DX
=Integral: (LNX) ^ 3D (- 1 / x)
=-(LNX) ^ 3 / x-integral: - 1 / XD (LNX) ^ 3
=-(LNX) ^ 3 / x-integral: - 3 (LNX) ^ 2 / x ^ 2DX
=-(LNX) ^ 3 / x + 3 integral: (LNX) ^ 2D (- 1 / x)
=-(LNX) ^ 3 / x-3 (LNX) ^ 2 / x + 3 integral: 1 / XD (LNX) ^ 2
=-(LNX) ^ 3 / x-3 (LNX) ^ 2 / x + 3 integral: 2lnx / x ^ 2DX
=-(LNX) ^ 3 / x-3 (LNX) ^ 2 / x + 3 integral: 2lnxd (- 1 / x)
=-(LNX) ^ 3 / x-3 (LNX) ^ 2 / x-6lnx / x + 3 integral: 2 / XD (LNX)
=-(lnx)^3/x-3(lnx)^2/x-6lnx/x-6/x+C
(C is a constant)
Idea: continuous integration by parts

How to find the integral of (LNX) ^ 2
Hope to list the detailed process

Use & nbsp; partial integral & nbsp;... & nbsp; but LS has error:
∫(lnx)^2dx = x(lnx)^2-∫x（2lnx）dx
But even so, it can't be done
This kind of trigonometric function has inverse logarithm,
Generally, we use substitution method to replace logarithm and inverse trigonometric function
See the figure below

Calculate a ^ 3 / a-1-a ^ 2-a-1
The process should be detailed

Is it a & sup3; / (A-1) - A & sup2; - A-1=+
Known:
a³-1
=(a-1)(a²+a+1)
Then:
a³/(a-1)
=(a³-1)/(a-1)+1/(a-1)
=a²+a+1+1/(a-1)
Substituting in the original formula, we get:
a³/(a-1)-a²-a-1
=a²+a+1+1/(a-1)-a²-a-1
=1/(a-1)

LIM (x → 2) radical x + denominator radical 2 / numerator 1 = what?

lim（x→2）√x+1/√2
=√2+√2/2
=3√2/2

How to calculate (a + 1) (a + 2) (a + 3) (a-6)? The method of immortality!

(a+1)(a+2)(a+3)(a-6)
=[(a+1)(a-6)] [(a+2)(a+3)]
=(a²-5a-6)（a²+5a+6）
=（a²）²-（5a+6）²
=a^4-25a²-60a-36

X→0,lim(sin1/x+cos1/x)^x

A:
lim(x→0) [sin(1/x)+cos(1/x) ]^x
=lim(x→0) { [ sin(1/x)+cos(1/x) ] ^2 }^(x/2)
=LIM (x → 0) [1 + sin (2 / x)] ^ (x / 2) because: 0

6 - (2 / 3-1 / 2) off formula calculation

6-(4/6-3/6)
6-7/6
36/6-7/6
30/6
5/1
five

How to prove LIM (x ^ 2 * sin1 / x) = 0, X → o?
Proof by pinching theorem

From X & sup2; → 0, X & sup2; is infinitesimal
Sin1 / X is a bounded function
Infinitesimal multiplied by bounded function is infinitesimal
So LIM (x ^ 2 * sin1 / x) = 0

A * b = a + B-1, if 3 * (- 2) = 3 + (- 2) - 1 = 0, calculate (- 1) * (- 3) * 2

=((-1)+(-3)-1)*2=(-5)+2-1=(-3)-1=-4