The general solution of the differential equation y '+ y + LNX = ax is?

The general solution of the differential equation y '+ y + LNX = ax is?

y'+y+lnx=ax
dy/dx=-y+ax-lnx
dy/dx=-y
lny=ln1/x+C0
y=C/x
Let y = C (x) / X
C'(x)/x=ax-lnx
dC(x)=ax^2dx-xlnxdx ∫xlnxdx=lnx*x^2/2-(1/2)∫xdx=x^2lnx/2-x^2/4+C
C(x)=ax^3/3-(x^2lnx)/2+x^2/4+C1
The general solution is y = ax ^ 2 / 3-xlnx / 2 + X / 4 + C1 / X

What is the integral of 1 / (LNX)?
Well, it's an indefinite integral
What is the original function of 1 / (LNX)?

This function is not a basic elementary function, so it cannot be restored,
It can only be used
We can use the convergence method of generalized integral,
For infinite generalized integral, ∫ (a ~ + ∞) f (x) DX, then make x ^ P (P > 1), find LIM (x →∞) (x ^ P) f (x), if the limit exists, then converge;
For the generalized integral of unbounded function, ∫ (a ~ b) f (x) DX (x = a is a singular point, also known as a defect), then (x-a) ^ P (0

What is the integral of LNX / x?

∫lnx/xdx
=∫lnxdlnx
=1/2(lnx)^2+c
C is a constant

Lim x tends to be 4, the denominator is √ (X-2) - √ 2, and the molecule is √ (2x + 1) - 3

Same as above: the numerator and denominator are multiplied by √ (X-2) + √ 2 and √ (2x + 1) + 3 at the same time
The obtained molecule is (2x-8) (√ X-2 + √ 2)
The denominator is (x-4) (√ 2x + 1 + 3)
Reductive decommensurate factor (x-4)
Then we take 4 into the formula and get the limit value of 2 √ 2 / 3

Calculation (a + 3) (A-1) + a (A-2)

=a^2-a+3a-3+a^2-2a
=2a^2-3

How to do LIM (x → 0) x ^ 2 / sin ^ 2 (x / 3) with X tending to 0 SiNx / x = 1

How to understand sin ^ 2

Calculation + (√ 3-1) 2 - (√ 3 - √ 2) (√ 3 + √ 2)

Original formula = 2 radical 3-2 - (radical 3) square + (radical 2) square = 2 radical 3-2-3 + 2 = 2 radical 3-3

LIM (H tends to 0) [sin (x + H) - SiNx] / h

LIM (H tends to 0) [sin (x + H) - SiNx] / h = (SiNx) '= cosx

Calculation: 1 / 3 + 2 / 3 + 3 / 3 + 4 / 3 +... + 10 / 3

Original form
=(1+2+3+…… +10）/3
=(1+10)×10÷2÷3
=55÷3
=55 / 3 (55 / 3)

Find I = Lim [1 / √ (4N ^ 2-1) + 1 / √ (4N ^ 2-2 ^ 2) +... + 1 / √ (4N ^ 2-N ^ 2)]

Original formula = Lim1 / n [1 / √ (4 - (1 / N) ^ 2) + 1 / √ (4N ^ 2 - (2 / N) ^ 2) +... + 1 / √ (4N ^ 2 - (n / N) ^ 2)]
=Integral of 1 / √ (4-x ^ 2) from 0 to 1
That is arcsin (x / 2) | 0,1 = pi / 6