General solution of differential equation x ^ 2dy + (y-2xy-x ^ 2) DX = 0

General solution of differential equation x ^ 2dy + (y-2xy-x ^ 2) DX = 0

∵x²dy+(y-2xy-x²)dx=0
==>E ^ (- 1 / x) dy / X & # 178; + (y-2xy-x & # 178;) e ^ (- 1 / x) DX / x ^ 4 = 0 (both ends of the equation multiply e ^ (- 1 / x) / x ^ 4)
==>e^(-1/x)dy/x²+y(1-2x)e^(-1/x)dx/x^4=e^(-1/x)dx/x²
==>e^(-1/x)dy/x²+yd[e^(-1/x)/x²]=e^(-1/x)d(-1/x)
==>d[ye^(-1/x)/x²]=d[e^(-1/x)]
==>Ye ^ (- 1 / x) / X & # 178; = e ^ (- 1 / x) + C (C is an integral constant)
==>ye^(-1/x)=x²[e^(-1/x)+C]
==>y=x²[1+Ce^(1/x)]
The general solution of the original equation is y = x & # 178; [1 + CE ^ (1 / x)] (C is an integral constant)

How to solve the differential equation y '(y ^ 2-x) = y
The reciprocal of Y times the square of Y minus x equals y

y`y^2 - y`x = y
→ y`y^2 = y + y`x
→ y`y^2 = (xy)`
→ y^2 dy = d(xy)
→ y^3 = 3xy + C
Do your homework by yourself

Solving the second order linear homogeneous differential equation with constant coefficients by y = e ^ x, y = e ^ (3x)

1 and 3 are the two roots of the characteristic equation of the homogeneous equation, so the characteristic equation is R ^ 2-4r + 3 = 0, so the obtained homogeneous equation is y '' - 4Y '+ 3Y = 0

LIM (n →∞) {1 + 2 / N} ^ kn = e ^ - 3?

lim（n→∞） {1+2/n}^kn =lim（n→∞） {1+2/n}^[(n/2)2k] =e^(2k)
e^(2k)=e^(-3)
2k=-3
k=-3/2

Mathematical sequence of senior one 1 + 1 / 2 + 1 / 3 + 1 / 4 + How to calculate the sum of + 1 / N series

This one has no general formula

Find the following limit LIM (n →∞) ∑ (upper n, lower k = 1) (1 / 1 + 2 +. + k)

Click to see the big picture

How to calculate the fractional sequence 1, 2 / 3, 5 / 9, (), 7 / 15, 4 / 9?

1、2/3、5/9、(6/12 )、7/15、4/9
That's 1 / 2
The above scores can be expressed as:
3/3,4/6/,5/9,6/12,7/15,8/18
That is, the numerator follows the + 1 rule and the denominator follows the + 3 rule

Let me ask you how to find this limit sum, LIM (n →∞) ∑ (k = 1, n) 1 / [(n ^ 2 + K ^ 2)] ^ & # 189;

Define by integral
The original formula = LIM (n →∞) ∑ (k = 1, n) 1 / [(n ^ 2 + K ^ 2)] ^ & # 189;
=lim（n→∞）∑（k=1,n） (1/[n^2(1+(k/n)^2）]^½)
=lim（n→∞）∑（k=1,n） (1/[1+(k/n)^2]^½)*(1/n)
=∫ [0,1] 1 / radical (1 + x ^ 2) DX
Let x = Tan T, t ∈ (- π / 2, π / 2)
dx=sec^2t dt
The original formula = ∫ [0,1] sec t DT
=ln|sect+tant|+C |[0,1]
=Ln | radical (1 + x ^ 2) + X | + C | [0,1]
Substitute 1 to get ln|1 + radical 2|
Substitute 0 to get ln1 = 0
So integral = ln (1 + radical 2)
LIM (n →∞) ∑ (k = 1, n) 1 / [(n ^ 2 + K ^ 2)] ^ & #189; = ln (1 + radical 2)

The general formula of the first four terms of 3,5,7,9 is

an=2*n+1

Find the limit LIM (n →∞) ∑ 1 / n [(K / 3) ∧ 3 + 1] k = 1 → n

It is not clear whether (K / 3) &# 179; + 1 is in numerator or denominator, but it is also applicable to the following conclusions
If Lim {n →∞} a [n] = C (also true for C = + ∞ or - ∞), then Lim {n →∞} 1 / N · ∑ {1 ≤ K ≤ n} a [k] = C
If (K / 3) & # 179; + 1 on the molecule:
From Lim {K →∞} (K / 3) & # 179; + 1 = + ∞, Lim {n →∞} 1 / N · ∑ {1 ≤ K ≤ n} ((K / 3) & # 179; + 1) = + ∞
If (K / 3) &# + 1 on the denominator:
From Lim {K →∞} 1 / ((K / 3) & # 179; + 1) = 0, Lim {n →∞} 1 / N · ∑ {1 ≤ K ≤ n} 1 / ((K / 3) & # 179; + 1) = 0