How to find the general solution of the differential equation y '= 3x ^ 2Y

How to find the general solution of the differential equation y '= 3x ^ 2Y

y' = dy/dx = 3x²y
=>(1 / y) dy = 3x & # 178; DX {integral on both sides}
=>Ln | y | = x & # 179; - ln | C | {the form of this arbitrary constant depends on the specific form}
=> ln|Cy| = x³
=> Cy = e^(x³)

General solution of differential equation y '+ [(2-3x ^ 2) / x ^ 3] y = 1

e^[∫(2-3x^2)/x^3dx]=e^(-1/x^2-3ln|x|)=e^(-1/x^2)/x^3
So e ^ (- 1 / x ^ 2) / x ^ 3 (y '+ (2-3x ^ 2) / x ^ 3 * y) = e ^ (- 1 / x ^ 2) / x ^ 3
(ye^(-1/x^2)/x^3)'=e^(-1/x^2)/x^3
Integral on both sides: ye ^ (- 1 / x ^ 2) / x ^ 3 = 1 / 2 ∫ e ^ (- 1 / x ^ 2) d (- 1 / x ^ 2) = e ^ (- 1 / x ^ 2) / 2 + C
y=x^3/2+Ce^(1/x^2)

A problem of differential equation. Finding the general solution of Y '' - 3Y = 3x ^ 2 + 1

The characteristic equation of the homogeneous equation y '' - 3Y = 0 is R & # 178; - 3 = 0, then r = ± √ 3
The general solution of the homogeneous equation y '' - 3Y = 0 is y = C1E ^ (√ 3x) + c2e ^ (- √ 3x) (C1, C2 are integral constants)
Let the special solution of the original equation be y = ax & # 178; + BX + C
∵y'=2Ax+B
y''=2A
Substituting into the original equation, 2a-3 (AX & # 178; + BX + C) = 3x & # 178; + 1
==>-3A=3,-3B=0,2A-3C=1
==>A=-1,B=0,C=-1
The special solution of the original equation is y = - X & # 178; - 1
So the general solution of the original equation is y = C1E ^ (√ 3x) + c2e ^ (- √ 3x) - X & # 178; - 1 (C1, C2 are integral constants)

There are three slideways in an amusement park, the first is short and steep, the second is long and slow, and the third is spiral. Xiaogang slides freely from the three slideways to the horizontal plane. Without friction, the following statement is correct
The kinetic energy of sliding down the three slides is equal

The kinetic energy is equal to the potential energy of Xiaogang from the top of the slide to the ground. The potential energy is w = GH. The height is constant, and Xiaogang's gravity is constant. Therefore, friction is not included,
The kinetic energy of sliding down the three slides is equal

Is the power consumption of 1000 W electrical appliances 1 degree per hour?

In fact, the input voltage can not be very stable and accurate in the rated voltage value, and the current is even more unstable (except for pure resistance circuit). For example, for a 50W fan, if you choose different gears, its actual power and power consumption will be different, In fact, it only works at about 1000W under full rated condition. The power consumption per hour is about 1000W

It is known that the solution set of inequality (3a-2b) with x greater than a-2b is less than half of X, and the solution set of inequality ax + B with X less than 0 is obtained

The solution set of (3a-2b) x greater than a-2b is x less than half
∴﹛3a-2b＜0
(a-2b)/(3a-2b)=½
∴a=-2b
∴a＜0,b＞0
ax+b＜0
ax＜-b
∴x＞-b/a
The solution of the inequality ax + b less than 0 about X is x ＞ - B / A

Resistivity of copper and aluminum at different temperatures
I want to convert the resistance value measured at the actual temperature into the standard resistance value of 20 degrees

AC power line refers to the low-voltage power line in power distribution engineering
1) The minimum cross section of conductor allowed by mechanical strength is selected
2) Select according to the allowable temperature rise
3) Select according to economic current density
4) Select according to allowable voltage loss
The low-voltage power line is commonly used in communication. Because of its large load current, it should be selected according to the heating (temperature rise) condition. Because if there is no limit, the power line is not suitable
With the increase of temperature, the insulation will be aging and damaged rapidly, and even lead to electrical fire
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For 220 V single phase AC
1: I = P / 220
2: Power line area s = I / 2.5 (mm2)
For 380V three-phase AC
1：
I = P / (380 * Γ 3 * power factor)
2: Sectional area of phase line s phase = I / 2.5 (mm2)
3: The section area of zero line s zero = 1.7 × S phase
Estimation of current carrying capacity of insulated conductor
Estimation formula:
Multiply it by nine at 2.5 and go up minus one
Multiply 35 by 3.5, double double group minus 0.5
Conditions are variable plus conversion, high temperature 10 fold copper upgrade
The number of pipes is 234, 876
explain:
(1) The formula in this section does not directly point out the current carrying capacity (safety current) of various insulated wires (rubber and plastic insulated wires), but "cross section multiplied by a certain multiple", which is obtained by mental calculation. It can be seen from table 53 that the multiple decreases with the increase of cross section
"2.5 mm, multiply by 9, subtract one from the top" means that the current carrying capacity of aluminum core insulated wires with various cross sections of 2.5 mm 'and below is about 9 times of the number of cross sections. For example, the current carrying capacity of 2.5 mm' conductor is 2.5 × 9 = 22.5 (a). The multiple relationship between the current carrying capacity and the number of cross sections of 4 mm 'and above is arranged along the line number, and the multiple is successively reduced by 1, that is 4 × 8, 6 × 7, 10 × 6, 16 × 5, 25 × 4
"35 times 3.5, double group minus 0.5" means that the current carrying capacity of 35mm conductor is 3.5 times of the number of cross sections, that is 35 × 3.5 = 122.5 (a). From the multiple relationship between the current carrying capacity and the number of cross sections of 50mm 'and above conductor, it becomes a group of two wire numbers, That is to say, the current carrying capacity of 50 and 70mm 'conductor is 3 times of the number of cross-sections; the current carrying capacity of 95 and 120mm' conductor is 2.5 times of the number of cross-sections, and so on
If the aluminum core insulated wire is exposed in the area where the ambient temperature is higher than 25 ℃ for a long time, the current carrying capacity of the wire can be calculated according to the above formula, and then 10% off. When the aluminum core insulated wire is not used, but the copper core insulated wire is used, Its current carrying capacity is slightly larger than that of aluminum wire of the same specification. According to the above formula, the current carrying capacity of one wire number larger than that of aluminum wire can be calculated. For example, the current carrying capacity of 16mm 'copper wire can be calculated as 25mm2 aluminum wire
1 economic section of conductor
The relationship curve between conductor section and annual expenditure is shown in Figure 1. Among them, curve 1 is the functional relationship curve between annual operation cost and conductor section; curve 2 is the functional relationship curve between investment and depreciation cost and conductor section; curve 3 is the relationship curve between conductor section and annual comprehensive expenditure
TZ = (c + C0) α · S + 3i2zd τ β × 10-3 (yuan / km) (1)
Where C -- annual maintenance cost coefficient
C0 -- capital repayment coefficient
α - price of conductor per unit cross-sectional area and length yuan / mm2 · km
S -- sectional area of conductor, mm2
IZD -- maximum current a
ρ - resistivity of conductor Ω · mm2 / km
τ - maximum load loss hours H
β - electricity price yuan / kW · H
In order to obtain the minimum conductor section of annual operation cost, the derivation of the above formula is obtained, and let = 0 get:
The economic section of conductor is s = SJ
2 economic current density
According to the definition of economic current density: J = get:
When the conductor material is fixed and the depreciation and maintenance rate is constant, the economic current density mainly depends on the regional electricity price and annual loss hours
Where K1 -- coefficient related to conductor price, depreciation and maintenance rate, K=
The value of γ is related to the electricity price β, and changes with the positive value. The electricity price varies with the voltage level and region, and the annual loss hours and load nature
For example: if the selling price of ACSR is 14500.00 yuan per ton, then α = 56 yuan / km · mm2; if the annual interest rate of conductor is 6.87%, and the repayment period is 20 years, then the annual repayment coefficient of capital is C0 = = 0.0934; the annual maintenance coefficient of conductor is 1.5%; the resistivity of aluminum is 30 Ω· mm2 / km
Then K1 = = 8.2
Substituting into equation (4), it is obtained that:
J=8.2γ(A/mm2) (5)
The relationship between economic current density and γ of ACSR is shown in Figure 2
According to the above analysis, the economic current density can not be determined by the maximum load utilization hours, for example, Tmax = 3000h, j = 1.65. When the current electricity price changes from 0.25 yuan to 0.80 yuan, and cos φ = 0.85, τ = 2300h, the economic current density is in the range of 0.35 ~ 0.19, which is about 4.7 ~ 8.5 times smaller than the original recommended value
3 Determination of economic current density of trunk line with branch line
The above discussion is based on the case that there is no branch line in a trunk line, and the load is concentrated at the end of the line. However, in engineering practice, most 10kV and low-voltage distribution networks belong to trunk lines with several branch loads, as shown in Figure 3. At this time, the economic current density is determined according to the head end current, and the section of the whole trunk line is bound to be too large, The premise of the study is that the power loss caused by the load concentrated at the end of the line is equal to the power loss caused by the trunk line with branch load
If the branch loads in the trunk line are equal and the electrical distance L is equal, the total active power loss is:

Where I -- load current at head end a
While the load is concentrated at the end, its active power loss is as follows:
Δρ=3I2 (7)
If the two equations are equal, the economic current density of trunk line with branch load is: JF = JKF ([SX (] 6n2 ﹥ n + 1) (2n + 1)
Figure 4 shows the relationship curve between K2 and the number of branch loads. According to equation (8), when n →∞, K2 =. For example: τ = 2500h, β = 0.45 yuan. If the load is concentrated at the end of the line, the economic current density j = K1, γ = 0.0298 × 8.2 = 0.244 (A / mm) 2; if the load is evenly distributed on the main line, the number n = 8, The economic current density of the main line is JF = k · J = 1.584 × 0.247 = 0.39 (A / mm2). When the load distribution in some lines is uneven or the difference is large, the economic current density of the main line can be calculated according to equations (6) and (7)
reference material: http://zhidao.baidu.com/question/26078424.html

In the sequence of 1 2 3 2 3 4 3 4 5 5 6 6 6 7 6 7 7 8., the hundredth number is ()

It is easy to observe that the above numbers are arranged according to every three consecutive natural numbers. For example, if the fifth number is 5 / 3 = 1.2, it means 1 + 2 = 3
By analogy, if the nth number is n / 3 plus the remainder, then the 100th number is 100 / 3 = 33.1
That is 33 + 1 = 34

The voltage at both ends of a resistor is 2V and the current intensity is 0.5A. If the voltage at both ends increases to 4V, the current passing through the resistor is ()
A.1A B.0.1A C.0.5A D.0.25A

A

x+786799/x=0.34x

x2+786799=0.34x2
0.66x2=-786799
x2=-786799/0.66
∵ the square of a number is not equal to a negative number
The equation has no solution