The general solution of differential equation d ^ 2Y / DX ^ 2 + W ^ 2Y = 0? Which (C, C1 and C2 are arbitrary constants) y = coswx y = csinwx y = c1coswx + c2sinwx Please explain a little bit

The general solution of differential equation d ^ 2Y / DX ^ 2 + W ^ 2Y = 0? Which (C, C1 and C2 are arbitrary constants) y = coswx y = csinwx y = c1coswx + c2sinwx Please explain a little bit

d^²y/dx²+w²y=0
There is an operator method in signal system
λ²+w²=0
λ = ± JW formula
y=C1coswx+C2sinwx

In this paper, we prove that y = C1 * e ^ x + C2 * e ^ (2x) (C1, C2 are arbitrary constants) is the general solution of the second order differential equation y '' - 3Y '+ 2Y = 0
Verify that y = C1 * e ^ x + C2 * e ^ (2x) (C1, C2 are arbitrary constants) is the general solution of the second order differential equation y '' - 3Y '+ 2Y = 0, and find the special solution of the equation satisfying the initial condition [y (0) = 0, y' (0) = 1]

First question:
∵y＝c1×e^x＋c2×e^（2x）,∴y′＝c1×e^x＋2c2×e^（2x）,y″＝c1×e^x＋4c2×e^（2x）.
∴y″－3y′＋2y
＝［c1×e^x＋4c2×e^（2x）］－3［c1×e^x＋2c2×e^（2x）］＋2［c1×e^x＋c2×e^（2x）］
＝（c1×e^x－3c1×e^x＋2c1×e^x）＋［4c2×e^（2x）－6c2×e^（2x）＋2c2×e^（2x）
＝0.
The general solution of the differential equation y ″ - 3Y ′ + 2Y = 0 is y = C 1 × e ^ x + C 2 × e ^ (2x)
Second question:
Let x = 0 in y = C1 × e ^ x + C2 × e ^ (2x), then C1 × e ^ 0 + C2 × e ^ 0 = C1 + C2 = 0
Let x = 0 in Y ′ = C1 × e ^ x + 2c2 × e ^ (2x), then C1 × e ^ 0 + 2c2 × e ^ 0 = C1 + 2c2 = 1
Simultaneous: C1 + C2 = 0, C1 + 2c2 = 1, it is easy to get: C1 = - 1, C2 = 1
The special solution of the differential equation satisfying the condition is y = - e ^ x + e ^ (2x)

Solving differential equation: y '& sup2; + 2XY' - X & sup2; - 4Y = 0
y’²+2xy’-x²-4y=0

y'^2+2xy'+x^2= 2x^2+4y
(y'+x)^2=2(x^2+2y)
Y '+ x = sqrt (2x ^ 2 + 4Y). Y' + x = - sqrt (2x ^ 2 + 4Y) deal with it by yourself
(y'+x)/sqrt(2x^2+4y) = 1
(sqrt(2x^2+4y))'= 2
sqrt(2x^2+4y) = 2x+2C
=> y = 1/2*x^2+2Cx+C^2

Why is the voltage in series circuit equal to u = U1 + U2

In a series circuit, there is a partial voltage,
That is, the total voltage is fixed, and the two resistors share the total voltage with the same current
Shunt circuit is a shunt circuit,
That is, the voltage is equal, the main current = the sum of the two branch currents

Two non-zero vectors a, B | a + B | = √ 3 | A-B |, in the same plane, find the value range of the angle between a and B

|A + B | = sqrt (3) | A-B |, so: | a + B | ^ 2 = 3 | A-B | ^ 2
That is: | a | ^ 2 + | B | ^ 2 + 2A · B = 3 (| a | ^ 2 + | ^ 2-2a · b)
That is: 4A · B = |a | ^ 2 + |b | ^ 2 ≥ 2 |a | * |b|
That is: cos ≥ 1 / 2, that is: ∈ [0, π / 3]

In the power supply circuit, the output voltage of the power supply is U1, the voltage lost on the line is U2, the voltage obtained by the electric appliance is U3, and the total current is I
In the power supply circuit, the output voltage of the power supply is U1, the voltage lost on the line is U2, the voltage obtained by the electric appliance is U3, the total current is I, and the total resistance of the transmission line is R. if the loss power of the line is calculated, the available formula is ()
A.I^2R B.U^2/2/R C.(U1-U3)^2/R D.I(U1-U3)
What is the relationship between U1, U2 and U3?
B should be U2 ^ 2 / R. sorry, wrong number

Loss power P = U2 * I, U2 = u1-u3, so D is correct
It can also be p = U2 ^ 2 / r = (u1-u3) ^ 2 / R, so C is also correct
It can also be p = I * I * r = I ^ 2 * r, so a is also correct
B more than one 2, incorrect
U1=U2+U3

If the angle of non-zero vector AB is 60 and | A-B | = 1, then the value range of | a + B |

|a-b|=1
|a-b|^2=1
(a-b) ^ 2 = a ^ 2 + B ^ 2-2ab = 1 = a ^ 2 + B ^ 2-2 | a | B | cos60 ° = | a | 2 + | B | 2 - | a | B | ≥ 2 | a | B | - | a | B | = | a | B |
|a||b|≤1
Then | a + B | ^ 2 = (a + b) ^ 2 = a ^ 2 + B ^ 2 + 2Ab = a ^ 2 + B ^ 2 + 2 | a | B | cos60 ° = | a | ^ 2 + | ^ 2 + | a | B | = | a | ^ 2 + | B | ^ 2 - | a | B | + 2 | a | ^ B ||
=1+2|a||b|≤1+2×1=3
And | a + B | ^ 2 = | a | ^ 2 + | B | ^ 2 + | a | B | = | a | ^ 2 + | B | ^ 2 - | a | B | + 2 | a | ^ 2 + | B | = 1
So there is 1 ≤| a + B ≤√ 3

U 1 = 1; U N = 2U (n-1) 1 / 1 u (n-1)
I want to know if the fixed point method or the feature follow the method 3Q… !

Do you understand the recurrence relation of sequence~

Given the vector p = A / | a | + B / | B |, where a and B are non-zero vectors, then the value range of | P |, is____

0=√（1²+1²-2×1×1×cos0°）≤|P|＝√（1²+1²-2cosθ）≤
√（1+1-2cos180°）＝2.
Then the value range of | P | is [0,2]

The known sequence an satisfies U1 = a (a is a positive number), u (n + 1) = - 1 / (UN) + 1, n = 1,2,3 When we ask what the value of n is, UN = a

u2=-1/u1+1=(a-1)/a
u3=-1/[(a-1)/a]+1=-1/(a-1)
u4=-1/[-1/(a-1)]+1=a
Periodic sequence
So when n = 3K + 1, K ∈ n, UN = a