When the root value of the second order constant coefficient differential equation is an imaginary number, why does the solution not contain the imaginary number I when it is represented by triangle?

When the root value of the second order constant coefficient differential equation is an imaginary number, why does the solution not contain the imaginary number I when it is represented by triangle?

Let me give you an example
For example: find the general solution of the equation y "+ y = 0
The characteristic equation corresponding to the equation is: x ^ 2 + 1 = 0 = > x = ± I
That is to say, the linear independent fundamental solution system of the equation is as follows:
u(x) = e^(i * x) = cos x + i * sin x,v(x) = e^(-i * x) = cos x - i * sin x
If we combine the linearly independent fundamental solutions, we will get the fundamental solutions,
u1(x) = [u(x) + v(x)] / 2 = cos x
v1(x) = [u(x) - v(x)] / (2i) = sin x
U1 (x), V1 (x) are still the basic solutions of the original equation
y = C1 * u1(x) + C2 * v1(x) = C1 * cos x + C2 * sin x.
In this way, there is no imaginary number I in the general solution
Given the function f (x) = 32X3 + 32x, then f (1101) + F (2101) + +f（100101）=______ .
F (x) + F (1-x) = 32X3 + 32x + 32 − 2x3 + 32 − 2x = 32X3 + 32x + 32 − 2x · 32x − 1 (3 + 32 − 2x) · 32x − 1 = 32X3 + 32x + 33 + 32x = 1, so f (1101) + F (100101) = f (2101) + F (99101) = =So f (1101) + F (2101) + +F (100101) = 50 × 1 = 50, so the answer is: 50
If the two intersection points of circle x ^ 2 + y ^ 2 + 2My + m + 6 = 0 and Y axis are on the same side of the origin, the value range of M is obtained
Why is m > 6?
x^2+y^2+2my+m+6=0
Let x = 0
have to
y²+2my+m+6=0
Because the intersection is on the same side of the origin, so
1.Δ=(2m)²-4(m+6)>0
m²-m-6>0
（m+2）（m-3）>0
M3
2.m+6>0
m>-6
therefore
The range of M is: - 6
Circle X & # 178; + Y & # 178; + 2My + m + 6 = 0
x²+(y+m)²-m²+m+6=0
The two intersections of the circle and the Y axis are on the same side of the origin
So when x = 0,
（y+m)²=m²-m+6>0
That is M & # 178; - M + 6 > 0
(m+2)(m-3)>0
So m3
x^2+y^2+2my+m+6=0
Point of intersection with y-axis
y^2+2my+m+6=0
The two intersections of y-axis are on the same side of the origin, and Y1 and Y2 are of the same sign
y1*y2>0
According to Weida's theorem
y1*y2=c/a=m+6>0
m>-6
x^2+y^2+2my+m+6=0
x^2+(y+m)^2=m^2-m-6
m^2-m-6>0
(m-3)(m+2)>0
m>3 m
Given the function f (x) = 32X3 + 32x, then f (1101) + F (2101) + +f（100101）=______ .
F (x) + F (1-x) = 32X3 + 32x + 32 − 2x3 + 32 − 2x = 32X3 + 32x + 32 − 2x · 32x − 1 (3 + 32 − 2x) · 32x − 1 = 32X3 + 32x + 33 + 32x = 1, so f (1101) + F (100101) = f (2101) + F (99101) = =So f (1101) + F (2101) + +F (100101) = 50 × 1 = 50, so the answer is: 50
The minimum value of function y = sin (x - π 6) cosx______ .
The minimum value of y = sin (x - π 6) cosx = (32sinx-12cosx) cosx = 32sinxcosx-12cos2x = 34sin2x − 14 (cos2x + 1) = 12sin (2x − π 6) - 14  y = sin (x - π 6) cosx is: − 12 − 14 = − 34, so the answer is: - 34
Given the function f (x) = 32X3 + 32x, then f (1101) + F (2101) + +f（100101）=______ .
F (x) + F (1-x) = 32X3 + 32x + 32 − 2x3 + 32 − 2x = 32X3 + 32x + 32 − 2x · 32x − 1 (3 + 32 − 2x) · 32x − 1 = 32X3 + 32x + 33 + 32x = 1, so f (1101) + F (100101) = f (2101) + F (99101) = =So f (1101) + F (2101) + +F (100101) = 50 × 1 = 50
Simplify the square of root 6sinx / 2cosx / 2 + root 2cosx / 2
The original formula = √ 6 / 2 * SiNx + √ 2 / 2 * (cosx + 1) = √ 2 [√ 3 / 2 * SiNx + 1 / 2 * cosx) + √ 2 / 2 = √ 2Sin (x + π / 6) + √ 2 / 2
(1) It is known that y = (2m-1) x ^ m ^ 2-3 is a positive proportional function, and Y decreases with the increase of X
(2) It is known that y = (2m-1) x ^ m ^ 2-3 is a positive proportion function, and the function image passes through the first and third quadrants to find the value of M
(3) It is known that y = (2m-1) x + m ^ 2-4 is a positive proportional function, and Y decreases with the increase of X
It's all up here,
It is proved by Lagrange mean value theorem that the absolute value of SiNx siny is less than or equal to the absolute value of X-Y
f(x)=sin(x)
Endpoints X and Y
sinx-siny=cos(ξ)*(x-y)≤x-y
Given that the coefficient of y = (2m-1) x is m05-3, which is a positive proportional function, and Y decreases with the increase of X, what is the value of M
m=-2 ∵m²-3=1
m²=4
M = plus or minus 2
∵ y decreases with the increase of X
So 2m-1 < 0
2m＜1
m＜1/2
∵ M = plus or minus 2, m < 1 / 2
Ψ M = 2 tut tut