For y '+ Y & # SiNx = 0

For y '+ Y & # SiNx = 0

y'/y^2=-sinx
Simultaneous integral of the above equation
-1 / y = cosx + C (C is constant)
y=-1/(cosx+c)

The general solution of Y '' - 2Y '+ y = SiNx + e ^ x

The characteristic equation of ∵ y '' - 2Y '+ y = 0 is R & # 178; - 2R + 1 = 0, then r = 1
The general solution of ∧ y '' - 2Y '+ y = 0 is
Y = (c1x + C2) e ^ x (C1, C2 are integral constants)
Let y '' - 2Y '+ y = SiNx + Xe ^ X be a special solution
y=Acosx+Bsinx+Cx³e^x
∴ y'=-Asinx+Bcosx+3Cx²e^x+Cx³e^x
y''=-Acosx-Bsinx+6Cxe^x+6Cx²e^x+Cx³e^x
Put it into the original equation
2Asinx-2Bcosx+6Cxe^x=sinx+xe^x
The solution is: a = 1 / 2, B = 0, C = 1 / 6
The special solution of ∧ y '' - 2Y '+ y = SiNx + Xe ^ x is
y=(1/2)cosx+(1/6)x³e^x
So the general solution of Y '' - 2Y '+ y = SiNx + Xe ^ x is
Y = (c1x + C2) e ^ x + (1 / 2) cosx + (1 / 6) x & # 179; e ^ x (C1, C2 are integral constants)

Find the general solution or special solution of Y '+ Y / x = SiNx / X (where y (PAI) = 1)
It's urgent. Thank you

y'+y/x=sinx/x
therefore
xdy+ydx=sinxdx
There are two points on both sides
xy=-cosx+C

It is known that X1 and X2 are the two roots of the equation x ^ 2-mx-4 = 0, and LG (x1 + x2) / (lgx1 + lgx2) = 2 is the value of M

X1 and X2 are the two roots of the equation x ^ 2-mx-4 = 0
According to Weida's theorem
x1+x2=m
x1x2=-4
lg(x1+x2)/(lgx1+lgx2)=2
x1+x2=2lg(x1x2）
x1+x2=(x1x2)²
m=16

Let the real roots of x ^ 2-4x = m ^ 2 + 1 = 0 be α, β, and determine the range of real number m such that the absolute value α + absolute value β ≤ 5

The real roots of x ^ 2-4x-m ^ 2 + 1 = 0 are α, β
Then α + β = 4 α * β = 1-m ^ 2
α^2+β^2=(α+β)^2-2αβ=16-2(1-M^2)=14+2M^2
Absolute value α + absolute value β = √ (I α I + I β I) ^ 2 = √ [α ^ 2 + 2I α β I + β ^ 2]
=√(14+2M^2+2I1-M^2I)≤5
That is 2m ^ 2 + 2i1-m ^ 2I ≤ 11
(1) When 1-m ^ 21, 2m ^ 2 + 2m ^ 2-2 ≤ 11, 4m ^ 2 ≤ 13 - √ 13 / 2 ≤ m ≤ √ 13 / 2
That is - √ 13 / 2 ≤ M

Given that K is a nonnegative real number, the equation of X is: ① X & # 178; - (K + 1) x + k = 0. ② KX & # 178; - (K + 2) x + k = 0. When finding the value of K, the two equations have the same real number root?

Equation 1 (x-1) (x-k) = 0, get x = 1, K if the common root is x = 1, substitute into equation 2, get k-k-2 + k = 0, get k = 2 if the common root is x = k, substitute into equation 2, get k ^ 3-K (K + 2) + k = 0, namely K (k ^ 2-k-1) = 0, get k = 0, (1 + √ 5) / 2, (1 - √ 5) / 2, because K is non negative, synthesize three K values: 2,0, (1 + √ 5) /

If the equation of X: X * 2 (the square of 2) - 2 / X / + 2 = m has exactly three different real solutions, please guess the value of X and verify it
There should be a process! Thank you! Please!

When x ≥ 0
(x-2)^2=m+2
x=±[√(m+2)]+2
When x ≤ 0
(x+2)^2=m+2
x=±[√(m+2)]-2
The only case where there are three roots is
-[√(m+2)]+2=[√(m+2)]-2=0
here
√(m+2)=2
m=2

If there are exactly three different real solutions to the equation X-2 / X / + 2 = m, please guess the value of M and verify it
Write clearly and answer`

Guess M = 2
M = 2 when 1 x = 0
M > = 1 when 2 x > 0
X = (2 + ⊿) / 2 or x = (2 - ⊿) / 2
3 x=1
X = (- 2 + ⊿) / 2 or x = (- 2 - ⊿) / 2
3 different real solutions, so ⊿ = 2
Here M = 2
Three different real number solutions are x = 0, x = 2, x = - 2

If the equation 9 ^ - absolute value (2-x) - 4 * 3 ^ - absolute value (2-x) + a = 0 has a real solution, find the value range of A

9^ ( - |2-x| ) - 4 * 3^( -|2-x| ) + a = 0
[ 3^ ( - |2-x| ) ]^2 - 4 * 3^( -|2-x| ) + a = 0
Let t = 3 ^ (- |2-x|)
t^2 - 4t + a = 0
(t-2)^2 - 4 + a=0
a = 4 - (t-2)^2
∵- |2-x| ≤ 0
∴0＜3^( - |2-x| ≤ 1
That is: 0 < T ≤ 1
∴-2 ＜ t-2 ≤ -1
4＞ (t-2)^2 ≥ 1
0＜ 4 - (t-2)^2 ≤ 3
That is: 0 < a ≤ 3

If the equation | x | X-2 = KX has three unequal real roots, then the value range of real number k is___ ．

When x ≥ 0: kx2-2kx = xkx2 - (2k + 1) x = 0 〈 X1 = 0, X2 = 2K + 1K ＞ 0 〉 K ＜ - 12 or K ＞ 0 when x ＜ 0: kx2-2kx = - xkx2 - (2k-1) x = 0 〉 x = 2k-1k ＜ 0 〉 K ＜ - 12 or K ＞ 0 when x ＜ 0: kx2-2kx = - xkx2 - (2k-1) x = 0 〉 x = 2k-1k ＜ 0 〉 K ＜ 12. In conclusion, the roots of the equation are positive and negative, one is 0, and the range of K is (0, 12)