Find y "- 2Y '+ 5Y = (e ^ x) sin (2x)

Find y "- 2Y '+ 5Y = (e ^ x) sin (2x)

The characteristic equation of homogeneous equation is R ^ 2-2r + 5 = 0r = (2 ± 2I) / 2 = 1 ± I (I is the virtual unit), so the homogeneous general solution is y = e ^ x (c1cosx + c2sinx). Let the non-homogeneous special solution be y = e ^ x (asin2x + bcos2x) y '= e ^ x (asin2x + bcos2x) + e ^ x (2acos2x-2bsin2x) = e ^ x [(a-2b) sin2x + (2a + b) cos2x] y' '

Given that y = x, y = e ^ x, y = e ^ - X are three solutions of a second order non-homogeneous linear differential equation, then the general solution of the differential equation is?

Let y = ex (c1sinx + c2cosx) (C1, C2 are arbitrary constants) be the general solution of a second order linear homogeneous differential equation with constant coefficients, then the equation is______ ．

From the form of the general solution, we can know that the two roots of the characteristic equation are R1, 2 = 1 ± I, thus we can know that the characteristic equation is (r-r1) (r-r2) = R2 - (R1 + R2) r + r1r2 = R2 - 2R + 2. Therefore, the differential equation is y ″ - 2Y ′ + 2Y = 0. So the answer is y ″ - 2Y ′ + 2Y = 0

How to calculate the number of monomials

The sum of exponents of all letters (except π) in a monomial

Let x ∈ [0, π / 2], if the equation 3sin (2 + π / 3) = a has two solutions, find the value range of real number a

The intersection of 3sin (2x + π / 3) and y-axis is (0,3 √ 3 / 2), and the maximum value is 3

The equation | X-8 | = a of X has two different real roots in the interval [7,9]. Find the value range of real number a

Let y = | X-8 |, the problem can be transformed into the intersection problem of y = | X-8 | and y = a in the interval [7,9]. It is easy to find that x = 8 is the axis of symmetry of y = | X-8 |. When x equals 7 or 9 is y = 1, we can see the value range of a (0,1] from the image

It is known that if √ (1-cos α) / (1 + cos α) = (COS α - 1) / sin α holds, the value range of α is

Because √ (1-cos α) / (1 + cos α)
＝√(1-cosα)²/(1－cos²α)
＝√(1-cosα)²/sin²α
＝(1-cosα)/|sinα|
So (1-cos α) / | sin α | = (COS α - 1) / sin α
Thus sin α

What is the value range of α if (1-cos α / 1 + cos α) = (COS α - 1) / sin α holds?

The root sign should be greater than zero, (COS α - 1) / sin α should be greater than zero,
cosa-1＜0,sinα＜0,
1 + Cosa ≠ 0, to sum up the solution is OK!

In (0,2 π), the value range of α which makes sin α > cos α hold is
(A)（π/4,π/2） (B) ( π/4,π ）
(C) (π/4,5π/4） (C) ( π/4,π ）∪（5π/4,3π/2）

C

Cos (α + β) can be obtained when α is an acute angle and sin α = three fifths, β is an obtuse angle and sin β = half
sin（α-β）

Because Sin & # 178; α + cos & # 178; α = 1,
So (3 / 5) &# 178; + cos & # 178; α = 1
COS²α=±4/5
Because α is an acute angle
therefore
COSα=4/5
And because Sin & # 178; β + cos & # 178; β = 1, sin β = 1 / 2
So (1 / 2) &# 178; + cos & # 178; β = 1
therefore
COSβ=±√3/2
Because β is an obtuse angle
therefore
COSβ=－√3/2
So cos (α + β) = cos α cos β - sin α sin β = 4 / 5 × (- 3 / 2) - 3 / 5 × 1 / 2
= =-(3+4√3)/10