Let y = E2x + (1 + x) ex be a special solution of the second order linear differential equation y ″ + α y ′ + β y = γ ex with constant coefficients. Try to determine the constants α, β and γ and find the general solution of the equation

Let y = E2x + (1 + x) ex be a special solution of the second order linear differential equation y ″ + α y ′ + β y = γ ex with constant coefficients. Try to determine the constants α, β and γ and find the general solution of the equation

Y ′ = 2e2x + (2 + x) ex, y ″ = 4e2x + (3 + x) ex are obtained from y = E2x + (1 + x) ex. by substituting y, y ′, y ″ into the original differential equation, we can get: (4 + 2 α + β) E2x +(1 + α + β) xex + (3 + 2 α + β - γ) ex = 0, because: y = E2x + (1 + x) ex is a special solution of the equation, so for any defined x, the formula is always valid, so: 4 + 2 α + β = 01 + α + β = 03 + 2 α + β - γ = 0. The solution is: α = - 3, β = 2, γ = - 1, so the specific expression of the original differential equation is: y ″ - 3Y ′ + 2Y = - ex, and the characteristic equation of the corresponding homogeneous equation is: λ 2-3 λ+ 2 = 0, get the eigenvalue as: λ 1 = 1, λ 2 = 2, corresponding to the general solution of homogeneous equation is. Y = c1ex + c2e2x, and because: the non-homogeneous term is - ex, and λ = 1 is the eigenvalue, so: let the special solution of the original differential equation be y * = axex, and substitute it into the original differential equation to get: a = 1, so: y * = xex. According to the structure theorem of the solution of linear differential equation, the general solution of the original equation is: y =. Y + y * = c1ex + c2e2x + xex

The special solution form of differential equation y "+ y = SiNx can be set as y * = x (dsinx + ecosx). Why are I and - I the characteristic roots of SiNx?

The characteristic equation is R & # 178; + 1 = 0
We obtain r = I, - I
Its corresponding characteristic term is c1sinx + c2cosx

Solution of Y '' (x) + y (x) = SiNx differential equation
Such as the title

1 general solution R ^ 2 + 1 = 0 C1 * SiNx + C2 * cosx2 special solution 1 / (d ^ 2 + 1) * SiNx = im (1 / (D + I) / (D-I) * exp (IX)) = im (exp (IX) / 2I / D * 1 = im (x * exp (IX) / (2I)) = im (x (cosx + isinx) / (2I)) = - x * cos (x) / 2 you can verify diff (- x * cos (x) / 2,2) + - x * cos (x) / 2) = sin (

How to solve LIM (when x tends to 0) (Tan 3x) / x?

TaNx is equivalent to X. when x tends to zero and the denominator is multiplied by 3, it is equal to 3

Calculation (- 1) - (- 2 / 3) * (+ 9 / 4)

Original formula = (- 1) - (- 3 / 2)
=-1 + 3 / 2
=1 / 2

How much does LIM (3x / Tan 3x) x approach 0?
The result is 3,

Is it true that you are wrong? No matter how you calculate it, the value of this limit is 1
lim(3x/tan3x)
=lim(3x/3x) (tan3x 3x ,（x-->0)）
=1 （x-->0)
It can also be dealt with in this way: the limit of 0 / 0 infinitive is determined by the law of Robida
lim(3x/tan3x）
=lim(3/3sec^2 x)
=lim cos^2 3x
=1

1.87 10 2 and 30 / 5 1.13 10 2.4 simple calculation

Original formula = (1.87 + 1.13) + (2.6 + 2.4) = 3 + 5 = 8

Lim molecule is y, denominator is (ln1 + y) = (x tends to 0) process

lim(y-->0) y/ln(1+y)
=lim(y-->0)1/[1/y*ln(1+y)]
=lim(y-->0)1/[ln(1+y)^(1/y)]
=LIM (T - > ∞) 1 / [ln (1 + 1 / T) ^ t] [let 1 / y = t]
=1/e

Calculate (a + 3) (A-1) - A (A-2)

(a+3)(a-1)-a(a-2)
=a^2+3a-a-3-a^2+2a
=4a-3

LIM (x → 4) numerator [√ (1 + 2x) - 3] denominator [√ X-2] calculation problem, to process, thank you

lim(x→4) [√(1+2x)-3] / [√x-2]=lim(x→4) [√(1+2x)-3] [√x+2]/ [√x-2][√x+2]=lim(x→4) 4[√(1+2x)-3]/(x-4)=lim(x→4) 4[√(1+2x)-3][√(1+2x)+3]/(x-4)[√(1+2x)+3]=lim(x→4) 8(x-4) / (x-4)[√(1+2x)+3]...