Finding the general solution of the second order constant coefficient non second order linear differential equation y '' + 3Y '+ 2Y = 3sinx

Finding the general solution of the second order constant coefficient non second order linear differential equation y '' + 3Y '+ 2Y = 3sinx

The characteristic equation is R ^ 2 + 3R + 2 = 0 (R + 2) (R + 1) = 0r = - 2, and the general solution of - 1 homogeneous equation is Y1 = C1E ^ (- 2x) + c2e ^ (- x). Let y * = asinx + bcosxy * '= acosx bsinx, y * "= - asinx bcosx be substituted into the original equation: - asinx bcosx + 3acosx-3bsinx + 2asinx + 2bcosx = 3sinx (a-3b) SiNx + (...)

Find the general solution of the second order linear non-homogeneous differential equation y '' - y = x ^ 2 with constant coefficients,

General solution of nonhomogeneous differential equation = general solution of corresponding homogeneous differential equation + special solution
The solution process is roughly divided into the following two steps
1. To find the general solution of the corresponding homogeneous differential equation y '' - y = 0... (1), the characteristic equation of equation (1) is R ^ 2-1 = 0, then r = 1, - 1, so the general solution of equation (1) is y = CE ^ x + de ^ (- x), C and D are the quantities to be solved. Here, we need two boundary conditions, I don't know if there are any, that is, f (0) = a, f '(0) = B, a and B are known, which are used to bring in the general solution to determine the quantities to be solved, C and D, otherwise it can't be solved
2. Assuming that the required conditions in the first step are known, we can find the special solution now. We construct a special solution with parameters (undetermined coefficient method), bring it into the original equation, and then we can solve the coefficient according to the comparison of similar terms. Here we construct the following undetermined special y = A0 + A1 * x + A2 * x ^ 2, bring it into the original equation, and then we can get A0, A1, A2, so we can find the special solution

What is the general solution of the differential equation y '= x ^ 2

The general solution of Y '= x ^ 2 is y = 1 / 3 x ^ 3 + C (C is a constant)
The general solution of Y '' - 3Y '= 0 is y = e ^ 3x + C or y = C (C is a constant)

Why is the series of ∑ 1 / N ^ 2 convergent
rymidni
1/n^2 = 2(1-1/n^2)
Where does this come from?

This is a question that several mathematicians once asked Euler
The results can be obtained by the Maclaurin expansion of sine
That is, Σ 1 / N ^ 2 = the square of Pie / 6

It is known that the function f (x) is an increasing function, the domain of definition is [0,3], and f (x-1)

Because the function f (x) is an increasing function
So 1-X-1
Because f (x) is defined as [0,3]
So 0=

Proving series ∑_ (n = 1) ^ ∞ &; (Sin &; (NA)) / N ^ 4 absolute convergence

|sin⁡(na)|

If a and B satisfy 3 √ a + 5 √ B = 7, then the value range of S = 2 √ A-3 ∣ B ∣?

√a≥0 √b ≥0
0≤5√b≤7
0≤√b≤7/5
3√a+5√b=7
√a=(7-5√b)/3
S=2√a-3∣b∣
=2(7-5√b)/3-3(√b)²
=-3(√b)²-10√b/3+14/3
=-3(√b+5/9)²+151/27
When √ B = 0, s has the maximum value Smax = 14 / 3
When √ B = 7 / 5, that is, when B = 49 / 25, s has the minimum value smin = - 147 / 25
The range of S is [- 147 / 25,14 / 3]

It is proved that LIM (2n)! / A ^ (n!) = 0 (a > 1)
How to explain (2n + 2) (2n + 1) / A ^ (n + 1) on the first floor

An = (2n)! / A ^ (n!) A1 = 2 / a it is easy to know that an > 0 and a (n + 1) / an = (2n + 2) (2n + 1) / A ^ (n + 1) exist n such that when n > n (large enough) a (n + 1) / an = (2n + 2) (2n + 1) / A ^ (n + 1) 1 = > A = 1 + Ba ^ (n + 1) = (1 + b) ^ (n + 1) = 1 + b * (n + 1) + B ^ 2 * (n + 1) n / 2 + B ^ 3 * (n + 1) n (n-1) / 6 +... (2n + 2) (2n + 1)

If Sn = nm, SM = Mn (m ≠ n), then the value range of Sn + m is______ ．

Because Sn = n (a1 + an) & nbsp; 2 = n [2A1 + (n − 1) D] 2 = nm ①, SM = m (a1 + AM) 2 = m [2A1 + (m − 1) D] 2 = Mn ②, ① - ② get: (n-m) D = 2 (n − m) Mn, from m ≠ n, get: D = 2Mn, substitute d into the solution of ① to get: A1 = 1MN, then Sn + M = (M + n) (a1 + am + n) & nbsp; 2 = (M + n) [2A1 + (M + n − 1)

Brother, using the necessary condition of series convergence, it is proved that LIM n →∞ / N ^ n = 0

an=n!/n^n
Then LIM (n →∞) a (n + 1) / an
=lim(n→∞){(n+1)!/[(n+1)^(n+1)]}/[n!/(n^n)]
=lim(n→∞)(n^n)/[(n+1)^n]
=lim(n→∞)= 1/[(1+1/n)^n]
=1/e