The general solution of Y '' '= SiNx

The general solution of Y '' '= SiNx

y'''=sinx
Integral gain
y''=-cosx+C1
Integral gain
y'=-sinx+C1x+C2
Integral gain
y=cosx+C1x²+C2x+C3
The general solution is y = cosx + c1x & # 178; + c2x + C3

The general solution of Y '+ y = SiNx + cosx

∵ the characteristic equation of homogeneous equation y '+ y = 0 is R + 1 = 0, then r = - 1 ∵ the general solution of the homogeneous equation is y = CE ^ (- x) (C is a constant) ∵ let the solution of the original equation be y = asinx + bcosx, and substitute it into the original equation, then (a-b) SiNx + (a + b) cosx = SiNx + cos

Find the general solution of the differential equation y ′ + ycosx = (LNX) e-sinx

The given equation is a first order linear differential equation, and P (x) = cosx, q (x) = (LNX) e-sinx, so the general solution of the original equation is y = e − & nbsp; P (x) DX [& nbsp; Q (x) e & nbsp; P (x) dxpdx + C] = e − & nbsp; cosxdx [& nbsp; (LNX) e − sinxe − & nbsp; cosxdxdx + C] = e-sinx (& nbsp; lnxdx + C) = e-sinx (xlnx-x + C)

If cos α ≤ sin α is known, then the range where the terminal edge of angle α falls in the first quadrant is ()

Solution:
According to the image
knowable
The range is [2 π K, π / 4 + 2K π] K ∈ Z

If the solution set of inequality ax & # 178; + 5x + b > 0 is 1 / 3 ＜ x ＜ 1 / 2, what is ab equal to

It can be seen from the image that when a ＜ 0, there exists a specific interval (1 / 3,1 / 2), and the inequality is greater than zero
So 1 / 3 and 1 / 2 are the solution sets of ax & # 178; + 5x + B = 0
The results are as follows:

If the domain of y = f (x) is [- 2,4], then what is the domain of function f (x-1)? If the domain of y = f (x) is [- 2,4], then the domain of function f (x-1) is?

(1) Y = f (x-1) can be divided into: y = f (T) t = X-1 because f (T) and f (x) are the same function, so, - 2 ≤ t ≤ 4, that is, - 2 ≤ X-1 ≤ 4-1 ≤ x ≤ 5. Therefore, the domain of definition of function f (x-1) is: [- 1,5] (2) the domain of value of F (x) is: [- 2,4], and the image of function f (x-1) is obtained by translating function f (x) one unit to the right

Finding the definition and range of y = 4x-3 / 3x + 2

y=4x-3/3x+2
Domain: X ≠ - 2 / 3
y=[4/3(3x+2)-17/3]/(3x+2)
=4/3-17/3(3x+2)
therefore
Y ≠ 4 / 3 --- that is the range

Find the value range of the following function [2] y = (2x & # 178; + 1) / (X & # 178; + 2) [3] f (x) = {① X & # 178; + 2x (x ≤ - 1) ② X / 2-1 (x > - 1)

【2】y=(2x²+1)/(x²+2)
y=(2x²+1)/(x²+2)=[2(x²+2)-3]/(x²+2)=2-3/(x²+2)
∵x²+2≥2
The value range is (- 3 / 2, + ∞)

The range of function y = 5 (1 / x)

(0,1) ∪ (1, + infinity);
If t = 1 / x, then t ≠ 0;
y=5^t(t≠0)；
According to the image;
range
(0,1) ∪ (1, + infinity);
Have questions to ask!

The range of function y = x + 1 (x ∈ R +) is ()
A（0,1） B（0,1] C(1,+∞) D[0,1]

C X & # 178; > = 0, so Y > = 1