LIM (f (1 + SiNx) - f (1)) / sinx-3 (f (1-sinx) - f (1)) / SiNx = 8

LIM (f (1 + SiNx) - f (1)) / sinx-3 (f (1-sinx) - f (1)) / SiNx = 8

According to the definition of derivative (omit △ x → 0 or SiNx → 0 below LIM): Lim [f (a + △ x) - f (a)] / △ x = f '(a), so the formula of the problem can be reduced to the original formula = Lim [f (1 + SiNx) - f (1)] / SiNx + 3 [f (1 + (- SiNx)) - f (1)] / (- SiNx) = f' (1) + 3F '(1) = 4f' (1) = 8, so f '(1) = 2

Find the value of X -- > 0, Lim [SiNx / x-x]
I get one, and the answer is half

If the answer is correct, the answer is wrong

If the nonzero function f (x) has f (a + b) = f (a) &; f (b) for any real number a and B, and if x < 0, f (x) > 1
It is proved that f (x) is a decreasing function on (- ∞, 0)

Let x 10 so f (x 1) > F (x 2) be a decreasing function Please accept!

On the problem of solving second order non homogeneous linear differential equation with constant coefficient in postgraduate entrance examination mathematics
It is known that the equation is a second-order non-homogeneous linear differential equation with constant coefficients, and it has two special solutions: Y1 = cos2x-1 / 4xsin2x, y2 = sin2x-1 / 4xsin2x. Now the expression of this equation is required. Let the general solution of this equation be y = C1 * cos2x + C2 * sin2x - 1 / 4xsin2x, where C1 and C2 are arbitrary constants, Cos2x and sin2x should be the special solution of the corresponding homogeneous differential equation, and - 1 / 4xsin2x should be a special solution of this equation, but the title does not give these two conditions,

y1 = cos(2x) - 1/4 sin(2x)
y2 = sin(2x) - 1/4 sin(2x)
Then y 1-y 2 = cos (2x) - sin (2x) is the solution of the corresponding homogeneous differential equation
Therefore, the two eigenvalues must be conjugate. The solution of the corresponding homogeneous differential equation must be C1 cos (2x) + C2 sin (2x)

Do you want a second order differential equation with constant coefficients for postgraduate entrance examination?
Homogeneous and nonhomogeneous

Remember the formula that is actually the most simple in all the chapters of advanced mathematics. The real problem is that you will have fun when you meet in the exam

Second order linear differential equations with constant coefficients
In the step of undetermined coefficient, if y * is set, can we find y "Y 'and substitute it into the original equation? If so, please look at this problem
y“+3y'+2y=（3x）Xe（-x）
It's a single one
Let y * = x (AX + b) e (- x) be substituted into the original equation
（2ax+2a+b）=3x
What's wrong with me getting the first and second derivatives and taking them in?

∵ the characteristic equation of the homogeneous equation y '' + 3Y '+ 2Y = 0 is R & sup2; + 3R + 2 = 0, then R1 = - 1, R2 = - 2 ∵ the general solution of the homogeneous equation is y = C1E ^ (- x) + c2e ^ (- 2x) (C1, C2 are integral constants). Let the special solution of the original equation be y = x (AX & sup2; + BX + C) e ^ (- x) ∵ y' = (3ax & sup2; + 2bx + C) e ^ (...)

(x+y)dx+(3x+3y-4)dy=0

Let u = x + y, then Du / DX = 1 + dy / DX, i.e. dy / DX = Du / DX - 1. The original equation can be changed to dy / DX = - (x + y) / (3x + 3y-4), i.e. Du / DX - 1 = - U / (3u-4). This is already a differential equation with separable variables. (3u-4) / (2u-4) Du = DX, integrating on both sides, and simplifying (U brings back x + y), x + 3Y + ln (x + Y-2) ^ 2

Have mathematics differential equation kind of question anxious inquiry
Dy / DX ytanx = cosx.. what happens when McLaughlin series is expanded to x ^ 2, if y equals negative half π and x equals 0? In addition, when x equals π and Y equals 0,
Don't mean, integrity! Please provide the next process

Solving differential equation dy / DX ytanx = cosx
Let u = ycosx, i.e. y = u / cosx, then
dy/dx=[cosx(du/dx)+usinx]/cos²x=(1/cosx)(du/dx)+(utanx)/cosx=(1/cosx)(du/dx)+ytanx
Substituting into the original equation, (1 / cosx) (DU / DX) = cosx
So Du / DX = cos & sup2; X; Du = cos & sup2; xdx
u=∫cos²dx=(1/2)∫(1+cos2x)dx=(1/4)∫(1+cos2x)d(2x)=(1/4)(2x+sin2x)=(x+sinxcosx)/2
So y = [(x + sinxcosx) / 2cosx + C

The problem of mathematical differential equation
The abscissa of the intersection of the tangent line and the transverse axis at any point P (x, y) on the curve is equal to half of the abscissa of the tangent point
Write the differential equation and tell me how to do it

Tangent slope k = f '(x0)
Tangent y-f (x0) = f '(x0) (x-x0)
When y = 0
x=x0-f(x0)/f'(x0)
So x-f (x) / F '(x) = x / 2
So f (x) / F '(x) = x / 2

On mathematical differential equation
Find the differential equation x * d2y / DX2 + dy / dx-x = 0, satisfy the initial condition x = 1, y = 5 / 4, x = 1, y derivative = 3 / 2

Let z = y ', have XZ' + z-x = 0
So (xz-x ^ 2 / 2) '= 0
xz-x^2/2=C1
Because when x = 1, z = 3 / 2, so C1 = 1
xy'-x^2/2=1
y'=x/2+1/x
y=x^2/4+lnx+C2
The condition is C2 = 1
therefore
y=x^2/4+lnx+1