Let F X = e Λ x-k / 2x Λ 2-x 1, if k = 0, find the minimum value of F X 2, if x ≥ 0 Let F X = e Λ x-k / 2x Λ 2-x If k = 0, find the minimum value of FX 2 if FX ≥ 1 when x ≥ 0, find the value range of real number K

Let F X = e Λ x-k / 2x Λ 2-x 1, if k = 0, find the minimum value of F X 2, if x ≥ 0 Let F X = e Λ x-k / 2x Λ 2-x If k = 0, find the minimum value of FX 2 if FX ≥ 1 when x ≥ 0, find the value range of real number K

(1) The function f (x) = ex – (K / 2) x2 – x, when k = 0, f (x) = ex – x, the derivative can be f '(x) = ex – 1;
1) When x < 0, ex < E0 = 1, so f '(x) = ex – 1 < 0, then f (x) decreases monotonically. Calculate f (0) = E0 – 0 = 1 (function f (x) decreases monotonically from + ∞ to 0 on X ∈ (- ∞, 0]);
2) When x > 0, ex > E0 = 1, so f '(x) = ex – 1 > 0, then f (x) monotonically increases (the function f (x) monotonically increases from 0 to + ∞) on X ∈ [0, + ∞);
In conclusion, when k = 0, if and only if x = 0, the minimum value of F (x) is 1
(2) Let f (x) = e ^ X - (K / 2) x ^ 2-x - 1, then f '(x) = e ^ x-kx-1,
When k - = 1, the curve y = e ^ X and the line y = KX + 1 are tangent to the point (0,1),
So K ≤ 1
(when x ≥ 0, f '(x) = e ^ x-kx-1 ≥ 0)

Find the general solution of differential equation: y '- (y') 3-y '= 0, where 3 is the third power,

Let y '= P, then y' '= PDP / dy
Substituting into the original equation, we get that PDP / DY-P & sup3; - P = 0 = = > P (DP / DY-P & sup2; - 1) = 0
∴p=0,dp/dy-p²-1=0
∵ when p = 0, dy / DX = 0 = = > y = C (C is an integral constant)
The test shows that y = C is the solution of the original equation
When DP / DY-P & sup2; - 1 = 0, there is DP / dy = P & sup2; + 1
==>dp/(p²+1)=dy
==>Arctap = y + C1 (C1 is the integral constant)
==>p=tan(y+C1)
==>dy=tan(y+C1)dx
==>cos(y+C1)dy/sin(y+C1)=dx
==>d(sin(y+C1))/sin(y+C1)=dx
==>Ln │ sin (y + C1) │ = x + ln │ C2 │ (C2 is an integral constant)
==>sin(y+C1)=C2e^x
Therefore, sin (y + C1) = c2e ^ x is also the solution of the original equation
So the general solution of the original differential equation is y = C, or sin (y + C1) = c2e ^ x (C, C1, C2 are integral constants)

Y '- 2Y / x = X3 (the third power of x) for solving first order linear differential equations

Divide both sides by x ^ 2
y'/(x^2)-(2/x^3)y=x
General division
(xy'-2y)/(x^3)=x
[y/(x^2)]'=x
integral
y/(x^2)=(1/2)x^2+C
y=(1/2)x^4+Cx^2

Let y '+ P (x) y = q (x) have two different solutions, a (x), B (x) and C are arbitrary constants?
The answer should be a (x) + C [a (x) - B (x)]. Why does a (x) - B (x) correspond to the general solution of the homogeneous differential equation y '+ P (x) y = 0 which is not always zero? How does the answer come out?
This is the 10th question of the third year of postgraduate entrance examination in 2006

Non homogeneous linear differential equation with constant coefficients Y "- 3Y '+ 2Y = x * e ^ X-2
General solution

It can be divided into homogeneous solution and special solution y '' - 3Y '+ 2Y = 0 characteristic equation: T ^ 2 - 3T + 2 = 0 = = > t = 1 or 2 = = > y = c1'e ^ x + c2'e ^ (2x) y = x (AX + b) e ^ - XY' = [- ax ^ 2 + (2a-b) x + b] e ^ - x y '' = [ax ^ 2 - (4a-b) x + 2a-2b] e ^ - x is brought into the equation

Is y ″ + 3Y ′ - 2Y = e ^ (x ^ 2) a second order differential equation?

This is a second order nonhomogeneous differential equation with constant coefficients

General solution of differential equation (1 + x ^ 2) y '+ 2XY = 1

The equation is as follows
[(1+x^2)y]'=1
Integral: (1 + x ^ 2) y = x + C
Then y = (x + C) / (1 + x ^ 2)

The general solution of differential equation 2XY - (x ^ 2 + y ^ 2) y '= 0

Let y = XT, then dy = XDT + TDx, let dy = XDT + TDx, then, by substituting into the original equation, we get the 2xydx - (x ^ 2 + y ^ 2) dy = 0 = = (2xydx - (x ^ 2 + y ^ 2 + y ^ 2) dy = 0 = (2xydx - (x ^ 2 + y ^ 2 + y ^ 2) dy = 0 = (2xydx - (x ^ 2 + y ^ 2 + y ^ 2) in the original equation, we get the 2xydx - (x - (x ^ 2 + 2 + 2 + y ^ 2 + y ^ 2) dy = 0 = 0 = (2x \\\\35\\35\\\\\\\\\\\\\\-t & # 178

The general solution of the differential equation y '= (y + xlnx) / X is ()
Please write the detailed process. Thank you

This problem belongs to the first order linear non-homogeneous differential equation, using the constant variation method. Please refer to section 4, Chapter 7, Volume I, advanced mathematics (Sixth Edition), Tongji University

XDY = (y + xlnx) DXG for general solution of differential equation

y'=(y+xlnx)/x
y'=(y/x)+lnx
Let y = Xu
Then y '= u + Xu'
Substitution equation: U + Xu '= u + LNX
xu'=lnx
du=lnxdx/x
du=lnxd(lnx)
Integral: u = (LNX) ^ 2 / 2 + C
That is y = Xu = x (LNX) ^ 2 / 2 + Cx