Aunt Li folded a rope in half and measured a 1-meter-long cloth. It turned out that it was 3 decimeters longer than this cloth. Do you know this one A simple algorithm

Aunt Li folded a rope in half and measured a 1-meter-long cloth. It turned out that it was 3 decimeters longer than this cloth. Do you know this one A simple algorithm

The length of the folded rope is 10 decimeters + 3 decimeters = 13 decimeters
So the whole length is 13x2 = 26 decimeters

The differential of function y = x ^ 2 at x = 1

ey/ex=2x
ey/ex|(x=1)=2*1=2
∴dy=ey/ex dx
=2dx

A round desktop radius of 40 cm, with a square tablecloth cover on the desktop, so that all around the hanging 5 cm, tablecloth area

40 * 2 + 5 + 5 = 90 cm, 90 * 90 = 8100 square cm

(1-x) y '' + XY '- y = 1 a solution of homogeneous equation is known and Y1 = x a general solution of differential equation is obtained

-1 is a special solution
Y = UX, y = u + Xu 'y = u' + 2U 'substituting (1-x) y' '+ XY' - y = 0
(1-x)(u’‘+2u‘)+x(u+xu‘)-ux=0
(1-x)u’‘+(x^2-2x+1+1)u‘=0
u'=ln(1-x)+x^2/2-x+C1
u=(x-1)ln(1-x)+(1-x)+x^2/2-x+C1x+C2
y=x[(x-1)ln(1-x)+(1-x)+x^2/2-x+C1x+C2]-1

The table top is 160cm long and 100cm wide. My mother is going to design a tablecloth, which is twice the size of the table
Why is the length of the desktop 160 + 2x and the width 100 + 2x?

The table cloth is the same length as the width, so it needs + 2x

Let Y1 = Xe ^ x + e ^ (2x), y2 = Xe ^ x + e ^ (2x) - e ^ (- x), Y3 = Xe ^ x + e ^ (- x) be the solution of a second order linear nonhomogeneous equation. Find the general solution of the equation
Why do we only write y1-y2 and y1-y3, but not y2-y3 when we write the solution of the corresponding homogeneous equation?

It can also be y2-y3 and y2-y1. That is to say, as long as the results of these three special solutions are not linearly correlated, they can be used as the structure of solutions of homogeneous equations. But because they are second-order equations, only two are needed, so y2-y3 is not needed

A table top is 160cm long and 100cm wide. Now we are going to design a tablecloth, which is twice the area of the table top, and the hanging edge of the tablecloth is the same width
What is the width of the side hanging down? (accurate to 0.1) the answer should be specific

(160+X)*(100+X)=160*100*2
The solution is x ≈ 51.4
Therefore, the width of droop around is 51.4 / 2 ≈ 25.7

Let the linearly independent functions Y1, Y2 and Y3 be the solutions of the second order non-homogeneous linear equation y ″ + P (x) y ′ + Q (x) y = f (x), C1 and C2 be arbitrary constants, then the general solution of the non-homogeneous equation is ()
A. c1y1+c2y2+y3B. c1y1+c2y2-（c1+c2）y3C. c1y1+c2y2-（1-c1-c2）y3D. c1y1+c2y2+（1-c1-c2）y3

Because: Y1, Y2, Y3 are linearly independent, so: y1-y3, y2-y3 are linearly independent. And because: the functions Y1, Y2, Y3 are the solutions of the second order non-homogeneous linear equation y ″ + P (x) y ′ + Q (x) y = f (x), so: C1 (y1-y3) + C2 (y2-y3) is the general solution of Y ″ + P (x) y ′ + Q (x) y = 0. According to the structure of the second order linear non-homogeneous differential equation, we can know: C1 (y1-y3) + C2 (Y2)- Y3) + Y3 = c1y1 + c2y2 + (1-c1-c2) Y3 is the general solution of Y ″ + P (x) y ′ + Q (x) y = f (x)

A dining table, the desktop is 160cm long, 100cm wide rectangle, now ready to design a tablecloth, the area is twice the desktop, and tablecloth around the hanging edge
What is the width of the hanging edge? (accurate to 0.1cm)
The length and width of a rectangular grassland are 20 m and 15 m respectively. There is a path with equal width around it. The area of the path is 246 M & sup 2;. The side length and width of the path are calculated

Let the side length of the path be X
（20+2X)X（15+2X）—20X15=246
Ask for the answer

It is known that the three special solutions of two medium linear homogeneous differential equation are Y1 = 1. Y2 = x, Y3 = x & # 179;, and the general solution is obtained

Since it is a second order linear homogeneous equation, there should be two homogeneous solutions, and y2-y1 = X-1 and y3-y1 = x ^ 3 - 1 are not related. Therefore, it can be used as the basic solution system. The general solution of the equation is
Y = C1 [X-1] + C2 [x ^ 3 - 1], C1 and C2 are arbitrary constants