There is a rectangular table with a length of 100cm and a width of 60cm. The area of a rectangular table cloth is twice the area of the table, and when it is laid on the table, the length of each side is equal. Find the length and width of the table cloth (accurate to 1cm) (solution of quadratic equation with one variable)

Setting: the length of tablecloth is x cm, (x > 0)
The area of the table is: 100 × 60 = 6000m & # 178; then the area of the tablecloth is s = 6000 × 2 = 12000m & # 178;
It can be seen from the above that the width of tablecloth is s / x = 12000 / X
X-100=12000/X-60
Convert to X & # 178; - 40x-12000 = 0

Let Y1 = 3 + x ^ 2, y2 = 3 + x ^ 2 + exp (- x) be two special solutions of a second order linear non-homogeneous differential equation, and one solution of the corresponding homogeneous equation is Y3 = x, then the general solution of the differential equation is

The expression of the general solution of the second order homogeneous linear differential equation A1 (x) d ^ 2Y / DX ^ 2 + A2 (x) dy / DX + A3 (x) y = 0 (1) the second order homogeneous nonlinear differential equation A1 (x) d ^ 2Y / DX ^ 2 + A2 (x) dy / DX + A3 (x) y = f (x) (2) (2) y = c1y0 (x) + c2y1 (x) + Cy2 (x) where Y0 (x) is a special solution of (2), Y1 (x) and y

A rectangular dining table, 3 / 2 meters long, 1 / 2 meters wide. To this table with a tablecloth, tablecloth spread over the table
All around even droop 2 / 1 meter, seek tablecloth area?

It is equal to the length and width of the tablecloth, which is 2 / 1 meter more than that of the table, that is, 2 meters in length and 1 meter in width

What is a linear differential equation with constant coefficients?

"Linear" means that the power of function y and its n-th derivative is 1;
"Constant coefficient" means that the coefficients before y and its n-th derivative are constant;
"Differential equation" is an equation composed of independent variable x, function y and its n-th derivative;
A combination is a linear differential equation with constant coefficients

To lay a square tablecloth on a square table which is a meter long on one side, the tablecloth is 0.1 meter larger than the tablecloth on the other side. How large is the tablecloth needed?
The answer is the square of (a + 0.2). Why add 0.2?

It's more than 0.1 meters on all sides, 0.1 meters on one side, so the two sides add up to 0.2 meters, so the side length of tablecloth is a + 0.2
I hope I can help you,

To solve the fourth order homogeneous linear differential equation with constant coefficients with special solutions of Y1 = e ^ x, y2 = Xe ^ x, Y3 = 3sinx, Y4 = 2cosx

∵ Y1 = e ^ x, y2 = Xe ^ x, Y3 = 3sinx, Y4 = 2cosx are four linearly independent special solutions of the equation ∵ the root of the characteristic equation of the equation is R1 = R2 = 1, R3 = I, R4 = - I = = > the characteristic equation of the equation is (R ^ 2 + 1) (R-1) ^ 2 = 0 = = > R ^ 4-2r ^ 3

To lay a square tablecloth on a square table with a meter long on one side, the four sides of the tablecloth all exceed 0.1 meter of the table top. How much tablecloth do you need

The area formula of a square is the length of the side multiplied by the length of the side
So the tablecloth area s = (a + 0.1) * (a + 0.1)

Finding the general solution of the second order non homogeneous linear differential equation with constant coefficients y '' - 10Y '+ 9y = e ^ 2x

In a square tabletop with a side length of 70 cm, a tablecloth is required to hang down 12 cm all around. How many square centimeters is the area of this tablecloth?

Each side drops 12 cm, that is, the side length increases by 2 12 cm, the side length of tablecloth is 70 + 12 + 12 = 94 cm
Area 94x94 = 8836 square centimeter

Let y = C1E ^ 2x + c2e ^ 3x be the general solution of a second order homogeneous linear differential equation with constant coefficients

Y "+ py ˊ + QY = 0 is a second order homogeneous linear differential equation with constant coefficients. Its characteristic equation is R & sup2; + PR + q = 0. When the characteristic equation has two unequal real roots, the general solution of the differential equation is y = C1E ^ Rix + c2e ^ r2x_ 1=2,r_ The differential equation is Y "- 5Y ˊ + 6y = 0