What is the general solution of the differential equation y '- y = 0? rt,

What is the general solution of the differential equation y '- y = 0? rt,

∵y'-y=0 ==>dy/y=dx
==>Ln | y | = x + ln | C | (C is an integral constant)
==>y=Ce^x
The general solution of the differential equation y '- y = 0 is y = CE ^ x (C is an integral constant)

General solution of differential equation y '= y

dy/dx=y
(1/y)dy=dx
After the points on both sides
Ln y = x + C
y=±e^(x+c)
So the general solution is y = CE ^ X

Find the general solution of the differential equation y "+ y = 0
Urgent need

The characteristic equation is p * P + 1 = 0
Then p = ± I
y=Acosx+Bsinx

11. Find the general solution of Y '- 5Y' - 6y = 0

The characteristic equation is R & # 178; - 5r-6 = 0
（r＋1）（r-6）=0
R = - 1 or 6
therefore
The general explanation is
Y = the - x power of C1E + the 6x power of c2e

8x + 6y = 580 6.5x + 4.5y = 460 how to solve both X and y?
8x + 6y = 580 6.5x + 4.5y = 460 how to solve both X and y?

Multiply both sides of the first formula by 3:
Formula 24x + 18y = 1740 a
Multiply both sides of the second equation by 4:
26x + 18y = 1840 formula B
Subtract a from B,
We get 2x = 100, that is, x = 50
Substituting x = 50 into the first formula (or any formula), we get 400 + 6y = 580, 6y = 180, y = 30

For the quadratic function y = x ^ 2-4x + 5, the maximum or minimum value of the function can be obtained under the following different conditions: (1) x takes any real number (2) - 1 ≤ x ≤ 1
（3）1≤x≤4

y=x^2-4x+5=（x-2)^2+1
1. When x takes any real number, y = (X-2) ^ 2 + 1 (X-2) ^ 2 ≥ 0, y ≥ 1, so the maximum value is + ∞ and the minimum value is 1
2. When - 1 ≤ x ≤ 1, the function of X decreases at (- ∞, 2), so the maximum value is x = - 1, y = (- 1-2) ^ 2 + 1 = 10
The minimum value is x = 1, y = (1-2) ^ 2 + 1 = 2
3.1 ≤ x ≤ 4, when x is (- ∞, 2), the function decreases, when x is (2, + ∞), the minimum value of function increase is x = 2, then y = 1
When x = 1, y = (1-2) ^ 2 + 1 = 2; when x = 4, y = (4-2) ^ 2 + 1 = 5
Therefore, the maximum value is 5 and the minimum value is 1

Find the tangent equation of the square of the curve y = root 2-x at the point (1,1)

Derivation

The equation of a circle solves the tangent equation of a circle
The equation of circle x ^ 2Y ^ 2-6x-4y + 12 = 0 to find the tangent equation of circle P (2,0)

X ^ 2 + y ^ 2-6x-4y + 12 = 0 (x-3) ^ 2 + (Y-2) ^ 2 = 1 Center (3,2) radius 1 (1) if the slope of the tangent line exists, let the slope of the tangent line be K, so y = K (X-2) KX - Y - 2K = 0 the distance from the center of the circle to the straight line is equal to the radius of the circle 1, so | 3k-2-2k | / root sign (K & sup2; + 1) = 1K = 3 / 4 (2) if the slope of the straight line is not

The equation for a circle with the center of the circle on y = - X and passing through two points (2,0), (0, - 4)

Let the center of the circle O (x, y), then y = - x; and because the square of the radius r of the circle R ^ 2 = (X-2) ^ 2 + (y-0) ^ 2 = (x-0) ^ 2 + (y + 4) ^ 2, the two equations can obtain x, y and radius r, which is so simple

The equation of a circle with two points a (- 1,4), B (3,2) and the center of the circle on the y-axis is______ ．

Let the center coordinate of the circle be o (0, b), then there is 1 + (B − 4) 2 = 9 + (2 − b) 2, and the solution is b = 1, the center coordinate of the circle is (0, 1), the radius is r = 1 + (1 − 4) 2 = 10, and the equation of the circle is: x2 + (Y-1) 2 = 10. So the answer is: x2 + (Y-1) 2 = 10