Solving differential equation: y '- y = SiNx

Solving differential equation: y '- y = SiNx

1. Find the homogeneous general solution
The characteristic equation corresponding to homogeneous equation y '- y = 0: M-1 = 0
The characteristic root: M = 1
So the general solution is y_ ＝c exp(x)
2. Special solution
Because the free term on the right side of the equation: sin (x), is the imaginary part of exp (IX)
We only need to find the special solution of Y '- y = exp (IX) and take its imaginary part
Let the special solution be y × = a exp (IX) and substitute it into the original equation
A＝1/(i-1)
So the special solution y × = exp (IX) / (i-1) = (- 1 / 2) × (I + 1) × exp (IX) = (- 1 / 2) × (I + 1) × [cos (x) + isin (x)]
Take its imaginary part y * = - [sin (x) + cos (x)] / 2
3. The complete solution of the equation is obtained
y=y_ + y*=c exp(x) - [sin(x)+cos(x)]/2

Solving differential equation y "+ y '= SiNx
Solving differential equations by RT

Let y '= P get p' + P = SiNx, and the general solution of P '+ P = 0 is p = a * e ^ {- x}. Let p' + P = SiNx be p = u * e ^ {- x}, where u is the function of X, Substituting u'e ^ {- x} = SiNx, u '= SiNx * e ^ {x} integral is u = [(SiNx cosx) / 2] * e ^ {x} + B. thus, the general solution of P' + P = SiNx is p = {[(SiNx cosx) / 2] * e ^ {x} + B} * e ^ {- x} = (SiNx cosx) / 2 + b * e ^ {- x} integral is y = (SiNx cosx + SiNx) / 2-B * e ^ {- x} + C

The solution of Y ″ + y = SiNx is given by differential equation

Theorem 3 gives a formula of special solution: y = y + y *; what we already know is Y "+ y = 0; this form is generally y = c1cosx + c2sinx; obviously, for y = SiNx + C (constant), it is a special solution, so

The equation of the circle passing through the point a (- 1,4) B (3,2) and the center of the circle on the y-axis
It is known that the curve is the locus of two points whose distance ratio of a (- 4,0) B (2,0) is 2. The curve equation is solved
Given that the circle C is tangent to the Y axis, the center of the circle is cut on the straight line y = x, the chord length is 2 times the root sign 7, and the equation of circle C is obtained

1. The vertical bisector equation of AB is: 2x - y + 2 = 0, the intersection point of the line and Y axis is p (0,2), which is the center of the circle, and the radius of the circle r = | PA | = √ 20, then the circular equation is: X & # 178; + (Y-2) &# 178; = 202, and the set point is Q (x, y), then: | QA |: | QB | = 2:1 = = = = > > > > > > [(x + 4) &# 178; + Y & # 178;]: [(...]

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

The circle passes through points a (4, - 2) and B (- 2,2), and the center of the circle is on the y-axis

The circle passes through points a (4, - 2) and B (- 2,2), and its center is on the y-axis
Then the center of the circle is on the intersection of the vertical line and the Y axis in ab
The vertical of AB is y-0 = (3 / 2) * (x-1)
Let x = 0 give y = - 3 / 2
So the center of the circle is (0, - 3 / 2)
So the radius is r = √ [(0 + 2) &# 178; + (- 3 / 2-2) &# 178;] = √ 65 / 2
So the equation of circle is X & # 178; + (y + 3 / 2) &# 178; = 65 / 4

The equation of a circle passing through two points a (- 1,4) B (3,2) with its center on the z-axis

Because the center of the circle is on the y-axis, then the center O is (0, y)
Then Ao = Bo
(-1+0)^2+(4-y)^2=(3-0)^2+(2-y)^2
1+16-8y+y^2=9+4-4y+y^2
4y=4
y=1
R^2=(-1+0)^2+(4-y)^2=1+9=10
So the circular equation is
x^2+(y-1)^2=10

Given that the circle passes through points a (3,1), B (- 1,3) and its center is on the straight line 3x-y-2 = 0, find the equation of the circle (solve the equation system in more detail)

Look below

Through point a (√ 3,0) and point B (0,1), and the center of the circle is on the straight line 3x-y = 1, the equation of the circle is?

If passing through AB, the center of the circle is on the vertical bisector of ab
AB midpoint (√ 3 / 2,1 / 2)
AB slope (1-0) / (0 - √ 3) = - 1 / √ 3
So AB vertical bisector slope = √ 3
So Y-1 / 2 = √ 3 (x - √ 3 / 2) = √ 3x-3 / 2
√3x-y-1=0
The center of the circle is on this line, and 3x-y = 1
So the intersection point is the center of the circle C (0, - 1)
r=BC=1-(-1)=2
x²+(y+1)²=4

Through point a (√ 3,0) and point B (0,1), and the center of the circle is on the line 3x-y = 1, the equation of the circle is?
Through point a (√ 3,0) and point B (0,1), and the center of the circle is on the line 3x-y = 1, the equation of the circle is?
The points AB are all on the circle, so the vertical bisector of AB must pass through the center of the circle
Calculate the slope of AB line k = - 1 / √ 3, AB midpoint (√ 3 / 2,1 / 2)
So the analytic formula of the middle vertical line is: √ 3x-y-1 = 0
The intersection of the middle vertical line and the straight line 3x-y-1 = 0 is the center of the circle
Find the center of the circle as (0, - 1)
The distance between the center of the circle and any point of AB is the radius
The radius r = 2 is obtained
So the equation passing through points a and B and the center of the circle on the straight line 3x-y-1 = 0 is as follows:
x^2+(y+1)^2=4
Why is r equal to 2 here? I don't understand it here

The center of the circle is C (0, - 1)
Then r = AC = BC
Just use the two-point distance formula