General solution of differential equation XDY + (y + SiNx) DX = 0~

General solution of differential equation XDY + (y + SiNx) DX = 0~

xdy+ydx=-sinxdx
d(xy)=-sinxdx
Integral on both sides: xy = cosx + C

General solution of differential equation XDY / DX + y = cosx

Answer: XDY / DX + y = cosx
xy'+y=cosx
(xy)'=cosx
xy=sinx+C
So:
The general solution is xy = SiNx + C

General solution of differential equation y = XDY / DX + (y ^ 2) (SiNx) ^ 2

Ans：y = 2x/(x - sinxcosx + C)
y = x * dy/dx + y²sin²x
-x * dy/dx + y = y²sin²x
-(dy/dx)/y² + 1/(xy) = (sin²x)/x
Let v = 1 / y, then DV / DX = DV / dy * dy / DX = D (1 / y) / dy * dy / DX = - 1 / Y & # 178; * dy / DX = - (dy / DX) / Y & # 178;
=> dv/dx + v/x = sin²x/x
Integral factor = e ^∫ (1 / x) DX = e ^ LNX = x, multiply X by both sides of the equation to get
x * dv/dx + v = sin²x
x * dv/dx + v * dx/dx = sin²x
d(x * v)/dx = sin²x
x * v = (1/2)∫ (1 - cos2x) dx = (1/2)(x - 1/2 * sin2x) + C = x/2 - 1/2 * sinxcosx + C
v = (x - sinxcosx + C)/(2x)
1/y = (x - sinxcosx + C)/(2x)
y = 2x/(x - sinxcosx + C)

The polar coordinates are used to calculate ∫ ∫ xydxdy, where D is the closed region bounded by X & # 178; + Y & # 178; = 1 and X & # 178; + Y & # 178; = 2x in the first quadrant
The answer is 9 / 16

The polar coordinate equation of X & # 178; + Y & # 178; = 1 is: r = 1 x & # 178; + Y & # 178; = 2x is: R & # 178; = 2rcos θ, i.e. r = 2cos θ 2cos θ = 1, then: cos θ = 1 / 2, θ = π / 3, please draw your own picture. Therefore, the area surrounded by the two curves can be divided into two parts, the first part is θ: 0--π / 3, R: 0-- > 1

If we know that the vertex of the parabola is m (- 1, - 2) and it passes through the point n (1,10), then the analytic expression of the quadratic function is?

Let the analytic formula be y = a (X-H) ² + K
h=-1,k=-2
So: y = a (x + 1) ² - 2 is substituted into n (1,10)
10=a(1+1)²-2
The solution is a = 3
∴y=3(x+1)²-2
=3x²+6x+4

Given that the parabola passes through three points a (2,6), B (- 1,0), C (3,0), then the analytic expression of quadratic function is? Its vertex coordinates are?
The detailed process is mainly the calculation after substitution

The parabola passes through B (- 1,0), C (3,0), and these two points are on the x-axis, so we can let y = a (x + 1) (x-3) substitute the point a (2,6) to get: 6 = a * 3 * - 1A = - 2, so y = - 2 (x ^ 2-2x-3) = - 2x ^ 2 + 4x + 6y = - 2x ^ 2 + 4x + 6 = - 2 (x ^ 2-2x) + 6 = - 2 [(x-1) ^ 2-1] + 6 = - 2 (x-1) ^ 2 + 8; its vertex

The analytic expression of quadratic function with the vertex of parabola being (3, - 1) and passing through point (2,3)

Y = 4 (x-3) Square-1

Differential equation dy / DX = (X & sup2; + Y & sup2;) / 2XY,

The original formula can be reduced to: dy / DX = 0.5 (x / y) + 0.5 (Y / x)
Let u = Y / x, then y = UX, dy / DX = XDU / DX + U
The original formula becomes: XDU / DX + U = 0.5 / U + 0.5u
After simplification, put U on the left and X on the right
[u/(1-u^2)]du=(1/2x)dx
The solution of the original equation is as follows
lnx+ln(1-u^2)=c
(C is a constant, u ^ 2 is the square of u ~)
Finally, you can substitute u = Y / X

dy/dx=2xy
How to find the general solution of this equation?

dy/2y = x*dx
1/2*ln(y) = 1/2*x^2+C
ln(y) = x^2+C
y = exp(x^2) + C

If x + y = 3, xy = 5, then find the value of x-xy-y, X-cube + y-cube

x^2-xy-y^2=(x-y)^2+xy=3^2+5=14
x^3+y^3=(x+y)(x^2-xy+y^2)=3*14=42