General solution of differential equation y '- y = cosx

General solution of differential equation y '- y = cosx

y=(1/2)(sinx-cosx) +ce^x

The asymptote equation of hyperbola x29-y216 = 1 is______ ．

∵ hyperbola equation x29-y216 = 1, ∵ A2 = 9, B2 = 16, that is, a = 3, B = 4, then the asymptote equation of hyperbola is y = ± 43x, so the answer is: y = ± 43x

How to solve polar coordinate equation: for example, why is the polar coordinate equation of x ^ 2 + y ^ 2 = 2Y r = 2Sin θ?

Let x = RCOs θ
y=rsinθ
Put it in, get it
(rcosθ)²+(rsinθ)²=2rsinθ
r²=2rsinθ
r=2sinθ

X ^ 2 = 2Y to polar coordinate equation; ρ = 2Sin θ to rectangular coordinate equation

Directly remember the formula P & # 178; = x & # 178; + Y & # 178; X = PCOS θ y = PSIN θ, then x ^ 2 = 2Y is (PCOS θ) &# 178; = 2psin θ, and simplify it to PCOS & # 178; θ = 2Sin θ ρ = 2Sin θ, and multiply both sides of P to get ρ & # 178; = 2psin θ, then change it to X & # 178; + Y & # 178; = 2y. what's not clear

Change X & # 178; + Y & # 178; - 5 √ (X & # 178; + Y & # 178;) - 5x = 0 to polar coordinate equation

In the transformation of polar coordinates and rectangular coordinates
x=rcosθ,y=rsinθ,r=√(x²+y²)
So the original will become
r²-5r-5rcosθ=0

How to get the circle (x-a) ^ 2 + y ^ 2 = a ^ 2 represented by polar coordinates r = 2acos θ?

x²＋y²=ρ²,x=ρcosθ,y=ρsinθ
Then: ρ = 2acos θ, ρ & # 178; = 2A ρ cos θ, then: X & # 178; + Y & # 178; = 2aX, that is: (x-a) &# 178; + Y & # 178; = A & # 178;

Ask a question about polar coordinates. What is not a circle in the following curves represented by polar coordinates?
A.ρ^2+2ρ(cosa+√3sina)=5; B.ρ^2-6ρcosa-4ρsina=0; C.ρ^2-ρcosa=1; D.ρ^2+2ρ(cosa+sina)=1

x=pcosa
y=psina
A:x^2+y^2+2x+2√3y=5
(x+1)^2+(y+√3)^2=9
So it's round
B:x^2+y^2-6x-4y=0
(x-3)^2+(y-2)^2=13
So it's round
C:x^2+y^2-x=1
(x-1/2)^2+y^2=5/4
So it's round
D:x^2+y^2+2x+2y=1
(x+1)^2+(y+1)^2=3
So it's round

What is the representation of polar coordinates?

Any point P is expressed as (R, t), the origin is or, the distance from P to o, and t is the angle between the ray OP and the polar axis (that is, the positive half axis of X axis, especially the polar axis)

Polar equation of y = x

θ = π / 4 and θ = 3 π / 4, where θ is the polar angle
(in polar coordinate system, θ = K (k is a constant) represents the ray whose inclination angle is θ and whose starting point is the origin of the coordinate.)

When a is equal to, there is no XY term in the formula x-3axy-3y + 2 / 3xy-5

X ^ 2-3axy-3y ^ 2 + 2 / 3xy-5, when there is no XY term, 3A = 2 / 3, a = 2 / 9
elite