Dy / DX + (2 / x) y = - x when y (2) = 0,

Dy / DX + (2 / x) y = - x when y (2) = 0,

y'+2y/x=－x,
xy'+2y=－x^2,
x^2*y'+2xy＝－x^3,
(x ^ 2Y) '= - (x ^ 4 / 4)', then
X ^ 2Y = - x ^ 4 / 4 + C, y (2) = 0
C = 2 ^ 4 / 4 = 4
y＝(4－x^4/4)/x^2
＝4/x^2－x^2/4.

Finding the special solution of dy / DX = x + y in x = 0y = 0

∵dy/dx=x+y
==>dy-ydx=xdx
==>E ^ (- x) dy Ye ^ (- x) DX = Xe ^ (- x) DX
==>d(ye^(-x))=d(-xe^(-x)-e^(-x))
==>Ye ^ (- x) = c-xe ^ (- x) - e ^ (- x) (C is a constant)
==>y=Ce^x-x-1
The general solution of the original equation is y = CE ^ x-x-1
∵y(0)=0
Then we substitute the general solution and get C = 1
So the special solution of the original equation satisfying the given initial condition is y = e ^ x-x-1

Special solutions of differential equation dy / DX = Y / X-1 / 2 (Y / x) ^ 3 when y (1) = 1
The answer is y = x / √ (LNX + 1), why does x under the root sign have no absolute value? The commentary explains that the solution of the differential equation is a differentiable function, of course, it is a continuous function. Because the original differential equation x is on the denominator, so x is not equal to 0, so the continuous interval of the solution of the differential equation y = y (x) cannot include the point x = 0, Therefore, the continuous interval of the solution should be x = 1 without x = 0, so x cannot be negative, so there is no sign of absolute value here
Can the solution of differential equation only be continuous in one interval? It can't be (a, b) and (C, d)? I'm afraid I'm wrong because I have to take an exam,

No, when differential equations are used to solve physical problems, X has a range. For example, when X represents the length, there must be x > 0. But if X represents the temperature, there may be both positive and negative. So when solving mathematical differential equations, it is meaningless to consider all of them

How to solve the differential product equation (1 + y) DX - (1-x) dy = 0,

(1+y)dx-(1-x)dy=0
(1+y)dx=(1-x)dy
∫ [1/(1-x)] dx = ∫ [1/(1+y) ]dy
-ln|1-x| = ln|1+y|
ln|1-x||1-y| = c'
(1-x)(1+y) = e^(c') = C