Solving the ordinary differential equation y + 2XY '+ (x ^ 2) y' ', etc Er, it's y + 2XY '+ (x ^ 2) y' '= 0

Solving the ordinary differential equation y + 2XY '+ (x ^ 2) y' ', etc Er, it's y + 2XY '+ (x ^ 2) y' '= 0

Y + 2XY '+ (x ^ 2) y' '= 0 let x = e ^ t, t = lnxy' (x) = y '(T) / X. XY' (x) = y '(T) y' '(x) = (y' '(T) - y' (T)) / x ^ 2. X ^ 2Y '' (x) = y '' (T) - y '(T) - y' (T) + 2Y '(T) + y = 0, y' '(T) + y' (T) + y = 0 the solution is y = e ^ (- t / 2) (c1cos (t √ 3 / 2) + C

Solving ordinary differential equation y '= 2 (Y-2) ^ 2 / (x + Y-1) ^ 2

By solving the equations Y-2 = 0, x + Y-1 = 0, we can get x = - 1, y = 2. By transforming x = x + 1, y = Y-2, then the original differential equation is changed into dy / DX = 2Y ^ 2 / (x + y) ^ 2 = 2 (Y / x) / [1 + (Y / x)] ^ 2. If u = Y / x, then u + Du / DX = 2U ^ 2 / (1 + U) ^ 2. By separating variables, we can get (1 + U) ^ 2 / [u (u ^ 2 + 1)] Du = - DX / x [1 / U + 2 / (...)

The solution of differential equation in Higher Mathematics
General solution of differential equation y'cos ^ 2x + y-tanx = 0

The equation is y '+ 1 / cos ^ 2x * y = TaNx / cos ^ 2x
∫dx/cos^2x=tanx ∫-dx/cos^2x=-tanx
e^(∫dx/cos^2x)=e^(tanx) e^(∫-dx/cos^2x)=e^(-tanx)
∫tanx*e^(tanx)dx/cos^2x=∫tanx*e^(tanx)d(tanx)=(tanx-1)*e(tanx)+C
The general solution is y = (tanx-1) + C * e (- TaNx)

The solution of differential equation in higher number
dx/dt=c1*et-x
Et is the t power of E,

Write in the customary form y '+ y = (C1) e ^ X
Then we can use the general solution of the first order differential equation with constant coefficients
e^（-x）[(1/2)C1e^(2x)+C]=(1/2)C1e^x+C e^(-x)

Fundamentals of differential calculus of multivariate functions
Using total differential to find the approximate value of (1.01) ^ (2.99)

Let z = f (x, y) = x ^ y, (x0, Y0) = (1,3)
Δx=0.01,Δy=-0.01
∂z/∂x=yx^(y-1) ;∂z/∂y=x^(y)lnx
f(1+Δx,3+Δy)≈f(x0,y0)+(∂z/∂x)Δx+(∂z/∂y)Δy=1+(3*0.01)+0=1.03

Problems in differential calculus of multivariate functions
It is proved that if point P is the aggregation point of domain D, then ᦉ 8707; {PN}, ᦉ 8704; n ∈ n +, PN ∈ D. has ∞
L im Pn=P.
n→∞
If P is a gathering point of D, then p can belong to or not belong to D. Xiang. (1) whether the proof is general if S0 containing P belongs to d ', and what if P does not belong to D?

For the region D, if any of the ε domains of point P contains points in D which are different from P, then p is called a gathering point of D. (1) let S0 with P belong to D, and divide it into two parts, where the sub domain with P is a sphere (circle) domain, and the diameter D1 is less than half of the diameter d0 of the inscribed sphere of S0. In the sub domain without p, take one point as P1, and the other sub domain contains P, (2) by performing the operation of (1) on S1, we get the point P2 and the sub domain S2 containing the point P, and the diameter D2 of the sphere (circle) domain S2 is less than half of the diameter D1 of S1. Then we get the point sequence {PN}, and every element in the domain sequence {Sn} contains the point P, and the sequence {Sn} is shrinking continuously, and the sequence shrinks to one point, and P is the intersection of {PN}
Lim Pn=P.
n→∞
_________
Question supplement: if P is a gathering point of D, then p can belong to or not belong to D. Xiang "(1) does the proof that S0 with P belongs to d 'have generality? What if P does not belong to D?
Then d 'is extended to d'. And d 'in the above proof can be replaced by D'

On differential calculus of multivariate functions
X / z = ln Z / y for DZ

I don't know if your ln Z / y is ln (Z / y) or (LN z) / y. anyway, from the former point of view, I can find the differential on both sides of the equation
(z·dx-x·dz)/z²=(y·dz-z·dy)·y/z
Just work out DZ

Consult the problem of differential calculus of multivariate function
What are the necessary and sufficient conditions for the equality of the second mixed partial derivatives in binary functions
What is the general situation when the second mixed partial derivatives are not equal?

Figuratively speaking, the necessary and sufficient condition is: the binary function should be continuous and smooth. Imagine a smooth plane in a three-dimensional coordinate system, just like the water surface, without creases, so that the second-order partial derivatives of the function are equal
When they are not equal, they are usually not smooth. For example, when two planes intersect a straight line, the second-order partial derivatives on that line are not equal. Of course, if a second-order derivative itself is meaningless, let alone

The problem of solving special solution of ordinary differential equation by differential operator method in Higher Mathematics
How to use the differential operator method to solve y '' + y = - xsinx? My result is always half of the answer
My solving process is: y = xsins / (square of D + 1) = x / 2 * xsinx / D = x * (SiNx xcosx) / 2

Two days ago, after Chen Wendeng's class, he said that he didn't want to use differential operator to solve problems, but he only wanted to use the general idea of solving problems in the book for postgraduate entrance examination

Finding the general solution of Euler equation
Let x > 0, the general solution of the differential equation x ^ 2Y '' - 2XY '+ 2Y = x + 2? I just can't figure out how to use the differential operator method to solve 2

Here I only answer your questions
On the left side, you can use the method of dealing with Euler equation to get a polynomial about D, divide it to the right side, divide it into two parts and solve it respectively (if you want to add it). For the good solution in front (since you know how to solve it by this method), there is a ready-made formula in the back, that is, treat 2 as a polynomial (this rule also has (Division))
Just figure it out for yourself