On the differential operator method The operator method is used to find the special solution, For example, y * = {1 / (1 + 2D + D ^ 2)} (x ^ 2 + X + 1) The question is how to get the quotient of D from 1 / (1 + 2D + D ^ 2) Hope to list the detailed division steps The first floor uses Taylor expansion to find the quotient? What if P (d) = 1 + D + D ^ 2

On the differential operator method The operator method is used to find the special solution, For example, y * = {1 / (1 + 2D + D ^ 2)} (x ^ 2 + X + 1) The question is how to get the quotient of D from 1 / (1 + 2D + D ^ 2) Hope to list the detailed division steps The first floor uses Taylor expansion to find the quotient? What if P (d) = 1 + D + D ^ 2

1 / D can be understood as indefinite integral operation
1 / P (d) f (x): when f (x) is a polynomial of degree m, change 1 / P (d) into a power series of ascending power, and use the first m + 1 term
1/(1＋2x＋x^2)＝1/(1＋x)^2＝－[1/(1＋x)]'＝－[1－x＋x^2－x^3＋.]'＝－(－1＋2x－3x^2＋...)＝1－2x＋3x^2＋.
So, y = 1 / (1 + 2D + D ^ 2) (x ^ 2 + X + 1) = [1-2D + 3D ^ 2] (x ^ 2 + X + 1)
＝(x^2＋x＋1)－2(x^2＋x＋1)'＋3(x^2＋x＋1)''
＝(x^2＋x＋1)－2(2x＋1)＋3×2
＝x^2－3x＋5

What's the difference between differential integral and indefinite integral

Differential is derivative
Derivative function
Integral is the inverse operation of differential
Primitive function of derivative function
Indefinite integral is the derivation of all the original functions. Variable upper limit integral is the solution of one of the original functions. Definite integral is the solution of the function value of the upper limit of the original function minus the function value of the lower limit. The physical meaning is the so-called area of the integrand in the interval and the X axis (the part under the horizontal axis is calculated as negative)

A problem of solving indefinite integral by using the method of making up differential
Instead of the integral symbol, ^, the power is expressed by #. In the book, there is a method to solve indefinite integral called "rounding up differentiation". I think the most difficult part of this method is to split # g (x) DX, which can not be evaluated by integral table, into # f [K (x)] * k (x) 'DX, which can be further divided, A problem: find # (AX + b) ^ 4DX. The first step of the splitting process in the book is not written out. It is directly: # (AX + b) ^ 4DX - > # (AX + b) ^ 4 * (1 / a) d (AX + b). I can't change this structure. If you understand this, please help me solve the intermediate process

In fact, the integration differentiation method is the substitution method, which is to replace the integrand function with an easily solvable integral form. For example, when solving (1 / x) lnxdx integral, because the derivative (or differential) of LNX is 1 / x, the original formula can be transformed into (LNX) d (LNX) under the integral sign, so as to get the result equal to (LNX) / 2 C

How to grasp the key point of the connection among differential, derivative, definite integral and indefinite integral in Higher Mathematics?
When learning advanced mathematics, it is often very confused, a lot of things are simply unexpected. Sometimes, it feels very simple. It is very confusing

Everything of calculus comes from differential, derivative is the ratio of differential, so derivative is also called derivative, integral is the sum of differential
There is no indefinite integral in the world. Indefinite integral is the process of finding the original function when people calculate definite integral
So the most important part of calculus is differential. In geometry, differential is the increment of tangent
Of course, the derivative table and integral table must also be memorized
Indefinite integral and definite integral look similar, but the investigation direction is different
Indefinite integral mainly investigates the idea of "cou" and the table of back integral; definite integral mainly investigates its properties
It may be simpler, but these are the core ideas of calculus, I hope you can understand

Mathematical analysis of differential problems of multivariate functions
What is the sufficient condition for differentiability of multivariate function? It is better to have a detailed explanation

Partial derivative continuous → differentiable → existence of partial derivative

How to solve tangent vector equation of curve equation and normal vector equation of surface equation?

For the tangent vector of a curve, if it is given by a parametric equation, the variable can be derived from the parameter separately. If it is given by a system of equations, the implicit function of other variables to a certain variable can generally exist. Therefore, at this time, other variables are regarded as functions of this variable, and the equations of the system of equations can be derived from this variable, and the derivative of other variables to the function of this variable can be solved, Because other variables take this variable as a parameter, the tangent vector equation can be given by the method of parametric equation, and then the tangent vector can be obtained by bringing the coordinates of this point in
For the normal vector of the surface equation, we only need to derive the equation for each variable, and then bring the coordinates of the point into the normal vector of the surface equation
I just need to review a few examples to understand each other

How to determine the tangent vector of space curve to find the corresponding tangent equation and normal plane

It's obvious that for the curve represented by the system of equations f (x, y, z) = 0, G (x, y, z) = 0, first determine a variable as a parameter and turn other variables into functions of this variable. For example, with X as a parameter, the system of equations is reduced to: x = x, y = y (x) z = Z (x). Therefore, the tangent vector at any point of the curve is

On the solution of tangent and normal plane of space curve
I know how to solve ordinary equations
But if the curve is determined by a set of equations (surface equation and plane equation)
I don't know how to ask for it
Hope to talk about the solution

We get 2T, 1, 1
So the tangent equation is 2T / (X-2) = 1 / (Y-1) = 1 / Z
The tangent is Y-2 + so the equation is Y-1 + 1
Choose me, don't know to ask again, I played for a long time

How to find the tangent equation of a function of one variable at one point
Given that the curve passes through point (1,1), if the intersection of the tangent line at any point P on the curve and the y-axis is denoted as Q, then the circle with PQ as the diameter passes through point F (1,0), and the curve is obtained

Let the curve be y = f (x), and the coordinates of point P on the curve be (x, y),
The tangent equation passing through point P is Y-Y = f '(x) (x-x), q is (0, y-xf' (x)),
PQ^2=x^2+x^2(f'(x))^2
The midpoint coordinates of PQ are: (x / 2, y-xf '(x) / 2),
Since the point F (1,0) is on the circle, (x / 2-1) ^ 2 + (y-xf '(x) / 2) ^ 2 = [x ^ 2 + x ^ 2 (f' (x)) ^ 2] / 4
Or: (X-2) ^ 2 + (2Y XF '(x)) ^ 2 = x ^ 2 + x ^ 2 (f' (x)) ^ 2
It is reduced to: - x + 1 + y ^ 2-xyf '(x) = 0
That is: y ^ 2-xyy '= X-1
2xy^2-2x^2*yy'=2x(x-1)
Or: [2Y ^ 2dx-2xydy] / x ^ 3 = - 2 (x-1) / x ^ 3DX
The general solution is: y ^ 2 / x ^ 2 = 2 / X-1 / x ^ 3 + C, or: y ^ 2 = 2x-1 / x + CX ^ 2
Because the curve passes through point (1,1), we substitute C = 0
The obtained curve is y ^ 2 = 2x-1 / X

The problem of partial derivative of binary function
Let u (x, y) have a second order continuous partial derivative, u XX = u YY, u (x, 2x) = x, u x (x, 2x) = x ^ 2, find u y (x, 2x), u XX (x, 2x) and u XY (x, 2x)
Thank you

No answer... Can you tell me what you think? thank you