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MindMap Gallery ordinary differential equations

ordinary differential equations

This is a mind map about ordinary differential equations, including: basic concepts, separable variable equations, homogeneous equations, first-order linear ordinary differential equations, Bernoulli's equation, reducible second-order differential equations, second-order linear Differential Equations.

Edited at 2022-02-01 18:07:51

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ordinary differential equations

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Outline

POWER SERIES
- 12
Uyie Fakhri
ordinary differential equations
- 61
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Differential Calculus of Functions of One Variable
- 60
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Differential calculus of multivariate functions
- 30
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limit and continuity
- 100
- 1
- 1
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Mathematics
- 31
3zz
Core knowledge module
- 14
- 1
Mason·Carter

ordinary differential equations

basic concept

differential equations

Equations that relate independent variables, dependent variables, and their derivatives or differentials

Ordinary Differential Equations (ODE)

The unknown function in question is a function of one variable

Partial Differential Equations

Ordinary differential equation form

Order of Ordinary Differential Equations

The order of an ordinary differential equation is determined by the order of the highest derivative (differential) in the equation.

Solutions to Ordinary Differential Equations

The formula for y that makes the ordinary differential equation equal to zero

General solutions to ordinary differential equations

The solution of the equation contains arbitrary constants that are not related to each other, and the number is equal to the order of the equation

Special solutions to ordinary differential equations

Taking a special value of any constant is generally derived from the initial value condition

initial value condition

An ordinary differential equation of order n contains an initial value condition

The actual meaning is generally the initial state of things

The initial value condition of a first-order ordinary differential equation is the value of the function value of the original function at a certain point.

initial value problem

Find the value of a solution to an ordinary differential equation that satisfies the initial value condition

separable variable equation

form

Features

The first derivative has been solved

The right end is the multiplication of functions about x and y respectively.

solution

1. Identify the type

2. Separate variables

3. Integrate both sides of the equal sign and add appropriate constants

4. Simplify and get the general solution

5. Determine whether there is an initial value condition

Yes, substitute it to get a special solution

None, examine the roots of this equation, which is also the solution of the equation

Then get the complete solution

homogeneous equation

form

Solution (variable substitution)

1. Order

2. Bring it into the equation to simplify and eliminate y

3. What is obtained is an ordinary differential equation of separable variables.

4. Solve according to the solution method of ordinary differential equations with separable variables

First order linear ordinary differential equation

form

inhomogeneous term

When , it is called a first-order non-homogeneous linear ordinary differential equation

Recorded as (1)

When , it is called a first-order homogeneous linear ordinary differential equation

Recorded as (2)

solution

First solve (2), and obtain its general solution by solving separable variable equations, which contains an arbitrary constant

constant change

c becomes u, u is a function of x

Substitute the "general explanation" after the change into (1)

There must be two items that can cancel each other

summary

Solve first level General solutions to linear ordinary differential equations

constant variation method

You can also directly apply the formula in subsequent problem solving

Directly apply the formula method

Bernoulli equation

form

α=0, is a first-order linear ordinary differential equation

α=1, is a separable variable type

Solution (variable substitution)

1. Identify the type

2. Divide both sides by y raised to the α power at the same time

3. Let y raised to the (α-1) power equal to z

4. Transform into a first-order linear ordinary differential equation of z with respect to x

5. Solve according to the method of solving first-order linear ordinary differential equations

Generally, the method with formula is used

Reducible second-order differential equations

direct derivative

form

Reduction method

Calculate the integral directly on both sides of the equal sign, remember to add the constant, because the integral sign here represents the original function, remember to have two constants to get the general solution

Do not show the y and eliminate the dy type

form

solution

1. Identify the type

2. Changing the element eliminates dy, because it is changing the element of the first-order differential. After changing the element, the order is reduced.

3. Identify type solution

Finally, remember to change the yuan back to the differential

Do not show x, eliminate dx type

form

solution

1. Identify the type

2. Exchange yuan to eliminate dx. After changing yuan, the level will be reduced.

3. Identify type solution

Second-order linear differential equation

form

f(x) is a non-homogeneous term

Second-order linear differential operator

L[cy]=cL[y]

L[y x]=L[y] L[x]

superposition principle

Assume that y1=y1(x), y2=y2(x) are both solutions to L[y]=0, c1 and c2 are constants, then c1y1 c2y2 is also a solution to L[y]=0

When , the two functions are linearly related, and c1 and c2 are related to each other.

When , the two functions are linearly independent, c1 and c2 are independent of each other, c1y1 c2y2 is the general solution of L[y]=0

Structural Theorem for General Solutions of Second-Order Homogeneous Linear Differential Equations

Assume y1 and y2 are two linearly independent solutions of L[y]=0, c1 and c2 are arbitrary constants, then the general solution of L[y] is y=c1y1 c2y2

Structural theorem for general solutions of second-order non-homogeneous linear differential equations

Suppose y`=y`(x) is a special solution of L[x]=f(x), and Y=c1y1 c2y2 is a general solution of L[y]=0. Then, the general solution of L[y]=f(x) is y` Y, c1, c1 are arbitrary constants

generalized superposition principle

Assume that y1 and y2 are special solutions of L[x]=f1(x) and L[x]=f2(x) respectively, then y1 y2 is the solution of L[x]=f1(x) f2(x).