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MindMap Gallery second order differential equation

second order differential equation

Still worried about learning second-order differential equations? This is a mind map about second-order differential equations, which summarizes knowledge about differential equations with constant coefficients, non-homogeneous equations, English examples, etc.

Edited at 2021-10-29 18:24:03

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Outline

second order differential equation

Homogeneous differential equations with constant coefficients

General formula:y^'' p(t)y^' q(t)y=0(1.1)

Solve for p(t) (q(t) is a constant)

Let y=e^rt

Find the roots r1 and r2 of the characteristic equation (where ar^2 br c=0, △>0)

General solution: y=c1*e^(r1*t) c2*e^(r2*t)

Solution

Vronsky method

differential operator

L[y(t)]=y'' (t) p*y' (t) q*y

The nature and structure of solutions

y'' p(t)*y'=g(t),y(t0)=y0,y'(t0)=y0'

Superposition principle: If y1 and y2 are two solutions of 1.1, then c1*y1 c2*y2

y1, y2 are two different solutions of 1.1, and define W[y1,y2]=

y1 and y2 are two different solutions of 1.1. y(t)=c1*y1 c2*y2 contains all the solutions of 1.1. The necessary and sufficient condition is that there is a point t0 so that the Wronskian determinant is not 0 at the t0 point.

If y1 and y2 satisfy the two solutions of condition 1.1, then y1 and y2 constitute its basic solution system.

Abel's theorem: y1 and y2 are two solutions of 1.1, then their Longfursky determinant is

Solving for characteristic roots as complex numbers

Euler's formula: e^i*t=cos t=sin t

Generally: e^(m i*u)*t=e^(m*t)(cos u*t i*sin u*t)

General solution form: y=e^m*t(c1*cos u*t c2*i*sin u*t)

The condition y^'' p(t)y^' q(t)y=0 has a pair of complex characteristic roots m i*u and m-i*u

Solving the characteristic roots as multiple roots

Condition: y^'' p(t)y^' The characteristic root of q(t)y=0 is a double root

General solution form: y=c1*e^(r*t) c2*e^(r*t)

non-homogeneous equations

General formula: L[y(t)]=y'' (t) p*y' (t) q*y=g(t)

solution structure

Theorem 1: L[y1]=g(t), L[y2]=0, then L[y1-y2]

Theorem 2: L[y1]=g(t), L[y2]=0, W[y1,y2]≠0, L[Y]=g(t), then if L[&]=g(t) , then &=c1*y1 c2*y2 Y

Find special solution method

undetermined coefficient method

If g(t)=k*e^(a*t) in the equation, then let Y=A*e^(a*t)

If g(t)=k*sin (B*t) or g(t)=k*cos (B*t), then let Y be the linear combination of sin (B*t) and cos(B*t)

If g(t) is a polynomial, then let Y be a polynomial of the same degree

Application examples

Spring mass damping system:m*u''=f(t)

f(t) composition

Gravity m*g, always downward

Spring force Fs=-k*(L u)

resistance

>0, u increases, f goes up

<0, u decreases, f goes down

external force

Undamped free vibration

m*u'' k*u=0, general solution

Movement characteristics

amplitude

Damped vibration

mu'' ru' ku=0,△=r^2-4k*m

solution form

△>0,

△=0,

△<0,

forced vibration

The external force is periodic excitation, mu'' r*u'(t)=F0cos wt

general form of solution