MindMap Gallery 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.
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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