Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Homogeneous coordinates wikipedia, lookup

Cubic function wikipedia, lookup

Quartic function wikipedia, lookup

Elementary algebra wikipedia, lookup

System of polynomial equations wikipedia, lookup

History of algebra wikipedia, lookup

System of linear equations wikipedia, lookup

Equation wikipedia, lookup

Transcript
```Non-Homogeneous Second
Order Differential Equations
Procedure for solving non-homogeneous second order differential equations:
y" p( x) y'q( x) y  g ( x)
1. Determine the general solution y h  C1 y( x)  C2 y( x) to a homogeneous
second order differential equation: y" p( x) y'q( x) y  0
2. Find the particular solution y p of the non-homogeneous equation, using
one of the methods below.
3. The general solution of the non-homogeneous equation is:
y( x)  C1 y( x)  C2 y( x)  y p where C1 and C2 are arbitrary constants.
METHODS FOR FINDING THE PARTICULAR SOLUTION (yp ) OF A NONHOMOGENOUS EQUATION
1. Write down g(x). Start taking derivatives of g(x). List all the
Undetermined
terms of g (x) and its derivatives while ignoring the coefficients.
Coefficients.
Keep taking the derivatives until no new terms are obtained.
Restrictions:
1. D.E must have constant
coefficients:
ay"by'c  g ( x)
2. g(x) must be of a
certain, “easy to guess”
form.
Variation of
Parameters.
2. Compare the listed terms to the terms of the homogeneous
solution. If one or more terms are repeating, then the recurring
expression needs to be modified by multiplying all the repeating
terms by x.
3. Based on step 1 and 2 create an initial guess for yp.
4. Take the 1st and the 2nd derivatives of yp. Plug into the
differential equation. Solve for the constants.
5. Plug the values of the constants into yp.
y 2 ( x) g ( x)
y ( x) g ( x)
dx  y 2  1
dx
W ( y1 , y 2 )( x)
W ( y1 , y 2 )( x)
where y1 and y2 are solutions to the homogeneous equation and
y y2
W ( y1 , y 2 )( x)  1
 y1 y 2  y 2 y1
y1 y 2
y p ( x)   y1 
The set of solutions is linearly independent in I if
W ( y1 , y 2 )( x)  0 for every x in the interval.
Or equivalently:
y p ( x)  v1 y1  v2 y 2
where y1 and y2 are solutions to the homogeneous equation and
v1 and v2 are unknown functions of x.
To determine v1 and v2, solve the following system of equations
for v′1 and v′2.
y1v1  y 2 v 2  0
y1v1  y 2 v 2  g ( x)
Integrate v′1 and v′2 to find v1 and v2.
Substitute v1 and v2 into y p ( x)  v1 y1  v2 y 2
www.rit.edu/asc
Example #1. Solve the differential equation:
Solution:
1. Homogeneous equation:
y   2 y   t  e t
y  2 y  0
Characteristic equation:
r 2  2r  0
r (r  2)  0
r  0, r  2
2t
 yh  C1  C2 e
2. Particular solution:
g (t )  t  e t
The constant is already in the
homogeneous solution.
Multiplying it by t will repeat the
terms of g(t). So we need to
modify both the constant and the t.
t, et
 Terms: C , e t
g ' (t )  1  e t
g " (t )  e t
et
No new terms.
y p  ( At  B)  Ce t



Initial guess of yp
Part of a homogeneous solution.
Both terms need to be modified
Modify yp:
y p  t ( At  B)  Ce t  At 2  Bt  Ce t
y p  2 At  B  Ce t
y" p  2 A  Ce t
Plug the yp and its derivatives into the original differential equation:
y   2 y   t  e t implies 2 A  Ce t  22 At  B  Ce t   t  e t
2 A  2B  4 At  Ce t  t  e t 
 C  1  C  1
 4A  1  A  
1
4
2 A  2B  0  B  
So y p  
1
4
t2 t
t
  e t   (t  1)  e t and the general solution is:
4 4
4
y  yh  y p
t
y  C1  C 2 e 2t  (t  1)  e t
4
www.rit.edu/asc
Example #2. Solve the differential equation: y  2 y  y 
1. Homogeneous equation:
et
t
y  2 y  y  0
Characteristic equation:
r 2  2r  1  0
(r  1) 2  0
r  1, r  1
t
t
 yh  C1e  C2 te
y1  e t and y 2  te t
Not an “easy to guess”
function. It is a quotient so the
derivatives will get more
complicated, making it
impossible to list all terms.
y'1  e t and y' 2  te t  e t
2. Particular solution:
W ( y1 , y 2 )( x)  det
y1
y1
y2
et
 det t
y 2
e


te t
 e t te t  e t  te t  e t  te 2t  e 2t  te 2t  e 2t
t
t
te  e
So
y p ( x)   y1 
 e t 
y 2 ( x) g ( x)
y ( x) g ( x)
dx  y 2  1
dx 
W ( y1 , y 2 )( x)
W ( y1 , y 2 )( x)
et
et
et 
t dt  te t
t dt  e t 1dt  te t 1dt  te t  te t ln t
2t
2t


t
e
e
te t 
y p (t )  te t  te t ln t and the general solution is:
y  C1e t  C2 te t  te t  te t ln t  C1e t  C3te t  te t ln t
You try it:
1.
y" y'2 y  sin 2 x
2. y"2 y' y  xe x
3. y" y  sec x
www.rit.edu/asc
Solutions:
3
1
sin 2 x  cos 2 x
20
20
1
#2: y  C1e x  C 2 xe x  x 3 e x
6
#3: y  C1 cos x  C2 sin x  cos x ln cos x   x sin x
#1: y  C1e  x  C 2 e 2 x 
www.rit.edu/asc
```
Related documents