Transcript Math 4B Second Order

Differential Equations
Second-Order Linear DEs
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
A Second-Order Linear Differential Equation
can always be put into the form:
y  p(t)y  q(t)  y  g(t)
The general solution will always have the form:
y(t)  yh  yp
yh is the solution to the corresponding
homogeneous equation, where g(t)=0
yp is a particular solution to the original DE.
There should be two independent solutions to the homogeneous
equation, and yh will be a linear combination of these two.
In fact, the two independent solutions form a basis for a 2dimensional solution space (an actual vector space like we saw
last quarter).
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here are a few examples of linear, second
order DEs with constant coefficients.
1) y  3y  2y  0
2) y 2y  y  0
3) y  y  0
4) y 4y  13y  0
These equations are all homogeneous. We will solve some
inhomogeneous equations later. Also, we will add some
initial conditions to these problems.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
1) y  3y  2y  0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
1) y  3y  2y  0
For these constant-coefficient equations we will essentially
turn this into an algebra problem. Start by writing down the
CHARACTERISTIC EQUATION.
r2  3r  2  0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
1) y  3y  2y  0
For these constant-coefficient equations we will essentially
turn this into an algebra problem. Start by writing down the
CHARACTERISTIC EQUATION.
r2  3r  2  0
Solve for the roots of this equation, then each root will
correspond to a solution of the DE.
(r  1)(r  2)  0
r  1 or r  2
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
1) y  3y  2y  0
For these constant-coefficient equations we will essentially
turn this into an algebra problem. Start by writing down the
CHARACTERISTIC EQUATION.
r2  3r  2  0
Solve for the roots of this equation, then each root will
correspond to a solution of the DE.
(r  1)(r  2)  0
r  1 or r  2
y1(t)  e1t ; y2(t)  e2t
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
1) y  3y  2y  0
For these constant-coefficient equations we will essentially
turn this into an algebra problem. Start by writing down the
CHARACTERISTIC EQUATION.
r2  3r  2  0
Solve for the roots of this equation, then each root will
correspond to a solution of the DE.
(r  1)(r  2)  0
r  1 or r  2
y1(t)  e1t ; y2(t)  e2t
The general solution is a linear combination of these:
yh(t)  c1et  c2e2t
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
1) y  3y  2y  0
y(0)  0; y(0)  2
Now add some initial conditions.
We will need to determine the constants in our general
solution that match up with the given conditions.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
1) y  3y  2y  0
y(0)  0; y(0)  2
Now add some initial conditions.
We will need to determine the constants in our general
solution that match up with the given conditions.
yh(t)  c1et  c2e2t  0  c1e0  c2e0

0
0
t
2
t
 (t)  c1e  2c2e  2  c1e  2c2e
yh
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
1) y  3y  2y  0
y(0)  0; y(0)  2
Now add some initial conditions.
We will need to determine the constants in our general
solution that match up with the given conditions.
yh(t)  c1et  c2e2t  0  c1e0  c2e0

0
0
t
2
t
2

c
e

2
c
e


yh(t)  c1e  2c2e 
1
2
1 1 c1 
1 2 c  

 2 
1 1 0
0 1 2 


0
1 1 0
2  1 2 2
 


c1  2
1 0  2

0 1 2 
c2  2


yh(t)  2et  2e2t
Here is our final solution.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
2) y 2y  y  0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
2) y 2y  y  0
Start with the characteristic equation:
r2  2r  1  0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
2) y 2y  y  0
Start with the characteristic equation:
r2  2r  1  0
(r  1)(r  1)  0
r  1
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
2) y 2y  y  0
Start with the characteristic equation:
r2  2r  1  0
(r  1)(r  1)  0
r  1
This time there is only one repeated root, so
it seems like there is only one solution:
y1  e t
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
2) y 2y  y  0
Start with the characteristic equation:
r2  2r  1  0
(r  1)(r  1)  0
r  1
This time there is only one repeated root, so
it seems like there is only one solution:
y1  e t
This is one of the solutions, but we need to find another
independent solution. The trick is to multiply this solution by t:
y2  tet
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
2) y 2y  y  0
Start with the characteristic equation:
r2  2r  1  0
(r  1)(r  1)  0
r  1
This time there is only one repeated root, so
it seems like there is only one solution:
y1  e t
This is one of the solutions, but we need to find another
independent solution. The trick is to multiply this solution by t:
y2  tet
Now the general solution is a linear combination of these:
yh  c1et  c2tet
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
2) y 2y  y  0
y(2)  2; y(2)  1
Now add some initial conditions.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
2) y 2y  y  0
y(2)  2; y(2)  1
Now add some initial conditions.
y  c1et  c2tet
y  c1et  c2(et  tet )
Calculate the derivative, then
plug in to find the constants.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
2) y 2y  y  0
y(2)  2; y(2)  1
Now add some initial conditions.
y  c1et  c2tet
y  c1et  c2(et  tet )
Calculate the derivative, then
plug in to find the constants.
2  c1e2  c2(2)e2
 1  c1e2  c2(e2  (2)e2 )
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
2) y 2y  y  0
y(2)  2; y(2)  1
Now add some initial conditions.
y  c1et  c2tet
y  c1et  c2(et  tet )
Calculate the derivative, then
plug in to find the constants.
2  c1e2  c2(2)e2
 1  c1e2  c2(e2  (2)e2 )
2e2  c1  2c2  c1  0

2
2
c

e
2
e  c1  c2 
y  e2tet  y  te2t
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
3) y  y  0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
3) y  y  0
r2  1  0
r  i
it
This time the roots are complex numbers.
y1  e ; y2  e
it
The complex exponential solutions are
perfectly fine, but we can find real-valued
solutions as well by using Euler’s formula.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
3) y  y  0
r2  1  0
r  i
it
This time the roots are complex numbers.
y1  e ; y2  e
it
ei  cos()  i sin()
The complex exponential solutions are
perfectly fine, but we can find real-valued
solutions as well by using Euler’s formula.
This is Euler’s formula.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
3) y  y  0
r2  1  0
r  i
it
This time the roots are complex numbers.
y1  e ; y2  e
it
The complex exponential solutions are
perfectly fine, but we can find real-valued
solutions as well by using Euler’s formula.
ei  cos()  i sin()
This is Euler’s formula.
y1  cos(t)  i sin(t)
y2  cos(t)  i sin(t))
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
3) y  y  0
r2  1  0
r  i
it
This time the roots are complex numbers.
y1  e ; y2  e
it
ei  cos()  i sin()
y1  cos(t)  i sin(t)
The complex exponential solutions are
perfectly fine, but we can find real-valued
solutions as well by using Euler’s formula.
This is Euler’s formula.
cos(-t)=cos(t) and sin(-t)=-sin(t)
y2  cos(t)  i sin(t)  y2  cos(t)  i sin(t)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
3) y  y  0
r2  1  0
r  i
it
This time the roots are complex numbers.
y1  e ; y2  e
it
ei  cos()  i sin()
y1  cos(t)  i sin(t)
The complex exponential solutions are
perfectly fine, but we can find real-valued
solutions as well by using Euler’s formula.
This is Euler’s formula.
cos(-t)=cos(t) and sin(-t)=-sin(t)
y2  cos(t)  i sin(t)  y2  cos(t)  i sin(t)
Any linear combination of these solutions will
also be a solution to the DE, specifically:
1 (y 
2 1
 i (y1
2
y2 )  cos(t)
 y2 )  sin(t)
yh  c1 cos(t)  c2 sin(t)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
3) y  y  0
y(0)  0; y(0)  2
Plug in the given values to find
the solution that matches up.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
3) y  y  0
y(0)  0; y(0)  2
Plug in the given values to find
the solution that matches up.
y  c1 cos(t)  c2 sin(t)
y  c1 sin(t)  c2 cos(t)
0  c1 cos(0)  c2 sin(0)  c1  0
2  c1 sin(0)  c2 cos(0)  c2  2
y  2 sin(t)
The solution is simply an
oscillation because the roots
of the characteristic equation
were completely imaginary.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
4) y 4y  13y  0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
4) y 4y  13y  0
r2  4r  13  0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
4) y 4y  13y  0
r2  4r  13  0
 4  42  4(1)(13)
r
2(1)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
4) y 4y  13y  0
r2  4r  13  0
 4  42  4(1)(13)
r
2(1)
r  2  3i
These roots have both a real
and imaginary part.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
4) y 4y  13y  0
r2  4r  13  0
 4  42  4(1)(13)
r
2(1)
r  2  3i
These roots have both a real
and imaginary part.
The corresponding general solution is:
yh  e2t (c1 cos(3t)  c2 sin(3t))
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
4) y 4y  13y  0
Now add some initial values
y(0)  2; y(0)  1
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
4) y 4y  13y  0
Now add some initial values
y(0)  2; y(0)  1
The general solution is:
y  e2t (c1 cos(3t)  c2 sin(3t))
Compute the first derivative, then plug in the
given values:
y  e2t (3c1 sin(3t)  3c2 cos(3t))  2e2t (c1 cos(3t)  c2 sin(3t))
y(0)  2  2  e0(c1 cos(0)  c2 sin(0))
2  c1
y(0)  1  1  e0 (3c1 sin(0)  3c2 cos(0))  2e0 (c1 cos(0)  c2 sin(0))
 1  3c2  2c1
 1  3c2  4  c2  1
y  e2t (2 cos(3t)  sin(3t))
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Now add some initial values
4) y 4y  13y  0
y(0)  2; y(0)  1
The solution is y  e2t (2 cos(3t)  sin(3t))
Using some trigonometry this can be
re-written as a single cosine function.
2t
y  5e
cos(3t  )
  tan1( 1 )  0.46 radians
2
Here is the general formula:
y  c1 cos(t)  c2 sin(t)  y  A cos(t  )
c
1 2
A  c12  c2
;


tan
( )
2
c1
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Now add some initial values
4) y 4y  13y  0
y(0)  2; y(0)  1
The solution is y  e2t (2 cos(3t)  sin(3t))
Using some trigonometry this can be
re-written as a single cosine function.
2t
y  5e
cos(3t  )
  tan1( 1 )  0.46 radians
2
The plot of the solution shows
the function oscillating
between the exponential
“envelope” and eventually
approaching 0.
2
1
0.5
1.0
1.5
2.0
2.5
3.0
1
2
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB