Outline
Week 12: Application of vector differential equations to electrical
networks; vector differential equations
Course Notes: 6.3, 6.4
Goals: be able to solve a linear system of differential equations; find
characteristics of electiral networks involving inductors and capacitors
using methods learned this term.
Differential Equations
We’re going to doing this in a linear-systems context soon.
y 0 (t) = λy (t),
λ constant
Differential Equations
We’re going to doing this in a linear-systems context soon.
y 0 (t) = λy (t),
Solutions: y (t) = Ce λt , constant C
λ constant
Differential Equations
We’re going to doing this in a linear-systems context soon.
y 0 (t) = λy (t),
λ constant
Solutions: y (t) = Ce λt , constant C
Example: a population’s growth is 0.3 times the number of individuals in
that population per year.
Differential Equations
We’re going to doing this in a linear-systems context soon.
y 0 (t) = λy (t),
λ constant
Solutions: y (t) = Ce λt , constant C
Example: a population’s growth is 0.3 times the number of individuals in
that population per year.
y 0 (t) = 0.3y (t)
Differential Equations
We’re going to doing this in a linear-systems context soon.
y 0 (t) = λy (t),
λ constant
Solutions: y (t) = Ce λt , constant C
Example: a population’s growth is 0.3 times the number of individuals in
that population per year.
y 0 (t) = 0.3y (t)
y (t) = Ce 0.3t
for some constant C
Differential Equations
We’re going to doing this in a linear-systems context soon.
y 0 (t) = λy (t),
λ constant
Solutions: y (t) = Ce λt , constant C
Example: a population’s growth is 0.3 times the number of individuals in
that population per year.
y 0 (t) = 0.3y (t)
y (t) = Ce 0.3t
for some constant C
At year t = 0, there are 100 individuals.
Differential Equations
We’re going to doing this in a linear-systems context soon.
y 0 (t) = λy (t),
λ constant
Solutions: y (t) = Ce λt , constant C
Example: a population’s growth is 0.3 times the number of individuals in
that population per year.
y 0 (t) = 0.3y (t)
y (t) = Ce 0.3t
for some constant C
At year t = 0, there are 100 individuals.
y (t) = 100e 0.3t
Differential Equations
Example: a radioactive substance decays at a rate of 2% of its mass every
year.
Differential Equations
Example: a radioactive substance decays at a rate of 2% of its mass every
year.
y 0 (t) = −0.02y (t)
Differential Equations
Example: a radioactive substance decays at a rate of 2% of its mass every
year.
y 0 (t) = −0.02y (t)
y (t) = Ce −0.02t
where C is the amount at t = 0
Systems of Linear Differential Equations
y10 (t) = ay1 (t) + by2 (t)
y20 (t) = cy1 (t) + dy2 (t)
Systems of Linear Differential Equations
y10 (t) = ay1 (t) + by2 (t)
y20 (t) = cy1 (t) + dy2 (t)
0 y (t)
y := 10
y2 (t)
0
a b
A=
c d
y0 = Ay
y (t)
y := 1
y2 (t)
Guessing Solutions: Eigenvectors
Differential Equation:
y0 = Ay
Let’s take a guess from our previous examples: what if
y = e λt x
for some constant λ and some vector x?
Guessing Solutions: Eigenvectors
Differential Equation:
y0 = Ay
Let’s take a guess from our previous examples: what if
y = e λt x
for some constant λ and some vector x?
Then:
So, if y0 = Ay:
Hence:
y0 = λe λt x
λe λt x = A(e λt x)
λx = Ax
so λ and x are an eigenvalue/eigenvector pair of A.
Guessing Solutions: Eigenvectors
Differential Equation:
y0 = Ay
Let’s take a guess from our previous examples: what if
y = e λt x
for some constant λ and some vector x?
y0 = λe λt x
Then:
λe λt x = A(e λt x)
So, if y0 = Ay:
Hence:
λx = Ax
so λ and x are an eigenvalue/eigenvector pair of A.
We’ve successfully guessed a solution!
y = e λt x
where λ, x are an eigenvalue/eigenvector pair
Systems of Linear Differential Equations: Adding Solutions
Adding Solutions
Suppose y1 and y2 are both solutions to the system of differential
equations Ay = y0 .
Systems of Linear Differential Equations: Adding Solutions
Adding Solutions
Suppose y1 and y2 are both solutions to the system of differential
equations Ay = y0 .
Then (y1 + y2 ) is also a solution.
Systems of Linear Differential Equations: Adding Solutions
Adding Solutions
Suppose y1 and y2 are both solutions to the system of differential
equations Ay = y0 .
Then (y1 + y2 ) is also a solution.
Further, (c1 y1 + c2 y2 ) is also a solution, for any constants c1 and c2 .
(Home exercise: prove this is true!)
Solutions to Systems of Linear Differential Equations
Theorem
Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk
and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck ,
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
is a solution to the equation y0 = Ay.
Solutions to Systems of Linear Differential Equations
Theorem
Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk
and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck ,
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
is a solution to the equation y0 = Ay.
Example: y0 = Iy, y ∈ R2
Solutions to Systems of Linear Differential Equations
Theorem
Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk
and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck ,
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
is a solution to the equation y0 = Ay.
2
Example: y0 = Iy,
y∈R
1
λ1 = 1, x1 =
;
0
0
λ2 = 1, x2 =
;
1
Solutions to Systems of Linear Differential Equations
Theorem
Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk
and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck ,
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
is a solution to the equation y0 = Ay.
2
Example: y0 = Iy,
y∈R
1
λ1 = 1, x1 =
;
0
0
λ2 = 1, x2 =
;
1
1
0
t
t
y(t) = c1 e
+ c2 e
.
0
1
Solutions to Systems of Linear Differential Equations
Theorem
Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk
and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck ,
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
is a solution to the equation y0 = Ay.
General Question: Is there a solution to y0 = Ay that also has y(0) = y0 ,
for some constant vector y0 ?
Solutions to Systems of Linear Differential Equations
Theorem
Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk
and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck ,
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
is a solution to the equation y0 = Ay.
General Question: Is there a solution to y0 = Ay that also has y(0) = y0 ,
for some constant vector y0 ?
Suppose it has the form above:
y(0) = c1 x1 + c2 x2 + · · · + ck xk
?
=? y0
Solutions to Systems of Linear Differential Equations
Theorem
Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk
and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck ,
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
is a solution to the equation y0 = Ay.
General Question: Is there a solution to y0 = Ay that also has y(0) = y0 ,
for some constant vector y0 ?
Suppose it has the form above:
y(0) = c1 x1 + c2 x2 + · · · + ck xk
?
=? y0
If the eigenvalues of A form a basis then there is always exactly one
solution to y0 = Ay with any desired initial condition y(0) = y0 , and it has
the form given above.
Solutions to Systems of Linear Differential Equations
Theorem
Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk
and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck ,
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
is a solution to the equation y0 = Ay.
2
Example: y0 = Iy,
y∈R
1
λ1 = 1, x1 =
;
0
0
λ2 = 1, x2 =
;
1
Solutions to Systems of Linear Differential Equations
Theorem
Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk
and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck ,
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
is a solution to the equation y0 = Ay.
2
Example: y0 = Iy,
y∈R
1
λ1 = 1, x1 =
;
0
0
λ2 = 1, x2 =
;
1
7
Suppose we have initial conditions y(0) =
.
−3
Solutions to Systems of Linear Differential Equations
Theorem
Suppose A is an n-by-n matrix with eigenvalues and vectors λ1 , λ2 , . . . , λk
and x1 , x2 , . . . , xk . Then for any choice of constants c1 , c2 , . . . , ck ,
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
is a solution to the equation y0 = Ay.
2
Example: y0 = Iy,
y∈R
1
λ1 = 1, x1 =
;
0
0
λ2 = 1, x2 =
;
1
7
Suppose we have initial conditions y(0) =
.
−3
t 1
0
7e
y = 7e t
− 3e t
=
.
0
1
−3e t
Example
Find the solution to the system of linear differential equations
y10 (t) = y1 (t) + 4y2 (t) + 5y3 (t)
y20 (t) =
2y2 (t) + 6y3 (t)
0
y3 (t) =
3y3 (t)
with initial condition
 
0
y(0) = 11
2
Example
Find the solution to the system of linear differential equations
y10 (t) = y1 (t) + 4y2 (t) + 5y3 (t)
y20 (t) =
2y2 (t) + 6y3 (t)
0
y3 (t) =
3y3 (t)
with initial condition


1 4 5
A = 0 2 6
0 0 3
 
0
y(0) = 11
2
solving y0 = Ay
Example
Find the solution to the system of linear differential equations
y10 (t) = y1 (t) + 4y2 (t) + 5y3 (t)
y20 (t) =
2y2 (t) + 6y3 (t)
0
y3 (t) =
3y3 (t)
with initial condition


1 4 5
A = 0 2 6
0 0 3
 
1
λ1 = 1, x1 = 0
0
 
0
y(0) = 11
2
solving y0 = Ay
 
4
λ2 = 2, x2 = 1
0
 
29
λ3 = 3, x3 = 12
2
The form of the solution will be:
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
That is:
 
 
 
1
4
29
y(t) = c1 e t 0 + c2 e 2t 1 + c3 e 3t 12
0
0
2
To find the constants c1 , c2 , c3 we solve:
 
 
 
 
0
1
4
29
11 = c1 0 + c2 1 + c3 12
2
0
0
2
So c1 = −25, c2 = −1, and c3 = 1. Our solution is:
 
 
 


1
4
29
−25e t − 4e 2t + 29e 3t

−e 2t + 12e 3t
y(t) = −25e t 0 − 1e 2t 1 + e 3t 12 = 
3t
0
0
2
2e
Example
Find the solution to the system of linear differential equations
y10 (t) = y1 (t)
y20 (t) = 3y1 (t) − y2 (t)
with initial condition
y(0) =
4
1
Example
Find the solution to the system of linear differential equations
y10 (t) = y1 (t)
y20 (t) = 3y1 (t) − y2 (t)
with initial condition
y(0) =
1 0
A=
3 −1
4
1
Example
Find the solution to the system of linear differential equations
y10 (t) = y1 (t)
y20 (t) = 3y1 (t) − y2 (t)
with initial condition
y(0) =
1 0
A=
3 −1
2
λ1 = 1, x1 =
3
4
1
0
λ2 = −1, x2 =
1
The form of the solution will be:
y(t) = c1 e λ1 t x1 + c2 e λ2 t x2 + · · · + ck e λk t xk
That is:
2
−t 0
y(t) = c1 e
+ c2 e
3
1
t
To find the constants c1 , c2 we solve:
2
0
4
= c1
+ c2
8
3
1
So c1 = 2, c2 = −5. Our solution is:
2
4e t
−t 0
y(t) = 2e
− 5e
=
3
1
6e t − 5e −t
t
Complex Eigenvalues
0 −2
y (t) =
y(t)
8 0
0
−4
y(0) =
12
Complex Eigenvalues
0 −2
y (t) =
y(t)
8 0
0
Eigenvalues: λ1 = 4i, λ2 = −4i
−4
y(0) =
12
Complex Eigenvalues
0 −2
y (t) =
y(t)
8 0
0
Eigenvalues: λ1 = 4i,
λ2 = −4i
i
−i
Eigenvectors: x1 =
, x2 =
2
2
−4
y(0) =
12
Complex Eigenvalues
0 −2
y (t) =
y(t)
8 0
0
−4
y(0) =
12
Eigenvalues: λ1 = 4i,
λ2 = −4i
i
−i
Eigenvectors: x1 =
, x2 =
2
2
General solution: y(t) = c1 e 4it x1 + c2 e −4it x2
for some constants c1 and c2 .
Complex Eigenvalues
0 −2
y (t) =
y(t)
8 0
0
−4
y(0) =
12
Eigenvalues: λ1 = 4i,
λ2 = −4i
i
−i
Eigenvectors: x1 =
, x2 =
2
2
General solution: y(t) = c1 e 4it x1 + c2 e −4it x2
for some constants c1 and c2 .
Particular solution: y(t) = (3 + 2i)e 4it x1 + (3 − 2i)e −4it x2
Complex Eigenvalues: Closer Look
Particular solution:
y(t) = (3 + 2i)e 4it x1 + (3 − 2i)e −4it x2
= (3 + 2i)[cos(4t) + i sin(4t)]x1 + (3 − 2i)[cos(−4t) + i sin(−4t)]x2
= (3 + 2i)[cos(4t) + i sin(4t)]x1 + (3 − 2i)[cos(4t) − i sin(4t)]x2
i
i
= (3 + 2i)[cos(4t) + i sin(4t)]
+ (3 − 2i)[cos(4t) − i sin(4t)]
2
2
= ···
−4 cos(4t) − 6 sin(4t)
=
12 cos(4t) − 8 sin(4t)
Complex Eigenvalues: Closer Look
Suppose c1 = a + bi = c2 . Then the general solution is:
y(t) = c1 e 4it x1 + c2 e −4it x2
= c1 e 4it x1 + c1 e −4it x2
= c1 e 4it x1 + c1 e 4it x1
= c1 e 4it x1 + c1 e 4it x1
−2 sin(4t)
−2 cos(4t)
=a
+b
4 cos(4t)
−4 sin(4t)
which is real.
Complex Eigenvalues: Closer Look
Suppose c1 = a + bi = c2 . Then the general solution is:
y(t) = c1 e 4it x1 + c2 e −4it x2
= c1 e 4it x1 + c1 e −4it x2
= c1 e 4it x1 + c1 e 4it x1
= c1 e 4it x1 + c1 e 4it x1
−2 sin(4t)
−2 cos(4t)
=a
+b
4 cos(4t)
−4 sin(4t)
which is real.
Whence c1 and c2 ? Are they always conjugates when y(0) is real?
Complex Eigenvalues: Closer Look
Suppose c1 = a + bi = c2 . Then the general solution is:
y(t) = c1 e 4it x1 + c2 e −4it x2
= c1 e 4it x1 + c1 e −4it x2
= c1 e 4it x1 + c1 e 4it x1
= c1 e 4it x1 + c1 e 4it x1
−2 sin(4t)
−2 cos(4t)
=a
+b
4 cos(4t)
−4 sin(4t)
which is real.
Whence c1 and c2 ? Are they always conjugates when y(0) is real?
c1 x1 + c2 x2 = y(0)
c1 x1 + c2 x1 = y(0)
If y(0) has real only entries and x1 does not, then c1 = c2 .
Complex Eigenvalues
1 1
A=
−1 1
Complex Eigenvalues
1 1
A=
−1 1
Eigenvalues: λ1 = 1 + i, λ2 = 1 − i
Complex Eigenvalues
1 1
A=
−1 1
Eigenvalues: λ1 = 1+ i, λ2 = 1 − i
−i
Eigenvectors: x1 =
,
1
Complex Eigenvalues
1 1
A=
−1 1
Eigenvalues: λ1 = 1+ i, λ2 = 1−i
−i
i
Eigenvectors: x1 =
, x2 =
1
1
Complex Eigenvalues
1 1
A=
−1 1
Eigenvalues: λ1 = 1+ i, λ2 = 1−i
−i
i
Eigenvectors: x1 =
, x2 =
1
1
One solution:
Complex Eigenvalues
1 1
A=
−1 1
Eigenvalues: λ1 = 1+ i, λ2 = 1−i
−i
i
Eigenvectors: x1 =
, x2 =
1
1
One solution:
(1+i)t −i
e
1
−i
=e t e it
1
−i
=e (cos(t) + i sin(t))
1
t
t
e sin t
−e cos t
= t
+i
e cos t
e t sin t
t
Complex Eigenvalues
1 1
A=
−1 1
Eigenvalues: λ1 = 1+ i, λ2 = 1−i
−i
i
Eigenvectors: x1 =
, x2 =
1
1
One solution:
(1+i)t −i
e
1
t
t
e sin t
−e cos t
= t
+i
e cos t
e t sin t
General Solution:
=c1
t
e t sin t
−e cos t
+ c2
e t cos t
e t sin t
Complex Eigenvalues
1
0
4
A=
−5 −2
Complex Eigenvalues
1
0
4
A=
−5 −2
Eigenvalues: λ1 = −1 + 12 i,
λ2 = −1 − 12 i
Complex Eigenvalues
1
0
4
A=
−5 −2
Eigenvalues: λ1 = −1 + 12 i,
2+i
Eigenvectors: x1 =
,
−10
λ2 = −1 − 12 i
2−i
x2 =
−10
Complex Eigenvalues
1
0
4
A=
−5 −2
Eigenvalues: λ1 = −1 + 12 i,
2+i
Eigenvectors: x1 =
,
−10
Choosing one:
1
λ2 = −1 − 12 i
2−i
x2 =
−10
e λ1 t x1 = e (−1+ 2 i)t x1 = e −t e it/2 x1 = e −t (cos(t/2) + i sin(t/2))x1
2+i
−t
= e (cos(t/2) + i sin(t/2))
−10
−t 2 cos(t/2) − sin(t/2)
−t cos(t/2) + sin(t/2)
=e
+ ie
−10 cos(t/2)
−10 sin(t/2)
Complex Eigenvalues
1
0
4
A=
−5 −2
Eigenvalues: λ1 = −1 + 12 i,
λ2 = −1 − 12 i
2+i
2−i
Eigenvectors: x1 =
,
x2 =
−10
−10
Choosing one:
λ1 t
−t 2 cos(t/2) − sin(t/2)
−t cos(t/2) + sin(t/2)
e x1 = e
+ ie
−10 cos(t/2)
−10 sin(t/2)
General solution:
c1 2 cos(t/2) − sin(t/2)
c2 cos(t/2) + sin(t/2)
+ t
−10 cos(t/2)
−10 sin(t/2)
et
e