Laplace Transforms for Systems of Differential Equationslogo1 New Idea An Example Double Check...
Transcript of Laplace Transforms for Systems of Differential Equationslogo1 New Idea An Example Double Check...
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New Idea An Example Double Check
Laplace Transforms for Systems ofDifferential Equations
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
The Laplace Transform of a System
1. When you have several unknown functions x,y, etc., thenthere will be several unknown Laplace transforms.
2. Transform each equation separately.3. Solve the transformed system of algebraic equations for
X,Y, etc.4. Transform back.5. The example will be first order, but the idea works for any
order.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
The Laplace Transform of a System1. When you have several unknown functions x,y, etc., then
there will be several unknown Laplace transforms.
2. Transform each equation separately.3. Solve the transformed system of algebraic equations for
X,Y, etc.4. Transform back.5. The example will be first order, but the idea works for any
order.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
The Laplace Transform of a System1. When you have several unknown functions x,y, etc., then
there will be several unknown Laplace transforms.2. Transform each equation separately.
3. Solve the transformed system of algebraic equations forX,Y, etc.
4. Transform back.5. The example will be first order, but the idea works for any
order.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
The Laplace Transform of a System1. When you have several unknown functions x,y, etc., then
there will be several unknown Laplace transforms.2. Transform each equation separately.3. Solve the transformed system of algebraic equations for
X,Y, etc.
4. Transform back.5. The example will be first order, but the idea works for any
order.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
The Laplace Transform of a System1. When you have several unknown functions x,y, etc., then
there will be several unknown Laplace transforms.2. Transform each equation separately.3. Solve the transformed system of algebraic equations for
X,Y, etc.4. Transform back.
5. The example will be first order, but the idea works for anyorder.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
The Laplace Transform of a System1. When you have several unknown functions x,y, etc., then
there will be several unknown Laplace transforms.2. Transform each equation separately.3. Solve the transformed system of algebraic equations for
X,Y, etc.4. Transform back.5. The example will be first order, but the idea works for any
order.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
So Everything Remains As It Was
Time Domain (t)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
So Everything Remains As It Was
Time Domain (t)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
So Everything Remains As It Was
Time Domain (t)
OriginalDE & IVP
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
So Everything Remains As It Was
Time Domain (t)
OriginalDE & IVP
-L
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
So Everything Remains As It Was
Time Domain (t)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
-L
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
So Everything Remains As It Was
Time Domain (t) Transform domain (s)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
-L
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
So Everything Remains As It Was
Time Domain (t) Transform domain (s)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
-L
Algebraic solution,partial fractions
?
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
So Everything Remains As It Was
Time Domain (t) Transform domain (s)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
Laplace transformof the solution
-L
Algebraic solution,partial fractions
?
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
So Everything Remains As It Was
Time Domain (t) Transform domain (s)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
Laplace transformof the solution
-
�
L
L −1
Algebraic solution,partial fractions
?
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
So Everything Remains As It Was
Time Domain (t) Transform domain (s)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
Laplace transformof the solutionSolution
-
�
L
L −1
Algebraic solution,partial fractions
?
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
1. Note that the second equation is not really a differentialequation.
2. This is not a problem. The transforms will work the sameway, ...
3. ... but the second equation also relates the initial values toeach other. So certain combinations of initial values willnot be possible.
4. The initial values in this example are allowed.5. The example itself is related to equations that come from
the analysis of two loop circuits. So systems such as thisone certainly arise in applications.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
1. Note that the second equation is not really a differentialequation.
2. This is not a problem. The transforms will work the sameway, ...
3. ... but the second equation also relates the initial values toeach other. So certain combinations of initial values willnot be possible.
4. The initial values in this example are allowed.5. The example itself is related to equations that come from
the analysis of two loop circuits. So systems such as thisone certainly arise in applications.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
1. Note that the second equation is not really a differentialequation.
2. This is not a problem. The transforms will work the sameway, ...
3. ... but the second equation also relates the initial values toeach other. So certain combinations of initial values willnot be possible.
4. The initial values in this example are allowed.5. The example itself is related to equations that come from
the analysis of two loop circuits. So systems such as thisone certainly arise in applications.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
1. Note that the second equation is not really a differentialequation.
2. This is not a problem. The transforms will work the sameway, ...
3. ... but the second equation also relates the initial values toeach other. So certain combinations of initial values willnot be possible.
4. The initial values in this example are allowed.5. The example itself is related to equations that come from
the analysis of two loop circuits. So systems such as thisone certainly arise in applications.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
1. Note that the second equation is not really a differentialequation.
2. This is not a problem. The transforms will work the sameway, ...
3. ... but the second equation also relates the initial values toeach other. So certain combinations of initial values willnot be possible.
4. The initial values in this example are allowed.
5. The example itself is related to equations that come fromthe analysis of two loop circuits. So systems such as thisone certainly arise in applications.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
1. Note that the second equation is not really a differentialequation.
2. This is not a problem. The transforms will work the sameway, ...
3. ... but the second equation also relates the initial values toeach other. So certain combinations of initial values willnot be possible.
4. The initial values in this example are allowed.5. The example itself is related to equations that come from
the analysis of two loop circuits. So systems such as thisone certainly arise in applications.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+12X−Y = 0 (×−3 and add)
Y(6s+4) =2
s+1+12 =
12s+14s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+12X−Y = 0 (×−3 and add)
Y(6s+4) =2
s+1+12 =
12s+14s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+1
2X−Y = 0 (×−3 and add)
Y(6s+4) =2
s+1+12 =
12s+14s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+12X−Y = 0
(×−3 and add)
Y(6s+4) =2
s+1+12 =
12s+14s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+12X−Y = 0 (×−3 and add)
Y(6s+4) =2
s+1+12 =
12s+14s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+12X−Y = 0 (×−3 and add)
Y
(6s+4) =2
s+1+12 =
12s+14s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+12X−Y = 0 (×−3 and add)
Y(6s
+4) =2
s+1+12 =
12s+14s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+12X−Y = 0 (×−3 and add)
Y(6s+4)
=2
s+1+12 =
12s+14s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+12X−Y = 0 (×−3 and add)
Y(6s+4) =2
s+1+12
=12s+14
s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+12X−Y = 0 (×−3 and add)
Y(6s+4) =2
s+1+12 =
12s+14s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
6x+6y′+ y = 2e−t, x(0) = 12x− y = 0, y(0) = 2
6X +6sY−12+Y =2
s+12X−Y = 0 (×−3 and add)
Y(6s+4) =2
s+1+12 =
12s+14s+1
Y =12s+14
(s+1)(6s+4)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2), A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2), A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)
s =−1 : 2 = A(−2), A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 :
2 = A(−2), A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2
= A(−2), A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2)
, A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2), A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2), A =−1
s =−23
:
6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2), A =−1
s =−23
: 6
= B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2), A =−1
s =−23
: 6 = B(
13
)
, B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2), A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2), A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4
=− 1s+1
+31
s+ 23
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2), A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
12s+14(s+1)(6s+4)
=A
s+1+
B6s+4
12s+14 = A(6s+4)+B(s+1)s =−1 : 2 = A(−2), A =−1
s =−23
: 6 = B(
13
), B = 18
Y(s) = − 1s+1
+18
6s+4=− 1
s+1+3
1s+ 2
3
y(t) = −e−t +3e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
2X−Y = 0
X =12
Y
=12
[− 1
s+1+3
1s+ 2
3
]= −1
21
s+1+
32
1s+ 2
3
x(t) = −12
e−t +32
e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
2X−Y = 0
X =12
Y
=12
[− 1
s+1+3
1s+ 2
3
]= −1
21
s+1+
32
1s+ 2
3
x(t) = −12
e−t +32
e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
2X−Y = 0
X =12
Y
=12
[− 1
s+1+3
1s+ 2
3
]= −1
21
s+1+
32
1s+ 2
3
x(t) = −12
e−t +32
e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
2X−Y = 0
X =12
Y
=12
[− 1
s+1+3
1s+ 2
3
]
= −12
1s+1
+32
1s+ 2
3
x(t) = −12
e−t +32
e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
2X−Y = 0
X =12
Y
=12
[− 1
s+1+3
1s+ 2
3
]= −1
21
s+1+
32
1s+ 2
3
x(t) = −12
e−t +32
e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Solve the Initial Value Problem6x+6y′+ y = 2e−t, 2x− y = 0, x(0) = 1, y(0) = 2
2X−Y = 0
X =12
Y
=12
[− 1
s+1+3
1s+ 2
3
]= −1
21
s+1+
32
1s+ 2
3
x(t) = −12
e−t +32
e−23 t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values:
Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!
2x− y = 0: Same as x =y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0:
Same as x =y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
.
Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y =
6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t
(−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3
+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6
−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1)
+ e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t
(9−12+3)= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9
−12+3)= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12
+3)= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations
logo1
New Idea An Example Double Check
Does x(t) =−12
e−t +32
e−23 t, y(t) =−e−t +3e−
23 t
Really Solve the Initial Value Problem6x+6y′+y = 2e−t, 2x−y = 0, x(0) = 1, y(0) = 2
Initial values: Look at x and y!2x− y = 0: Same as x =
y2
. Look at x and y!
6x+6y′+ y = 6[−1
2e−t +
32
e−23 t]
+6[e−t−2e−
23 t]
+[−e−t +3e−
23 t]
= e−t (−3+6−1) + e−23 t (9−12+3)
= 2e−t √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms for Systems of Differential Equations