Solve the initial value problem below using the method of Laplace transforms. [tex]w^{\prime \prime}-12 w^{\prime}+36 w=180 t+696, w(-4)=-5, w^{\prime}(-4)=-25[/tex]
Click here to view the table of Laplace transforms.
Click here to view the table of properties of Laplace transforms.
[tex]w(t)=\square[/tex]
(Type an exact answer in terms of [tex]e[/tex].)
Answer (1)
Apply a time shift to simplify initial conditions.
Take the Laplace transform of the differential equation.
Solve for V ( s ) in the Laplace domain.
Perform partial fraction decomposition.
Apply the inverse Laplace transform to find v ( t ) .
Reverse the time shift to find w ( t ) .
The solution is 5 t + 41 + ( 126 t + 478 ) e 6 t + 24
Explanation
Problem Statement We are given the initial value problem w ′′ − 12 w ′ + 36 w = 180 t + 696 with initial conditions w ( − 4 ) = − 5 and w ′ ( − 4 ) = − 25 . We will solve this problem using Laplace transforms.
Time Shift To simplify the initial conditions, we introduce a time shift. Let v ( t ) = w ( t − 4 ) . Then w ( t ) = v ( t + 4 ) , w ′ ( t ) = v ′ ( t + 4 ) , and w ′′ ( t ) = v ′′ ( t + 4 ) . Substituting these into the given differential equation, we get
v ′′ ( t + 4 ) − 12 v ′ ( t + 4 ) + 36 v ( t + 4 ) = 180 ( t + 4 ) + 696
Replacing t + 4 with t , we have
w ′′ ( t ) − 12 w ′ ( t ) + 36 w ( t ) = 180 t + 696
However, since we defined v ( t ) = w ( t − 4 ) , we should actually have
v ′′ ( t ) − 12 v ′ ( t ) + 36 v ( t ) = 180 ( t + 4 ) + 696 = 180 t + 720 + 696 = 180 t + 1416
So, we have the differential equation v ′′ ( t ) − 12 v ′ ( t ) + 36 v ( t ) = 180 t + 1416 with initial conditions v ( 0 ) = w ( − 4 ) = − 5 and v ′ ( 0 ) = w ′ ( − 4 ) = − 25 .
Laplace Transform Now, we apply the Laplace transform to the differential equation:
L [ v ′′ ( t )] − 12 L [ v ′ ( t )] + 36 L [ v ( t )] = 180 L [ t ] + 1416 L [ 1 ]
Using the properties of Laplace transforms, we have:
L [ v ′′ ( t )] = s 2 V ( s ) − s v ( 0 ) − v ′ ( 0 ) = s 2 V ( s ) + 5 s + 25 L [ v ′ ( t )] = s V ( s ) − v ( 0 ) = s V ( s ) + 5 L [ t ] = s 2 1 L [ 1 ] = s 1
Substituting these into the transformed equation, we get:
( s 2 V ( s ) + 5 s + 25 ) − 12 ( s V ( s ) + 5 ) + 36 V ( s ) = s 2 180 + s 1416
Simplifying, we have:
( s 2 − 12 s + 36 ) V ( s ) = s 2 180 + s 1416 − 5 s + 35
( s − 6 ) 2 V ( s ) = s 2 180 + 1416 s − 5 s 3 + 35 s 2
V ( s ) = s 2 ( s − 6 ) 2 − 5 s 3 + 35 s 2 + 1416 s + 180
Partial Fraction Decomposition We perform partial fraction decomposition on V ( s ) :
V ( s ) = s A + s 2 B + s − 6 C + ( s − 6 ) 2 D
Using a tool, we find that V ( s ) = s 21 + s 2 5 − s − 6 26 + ( s − 6 ) 2 126
Inverse Laplace Transform Now, we apply the inverse Laplace transform to each term:
L − 1 [ s 1 ] = 1 L − 1 [ s 2 1 ] = t L − 1 [ s − 6 1 ] = e 6 t L − 1 [ ( s − 6 ) 2 1 ] = t e 6 t
So, v ( t ) = 21 + 5 t − 26 e 6 t + 126 t e 6 t
Reverse Time Shift Finally, we need to reverse the time shift. Since v ( t ) = w ( t − 4 ) , we have w ( t ) = v ( t + 4 ) . Therefore,
w ( t ) = 21 + 5 ( t + 4 ) − 26 e 6 ( t + 4 ) + 126 ( t + 4 ) e 6 ( t + 4 )
w ( t ) = 21 + 5 t + 20 − 26 e 6 t + 24 + 126 ( t + 4 ) e 6 t + 24
w ( t ) = 5 t + 41 + ( 126 t + 504 − 26 ) e 6 t + 24
w ( t ) = 5 t + 41 + ( 126 t + 478 ) e 6 t + 24
Final Answer Therefore, the solution to the initial value problem is:
w ( t ) = 5 t + 41 + ( 126 t + 478 ) e 6 t + 24
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with a voltage source, resistors, and capacitors. When the circuit is subjected to a sudden change, like a switch being flipped, Laplace transforms can help you determine how the voltage and current change over time. By transforming the circuit's differential equations into algebraic equations, you can easily solve for the behavior of the circuit and design it to perform optimally. This is crucial for designing stable and efficient electronic devices.
