Consider the following inverse forms:
[tex]$L ^{-1}\left{\frac{s-a}{(s-a)^2+b^2}\right}=e^{a t} \cos (b t)$[/tex]
and
[tex]$L ^{-1}\left{\frac{b}{(s-a)^2+b^2}\right}=e^{a t} \sin (b t)$[/tex]
Use the Laplace transform and these inverses to solve the given initial-value problem.
[tex]$y^{\prime}+y=e^{-3 t} \cos (2 t), \quad y(0)=0$[/tex]
[tex]$y(t)= \square$[/tex]
Answer (2)
Apply Laplace transform to the differential equation y ′ + y = e − 3 t cos ( 2 t ) with initial condition y ( 0 ) = 0 .
Solve for Y ( s ) in the Laplace domain: Y ( s ) = ( s + 1 ) (( s + 3 ) 2 + 4 ) s + 3 .
Perform partial fraction decomposition to get Y ( s ) = s + 1 1/4 − ( s + 3 ) 2 + 4 ( 1/4 ) s + 1/4 .
Apply the inverse Laplace transform to find the solution: y ( t ) = 4 1 e − t + 4 1 e − 3 t ( sin ( 2 t ) − cos ( 2 t )) .
4 1 e − t + 4 1 e − 3 t ( sin ( 2 t ) − cos ( 2 t ))
Explanation
Problem Setup We are given the differential equation y ′ + y = e − 3 t cos ( 2 t ) with the initial condition y ( 0 ) = 0 . We will use Laplace transforms to solve this initial value problem.
Applying Laplace Transform First, we apply the Laplace transform to both sides of the differential equation:
L { y ′ + y } = L { e − 3 t cos ( 2 t )}
Using the linearity property of the Laplace transform, we have:
L { y ′ } + L { y } = L { e − 3 t cos ( 2 t )}
Using Laplace Transform Properties and Initial Condition We know that L { y ′ } = s Y ( s ) − y ( 0 ) , where Y ( s ) = L { y ( t )} . Also, L { y ( t )} = Y ( s ) . Thus,
s Y ( s ) − y ( 0 ) + Y ( s ) = L { e − 3 t cos ( 2 t )}
Since y ( 0 ) = 0 , we have:
s Y ( s ) + Y ( s ) = L { e − 3 t cos ( 2 t )}
Finding Laplace Transform of the Forcing Function We are given that L − 1 { ( s − a ) 2 + b 2 s − a } = e a t cos ( b t ) . Therefore, L { e a t cos ( b t )} = ( s − a ) 2 + b 2 s − a .
In our case, we have e − 3 t cos ( 2 t ) , so a = − 3 and b = 2 . Thus,
L { e − 3 t cos ( 2 t )} = ( s − ( − 3 ) ) 2 + 2 2 s − ( − 3 ) = ( s + 3 ) 2 + 4 s + 3
Solving for Y(s) Substituting this into our equation, we get:
s Y ( s ) + Y ( s ) = ( s + 3 ) 2 + 4 s + 3
Factoring out Y ( s ) , we have:
Y ( s ) ( s + 1 ) = ( s + 3 ) 2 + 4 s + 3
Solving for Y ( s ) , we get:
Y ( s ) = ( s + 1 ) (( s + 3 ) 2 + 4 ) s + 3
Partial Fraction Decomposition Now we need to find the inverse Laplace transform of Y ( s ) . We can use partial fraction decomposition:
( s + 1 ) (( s + 3 ) 2 + 4 ) s + 3 = s + 1 A + ( s + 3 ) 2 + 4 B s + C
Multiplying both sides by ( s + 1 ) (( s + 3 ) 2 + 4 ) , we get:
s + 3 = A (( s + 3 ) 2 + 4 ) + ( B s + C ) ( s + 1 )
s + 3 = A ( s 2 + 6 s + 13 ) + B s 2 + B s + C s + C
s + 3 = ( A + B ) s 2 + ( 6 A + B + C ) s + ( 13 A + C )
Comparing coefficients, we have:
A + B = 0
6 A + B + C = 1
13 A + C = 3
From the first equation, B = − A . Substituting into the second equation, we get 6 A − A + C = 1 , so 5 A + C = 1 . We also have 13 A + C = 3 . Subtracting the first equation from the second, we get 8 A = 2 , so A = 4 1 . Then B = − 4 1 . Substituting A = 4 1 into 13 A + C = 3 , we get 4 13 + C = 3 , so C = 3 − 4 13 = 4 12 − 13 = − 4 1 .
Thus, A = 4 1 , B = − 4 1 , and C = − 4 1 .
Inverse Laplace Transform So, we have:
Y ( s ) = s + 1 4 1 + ( s + 3 ) 2 + 4 − 4 1 s − 4 1 = 4 1 s + 1 1 − 4 1 ( s + 3 ) 2 + 4 s + 1
Y ( s ) = 4 1 s + 1 1 − 4 1 ( s + 3 ) 2 + 4 s + 3 − 2 = 4 1 s + 1 1 − 4 1 ( s + 3 ) 2 + 4 s + 3 + 4 1 ( s + 3 ) 2 + 4 2
Now we take the inverse Laplace transform:
y ( t ) = L − 1 { 4 1 s + 1 1 } − L − 1 { 4 1 ( s + 3 ) 2 + 4 s + 3 } + L − 1 { 4 1 ( s + 3 ) 2 + 4 2 }
y ( t ) = 4 1 e − t − 4 1 e − 3 t cos ( 2 t ) + 4 1 e − 3 t sin ( 2 t )
Final Solution Therefore, the solution to the initial value problem is:
y ( t ) = 4 1 e − t − 4 1 e − 3 t cos ( 2 t ) + 4 1 e − 3 t sin ( 2 t )
y ( t ) = 4 1 e − t + 4 1 e − 3 t ( sin ( 2 t ) − cos ( 2 t ))
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you're designing a circuit with a resistor, capacitor, and inductor, and you need to determine the current flow when a voltage source is applied. By using Laplace transforms, you can convert the differential equations that describe the circuit's behavior into algebraic equations, making them much easier to solve. This allows you to predict the circuit's response to different inputs and optimize its performance.
The solution to the initial value problem y ′ + y = e − 3 t cos ( 2 t ) , y ( 0 ) = 0 is y ( t ) = 4 1 e − t + 4 1 e − 3 t ( sin ( 2 t ) − cos ( 2 t )) . This uses Laplace transforms to simplify the solution. The method involves applying the transform, solving for Y(s), and performing inverse transforms.
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