Use the Laplace transform to solve the given initial-value problem.
[tex]2 y^{\prime \prime}+3 y^{\prime \prime}-3 y^{\prime}-2 y=e^{-t}, \quad y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=1[/tex]
[tex]y(t)=[/tex]
Answer (2)
Apply Laplace transform to the differential equation, using initial conditions to simplify.
Solve for Y ( s ) in the transformed equation.
Perform partial fraction decomposition on Y ( s ) .
Find the inverse Laplace transform to obtain the solution: y ( t ) = − 2 1 e − t + 18 5 e t − 9 8 e − 2 t + 9 1 e − 2 t
Explanation
Problem Analysis We are given the third-order linear homogeneous differential equation with constant coefficients: 2 y ′′′ + 3 y ′′ − 3 y ′ − 2 y = e − t with initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 , and y ′′ ( 0 ) = 1 . Our goal is to solve this initial value problem using the Laplace transform.
Applying Laplace Transform First, we apply the Laplace transform to both sides of the differential equation. Using the linearity property of the Laplace transform, we have: L { 2 y ′′′ + 3 y ′′ − 3 y ′ − 2 y } = L { e − t } 2 L { y ′′′ } + 3 L { y ′′ } − 3 L { y ′ } − 2 L { y } = L { e − t }
Laplace Transform Properties Now, we use the properties of the Laplace transform for derivatives. Let Y ( s ) = L { y ( t ) } . Then: L { y ′ ( t ) } = s Y ( s ) − y ( 0 ) $ L { y ′′ ( t ) } = s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) $ \mathcal{L}\left{y^{\prime \prime \prime}(t)\right} = s^3Y(s) - s^2y(0) - sy^{\prime}(0) - y^{\prime \prime}(0) Also, we know that L { e − t } = s + 1 1 .
Substituting Initial Conditions Substituting the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 , and y ′′ ( 0 ) = 1 into the Laplace transforms of the derivatives, we get: L { y ′ ( t ) } = s Y ( s ) $ L { y ′′ ( t ) } = s 2 Y ( s ) $ \mathcal{L}\left{y^{\prime \prime \prime}(t)\right} = s^3Y(s) - 1 S u b s t i t u t in g t h ese in t o t h e t r an s f or m e d e q u a t i o n , w e ha v e : 2(s^3Y(s) - 1) + 3(s^2Y(s)) - 3(sY(s)) - 2Y(s) = \frac{1}{s+1}
Solving for Y(s) Now, we solve for Y ( s ) : 2 s 3 Y ( s ) − 2 + 3 s 2 Y ( s ) − 3 s Y ( s ) − 2 Y ( s ) = s + 1 1 ( 2 s 3 + 3 s 2 − 3 s − 2 ) Y ( s ) = s + 1 1 + 2 Y ( s ) = ( s + 1 ) ( 2 s 3 + 3 s 2 − 3 s − 2 ) 1 + 2 s 3 + 3 s 2 − 3 s − 2 2 Y ( s ) = ( s + 1 ) ( 2 s 3 + 3 s 2 − 3 s − 2 ) 1 + 2 ( s + 1 ) = ( s + 1 ) ( 2 s 3 + 3 s 2 − 3 s − 2 ) 2 s + 3 Notice that 2 s 3 + 3 s 2 − 3 s − 2 = ( s − 1 ) ( 2 s 2 + 5 s + 2 ) = ( s − 1 ) ( 2 s + 1 ) ( s + 2 ) . Therefore, Y ( s ) = ( s + 1 ) ( s − 1 ) ( 2 s + 1 ) ( s + 2 ) 2 s + 3
Partial Fraction Decomposition Next, we perform partial fraction decomposition on Y ( s ) . Let Y ( s ) = s + 1 A + s − 1 B + 2 s + 1 C + s + 2 D Then 2 s + 3 = A ( s − 1 ) ( 2 s + 1 ) ( s + 2 ) + B ( s + 1 ) ( 2 s + 1 ) ( s + 2 ) + C ( s + 1 ) ( s − 1 ) ( s + 2 ) + D ( s + 1 ) ( s − 1 ) ( 2 s + 1 ) To find A, let s = − 1 : 2 ( − 1 ) + 3 = A ( − 2 ) ( − 1 ) ( − 1 + 2 ) ⟹ 1 = − 2 A ⟹ A = − 2 1 To find B, let s = 1 : 2 ( 1 ) + 3 = B ( 2 ) ( 3 ) ( 3 ) ⟹ 5 = 18 B ⟹ B = 18 5 To find C, let s = − 2 1 : 2 ( − 2 1 ) + 3 = C ( 2 1 ) ( − 2 3 ) ( − 2 1 + 2 ) ⟹ 2 = C ( 2 1 ) ( − 2 3 ) ( 2 3 ) ⟹ 2 = − C 8 9 ⟹ C = − 9 16 To find D, let s = − 2 : 2 ( − 2 ) + 3 = D ( − 1 ) ( − 3 ) ( 2 ( − 2 ) + 1 ) ⟹ − 1 = D ( − 1 ) ( − 3 ) ( − 3 ) ⟹ − 1 = − 9 D ⟹ D = 9 1 Therefore, Y ( s ) = − 2 ( s + 1 ) 1 + 18 ( s − 1 ) 5 − 9 ( 2 s + 1 ) 16 + 9 ( s + 2 ) 1 = − 2 ( s + 1 ) 1 + 18 ( s − 1 ) 5 − 9 ( s + 2 1 ) 8 + 9 ( s + 2 ) 1
Inverse Laplace Transform Finally, we find the inverse Laplace transform of each term: y ( t ) = − 2 1 e − t + 18 5 e t − 9 8 e − 2 1 t + 9 1 e − 2 t Thus, y ( t ) = − 2 1 e − t + 18 5 e t − 9 8 e − 2 t + 9 1 e − 2 t
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you're designing a control system for a motor, and you need to predict how the motor's speed will respond to different inputs. By using Laplace transforms, you can convert the differential equations that describe the motor's behavior into algebraic equations, making them much easier to solve. This allows you to quickly determine the system's stability and optimize its performance, ensuring the motor operates efficiently and reliably.
The solution to the initial-value problem is found by applying the Laplace transform and solving for Y(s) using initial conditions. Through partial fraction decomposition and using inverse Laplace transforms, we derive the final solution. The resulting function that satisfies the given differential equation and initial conditions is y ( t ) = − 2 1 e − t + 18 5 e t − 9 8 e − 2 t + 9 1 e − 2 t .
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