Solve the third-order initial value problem below using the method of Laplace transforms.
[tex]y^{\prime \prime \prime}+5 y^{\prime \prime}+2 y^{\prime}-8 y=-32, y(0)=6, y^{\prime}(0)=-13, y^{\prime \prime}(0)=77[/tex]
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Answer (2)
Apply Laplace transform to the given differential equation.
Substitute initial conditions and solve for Y ( s ) in the Laplace domain.
Perform partial fraction decomposition on Y ( s ) .
Find the inverse Laplace transform to obtain the solution: y ( t ) = 6 e − 4 t − 5 e − 2 t + e t + 4 .
Explanation
Problem Setup We are given a third-order initial value problem:
y ′′′ + 5 y ′′ + 2 y ′ − 8 y = − 32 , y ( 0 ) = 6 , y ′ ( 0 ) = − 13 , y ′′ ( 0 ) = 77
We will solve this problem using Laplace transforms.
Applying Laplace Transform First, we apply the Laplace transform to both sides of the differential equation. Using the properties of Laplace transforms, we have:
L y ′′′ = s 3 Y ( s ) − s 2 y ( 0 ) − s y ′ ( 0 ) − y ′′ ( 0 ) L y ′′ = s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) L y ′ = s Y ( s ) − y ( 0 ) L y = Y ( s ) L − 32 = − s 32
Substituting the initial conditions y ( 0 ) = 6 , y ′ ( 0 ) = − 13 , and y ′′ ( 0 ) = 77 , we get:
L y ′′′ = s 3 Y ( s ) − 6 s 2 + 13 s − 77 L y ′′ = s 2 Y ( s ) − 6 s + 13 L y ′ = s Y ( s ) − 6
Substituting into the Equation Now, we substitute these expressions into the Laplace transformed differential equation:
( s 3 Y ( s ) − 6 s 2 + 13 s − 77 ) + 5 ( s 2 Y ( s ) − 6 s + 13 ) + 2 ( s Y ( s ) − 6 ) − 8 Y ( s ) = − s 32
Combining terms, we have:
( s 3 + 5 s 2 + 2 s − 8 ) Y ( s ) − 6 s 2 + 13 s − 77 − 30 s + 65 + 2 s − 12 = − s 32
( s 3 + 5 s 2 + 2 s − 8 ) Y ( s ) − 6 s 2 − 15 s − 24 = − s 32
Solving for Y(s) Next, we solve for Y ( s ) :
( s 3 + 5 s 2 + 2 s − 8 ) Y ( s ) = 6 s 2 + 15 s + 24 − s 32
Y ( s ) = s 3 + 5 s 2 + 2 s − 8 6 s 2 + 15 s + 24 − s ( s 3 + 5 s 2 + 2 s − 8 ) 32
Y ( s ) = s ( s 3 + 5 s 2 + 2 s − 8 ) 6 s 3 + 15 s 2 + 24 s − 32
We can factor the denominator as s ( s − 1 ) ( s + 2 ) ( s + 4 ) . Thus,
Y ( s ) = s ( s − 1 ) ( s + 2 ) ( s + 4 ) 6 s 3 + 15 s 2 + 24 s − 32
Partial Fraction Decomposition Now, we perform partial fraction decomposition on Y ( s ) . From the tool, we have:
Y ( s ) = s + 4 6 − s + 2 5 + s − 1 1 + s 4
Inverse Laplace Transform Finally, we find the inverse Laplace transform of each term:
L − 1 s + 4 6 = 6 e − 4 t L − 1 s + 2 − 5 = − 5 e − 2 t L − 1 s − 1 1 = e t L − 1 s 4 = 4
Combining these, we get the solution:
y ( t ) = 6 e − 4 t − 5 e − 2 t + e t + 4
Final Answer Therefore, the solution to the initial value problem is:
y ( t ) = 6 e − 4 t − 5 e − 2 t + e t + 4
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you're designing a control system for a robot arm. The system's behavior can be modeled using differential equations, and Laplace transforms can simplify the process of finding the system's response to different inputs. This allows engineers to predict and optimize the arm's movements, ensuring it performs its tasks accurately and efficiently. Moreover, it helps in designing stable control systems, preventing oscillations or instability in the robot's motion.
Using Laplace transforms, we found the solution to the initial value problem as y ( t ) = 6 e − 4 t − 5 e − 2 t + e t + 4 . This involved transforming the equation, applying initial conditions, solving for Y ( s ) , and performing an inverse transform. The final answer effectively models the behavior defined by the original differential equation.
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