Use the Laplace transform to solve the given initial-value problem.
[tex]y^{\prime \prime}-5 y^{\prime}=8 e^{4 t}-4 e^{-t}, \quad y(0)=1, y^{\prime}(0)=-1[/tex]
[tex]y(t)=[/tex]
Answer (2)
Apply Laplace transform to the differential equation, incorporating initial conditions.
Solve for Y ( s ) in the transformed equation.
Perform partial fraction decomposition on Y ( s ) .
Apply the inverse Laplace transform to obtain the solution: y ( t ) = 5 12 + 15 19 e 5 t − 2 e 4 t − 3 2 e − t
Explanation
Problem Setup We are given the initial-value problem:
y ′′ − 5 y ′ = 8 e 4 t − 4 e − t , y ( 0 ) = 1 , y ′ ( 0 ) = − 1
Our objective is to solve this problem using Laplace transforms.
Applying Laplace Transform First, we apply the Laplace transform to both sides of the differential equation. Let Y ( s ) be the Laplace transform of y ( t ) . Using the properties of Laplace transforms, we have:
L { y ′′ ( t ) } = s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) = s 2 Y ( s ) − s ( 1 ) − ( − 1 ) = s 2 Y ( s ) − s + 1 L { y ′ ( t ) } = s Y ( s ) − y ( 0 ) = s Y ( s ) − 1 L { e 4 t } = s − 4 1 L { e − t } = s + 1 1
Substituting these into the Laplace transformed differential equation, we get:
s 2 Y ( s ) − s + 1 − 5 ( s Y ( s ) − 1 ) = 8 ( s − 4 1 ) − 4 ( s + 1 1 )
Simplifying, we have:
s 2 Y ( s ) − s + 1 − 5 s Y ( s ) + 5 = s − 4 8 − s + 1 4 ( s 2 − 5 s ) Y ( s ) − s + 6 = s − 4 8 − s + 1 4
Solving for Y(s) Next, we solve for Y ( s ) :
( s 2 − 5 s ) Y ( s ) = s − 4 8 − s + 1 4 + s − 6 Y ( s ) = s 2 − 5 s 1 ( s − 4 8 − s + 1 4 + s − 6 ) Y ( s ) = s ( s − 5 ) 1 ( ( s − 4 ) ( s + 1 ) 8 ( s + 1 ) − 4 ( s − 4 ) + ( s − 6 ) ( s − 4 ) ( s + 1 ) ) Y ( s ) = s ( s − 5 ) 1 ( ( s − 4 ) ( s + 1 ) 8 s + 8 − 4 s + 16 + ( s − 6 ) ( s 2 − 3 s − 4 ) ) Y ( s ) = s ( s − 5 ) 1 ( ( s − 4 ) ( s + 1 ) 4 s + 24 + s 3 − 3 s 2 − 4 s − 6 s 2 + 18 s + 24 ) Y ( s ) = s ( s − 5 ) 1 ( ( s − 4 ) ( s + 1 ) s 3 − 9 s 2 + 18 s + 48 ) Y ( s ) = s ( s − 5 ) ( s − 4 ) ( s + 1 ) s 3 − 9 s 2 + 18 s + 48
Partial Fraction Decomposition Now, we perform partial fraction decomposition on Y ( s ) .
Y ( s ) = s ( s − 5 ) ( s − 4 ) ( s + 1 ) s 3 − 9 s 2 + 18 s + 48 = s A + s − 5 B + s − 4 C + s + 1 D
After performing partial fraction decomposition, we find:
Y ( s ) = s 12/5 + s − 5 19/15 + s − 4 − 2 + s + 1 − 2/3
Inverse Laplace Transform Finally, we apply the inverse Laplace transform to each term to find y ( t ) :
y ( t ) = L − 1 { s 12/5 } + L − 1 { s − 5 19/15 } + L − 1 { s − 4 − 2 } + L − 1 { s + 1 − 2/3 } y ( t ) = 5 12 + 15 19 e 5 t − 2 e 4 t − 3 2 e − t
Final Solution Therefore, the solution to the initial-value problem is:
y ( t ) = 5 12 + 15 19 e 5 t − 2 e 4 t − 3 2 e − t
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with a voltage source, resistors, and capacitors. By using Laplace transforms, you can convert the differential equations that describe the circuit's behavior into algebraic equations, making it much easier to solve for the current and voltage at different points in the circuit. This helps engineers design and optimize circuits for various applications, from simple filters to complex control systems.
We solved the differential equation using Laplace transforms, applying the initial conditions. After transforming and performing partial fraction decomposition, we obtained an expression for Y ( s ) . Applying the inverse Laplace transform, we found the solution: y ( t ) = 5 12 + 15 19 e 5 t − 2 e 4 t − 3 2 e − t .
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