An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (1)
Apply Laplace transform to the given differential equation.
Substitute initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = 9 .
Solve for Y ( s ) in the Laplace domain.
Apply inverse Laplace transform to find y ( t ) = − 5 t + 6 t e − 2 t + 2 e 2 t − 2 e − 2 t .
Explanation
Problem Statement We are given the initial value problem: y ′′ − 4 y = 20 t − 24 e − 2 t , y ( 0 ) = 0 , y ′ ( 0 ) = 9 We will solve this using Laplace transforms.
Laplace Transform Taking the Laplace transform of both sides of the differential equation, we have: L { y ′′ − 4 y } = L { 20 t − 24 e − 2 t } Using the properties of Laplace transforms, we get: L { y ′′ } − 4 L { y } = 20 L { t } − 24 L { e − 2 t } [ s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 )] − 4 Y ( s ) = 20 ( s 2 1 ) − 24 ( s + 2 1 ) where Y ( s ) = L { y ( t )} .
Substituting Initial Conditions and Solving for Y(s) Substituting the initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = 9 , we have: s 2 Y ( s ) − s ( 0 ) − 9 − 4 Y ( s ) = s 2 20 − s + 2 24 ( s 2 − 4 ) Y ( s ) − 9 = s 2 20 − s + 2 24 ( s 2 − 4 ) Y ( s ) = s 2 20 − s + 2 24 + 9 Y ( s ) = s 2 ( s 2 − 4 ) 20 − ( s + 2 ) ( s 2 − 4 ) 24 + s 2 − 4 9 Y ( s ) = s 2 ( s − 2 ) ( s + 2 ) 20 − ( s + 2 ) ( s − 2 ) ( s + 2 ) 24 + ( s − 2 ) ( s + 2 ) 9 Y ( s ) = s 2 ( s − 2 ) ( s + 2 ) 20 − ( s + 2 ) 2 ( s − 2 ) 24 + ( s − 2 ) ( s + 2 ) 9
Inverse Laplace Transform We can simplify this expression by finding a common denominator: Y ( s ) = s 2 ( s − 2 ) ( s + 2 ) 2 20 ( s + 2 ) − 24 s 2 + 9 s 2 ( s + 2 ) 2 However, let's use the result from the python calculation tool: y ( t ) = − 5 t ⋅ He a v i s i d e ( t ) + 6 t e − 2 t ⋅ He a v i s i d e ( t ) + 2 e 2 t ⋅ He a v i s i d e ( t ) − 2 e − 2 t ⋅ He a v i s i d e ( t ) Since we are looking for y ( t ) for 0"> t > 0 , we can ignore the Heaviside function. Therefore, y ( t ) = − 5 t + 6 t e − 2 t + 2 e 2 t − 2 e − 2 t y ( t ) = − 5 t + 6 t e − 2 t + 2 ( e 2 t − e − 2 t ) y ( t ) = − 5 t + 6 t e − 2 t + 4 sinh ( 2 t )
Final Answer Thus, the solution to the initial value problem is: y ( t ) = − 5 t + 6 t e − 2 t + 2 e 2 t − 2 e − 2 t y ( t ) = − 5 t + 6 t e − 2 t + 4 sinh ( 2 t )
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with a sudden surge of voltage. 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 how the current and voltage will change over time, helping you design circuits that can handle these surges without failing.
