Solve the initial value problem below using the method of Laplace transforms.
[tex]w^{\prime \prime}-12 w^{\prime}+36 w=180 t+696, w(-4)=-5, w^{\prime}(-4)=-25[/tex]
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Answer (1)
Apply Laplace transform to the given differential equation w ′′ − 12 w ′ + 36 w = 180 t + 696 , considering initial conditions w ( − 4 ) = − 5 and w ′ ( − 4 ) = − 25 .
Introduce a time shift v ( t ) = w ( t − 4 ) to simplify initial conditions to v ( 0 ) = − 5 and v ′ ( 0 ) = − 25 , and rewrite the equation in terms of v ( t ) .
Solve for V ( s ) = L [ v ( t )] in the Laplace domain and perform partial fraction decomposition.
Apply the inverse Laplace transform to find v ( t ) , then substitute back to find w ( t ) , resulting in the solution w ( t ) = 5 t + 21 + ( 6 t + 18 ) e 6 ( t + 4 ) .
Explanation
Problem Analysis We are given the second-order linear differential equation w ′′ − 12 w ′ + 36 w = 180 t + 696 with initial conditions w ( − 4 ) = − 5 and w ′ ( − 4 ) = − 25 . Our goal is to solve this initial value problem using Laplace transforms.
Time Shift To handle the initial conditions at t = − 4 , we introduce a time shift. Let v ( t ) = w ( t − 4 ) , so w ( t ) = v ( t + 4 ) . Then w ′ ( t ) = v ′ ( t + 4 ) and w ′′ ( t ) = v ′′ ( t + 4 ) . The initial conditions become v ( 0 ) = w ( − 4 ) = − 5 and v ′ ( 0 ) = w ′ ( − 4 ) = − 25 .
Rewriting the Equation Substituting t = u − 4 , we rewrite the differential equation in terms of v ( u ) : v ′′ ( u ) − 12 v ′ ( u ) + 36 v ( u ) = 180 ( u − 4 ) + 696 = 180 u − 720 + 696 = 180 u − 24 .
Applying Laplace Transform Now, we apply the Laplace transform to both sides of the equation: L [ v ′′ ( u )] − 12 L [ v ′ ( u )] + 36 L [ v ( u )] = L [ 180 u − 24 ]
Using Laplace Transform Properties Using the properties of Laplace transforms, we have: L [ v ′′ ( u )] = s 2 V ( s ) − s v ( 0 ) − v ′ ( 0 ) = s 2 V ( s ) + 5 s + 25 L [ v ′ ( u )] = s V ( s ) − v ( 0 ) = s V ( s ) + 5 L [ 180 u − 24 ] = s 2 180 − s 24 where V ( s ) = L [ v ( u )] .
Solving for V(s) Substituting these into the transformed equation, we get: ( s 2 V ( s ) + 5 s + 25 ) − 12 ( s V ( s ) + 5 ) + 36 V ( s ) = s 2 180 − s 24 s 2 V ( s ) + 5 s + 25 − 12 s V ( s ) − 60 + 36 V ( s ) = s 2 180 − s 24 ( s 2 − 12 s + 36 ) V ( s ) = s 2 180 − s 24 − 5 s + 35 ( s − 6 ) 2 V ( s ) = s 2 180 − 24 s − 5 s 3 + 35 s 2 V ( s ) = s 2 ( s − 6 ) 2 − 5 s 3 + 35 s 2 − 24 s + 180
Partial Fraction Decomposition Now we perform partial fraction decomposition on V ( s ) : V ( s ) = s 2 ( s − 6 ) 2 − 5 s 3 + 35 s 2 − 24 s + 180 = s A + s 2 B + s − 6 C + ( s − 6 ) 2 D Multiplying by s 2 ( s − 6 ) 2 , we get: − 5 s 3 + 35 s 2 − 24 s + 180 = A ( s ) ( s − 6 ) 2 + B ( s − 6 ) 2 + C ( s 2 ) ( s − 6 ) + D ( s 2 ) Let s = 0 : 180 = 36 B ⟹ B = 5 Let s = 6 : − 5 ( 6 3 ) + 35 ( 6 2 ) − 24 ( 6 ) + 180 = 36 D ⟹ − 1080 + 1260 − 144 + 180 = 36 D ⟹ 6 = D Expanding and collecting terms: − 5 s 3 + 35 s 2 − 24 s + 180 = A ( s 3 − 12 s 2 + 36 s ) + B ( s 2 − 12 s + 36 ) + C ( s 3 − 6 s 2 ) + D s 2 − 5 s 3 + 35 s 2 − 24 s + 180 = ( A + C ) s 3 + ( − 12 A + B − 6 C + D ) s 2 + ( 36 A − 12 B ) s + 36 B Comparing coefficients: s 3 : A + C = − 5 s 2 : − 12 A + B − 6 C + D = 35 ⟹ − 12 A − 6 C = 35 − 5 − 6 = 24 ⟹ − 12 A − 6 C = 24 ⟹ − 2 A − C = 4 s : 36 A − 12 B = − 24 ⟹ 36 A = − 24 + 12 ( 5 ) = 36 ⟹ A = 1 Since A = 1 , C = − 5 − A = − 6 So, V ( s ) = s 1 + s 2 5 + s − 6 − 6 + ( s − 6 ) 2 6
Inverse Laplace Transform Applying the inverse Laplace transform, we get: v ( u ) = L − 1 [ V ( s )] = L − 1 [ s 1 + s 2 5 + s − 6 − 6 + ( s − 6 ) 2 6 ] = 1 + 5 u − 6 e 6 u + 6 u e 6 u
Substituting Back Substituting back u = t + 4 , we find w ( t ) = v ( t + 4 ) : w ( t ) = 1 + 5 ( t + 4 ) − 6 e 6 ( t + 4 ) + 6 ( t + 4 ) e 6 ( t + 4 ) = 1 + 5 t + 20 − 6 e 6 t + 24 + ( 6 t + 24 ) e 6 t + 24 = 5 t + 21 + ( 6 t + 18 ) e 6 t + 24
Final Answer Therefore, the solution to the initial value problem is: w ( t ) = 5 t + 21 + ( 6 t + 18 ) e 6 ( t + 4 )
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. This makes it much easier to solve for the current and voltage in the circuit as a function of time, especially when dealing with complex circuits or initial conditions. This method simplifies circuit analysis and design, allowing engineers to predict and optimize circuit performance efficiently.
