Solve the initial value problem below using the method of Laplace transforms.

[tex]w^{\prime \prime}-6 w^{\prime}+9 w=45 t+60, w(-1)=2, w^{\prime}(-1)=1[/tex]

Apply a time shift v ( t ) = w ( t + 1 ) to simplify initial conditions.
Take the Laplace transform of the differential equation and solve for V ( s ) .
Perform partial fraction decomposition to find the inverse Laplace transform v ( t ) .
Substitute back to find w ( t ) , resulting in ( 27 t + 24 ) e 3 t + 3 + 5 t + 10 ​ .

Explanation

Problem Analysis We are given the second-order linear differential equation w ′′ − 6 w ′ + 9 w = 45 t + 60 with initial conditions w ( − 1 ) = 2 and w ′ ( − 1 ) = 1 . Our goal is to solve this initial value problem using Laplace transforms.

Time Shift To simplify the initial conditions, we introduce a time shift. Let v ( t ) = w ( t − 1 ) . Then, v ′ ( t ) = w ′ ( t − 1 ) and v ′′ ( t ) = w ′′ ( t − 1 ) . The initial conditions become v ( 0 ) = w ( − 1 ) = 2 and v ′ ( 0 ) = w ′ ( − 1 ) = 1 .

Rewriting the DE We rewrite the differential equation in terms of v ( t ) : v ′′ ( t ) − 6 v ′ ( t ) + 9 v ( t ) = 45 ( t − 1 ) + 60 = 45 t − 45 + 60 = 45 t + 15 .

Applying Laplace Transform Now, we apply the Laplace transform to the differential equation: L [ v ′′ ( t ) − 6 v ′ ( t ) + 9 v ( t )] = L [ 45 t + 15 ] . Using the properties of Laplace transforms, we have: s 2 V ( s ) − s v ( 0 ) − v ′ ( 0 ) − 6 ( s V ( s ) − v ( 0 )) + 9 V ( s ) = s 2 45 ​ + s 15 ​ , where V ( s ) = L [ v ( t )] .

Substituting Initial Conditions and Solving for V(s) Substituting the initial conditions, we get: s 2 V ( s ) − 2 s − 1 − 6 ( s V ( s ) − 2 ) + 9 V ( s ) = s 2 45 ​ + s 15 ​ .
Simplifying, we have: ( s 2 − 6 s + 9 ) V ( s ) = s 2 45 ​ + s 15 ​ + 2 s − 1 + 12 = s 2 45 ​ + s 15 ​ + 2 s + 11 .
Thus, V ( s ) = s 2 ( s − 3 ) 2 45 ​ + s ( s − 3 ) 2 15 ​ + ( s − 3 ) 2 2 s ​ + ( s − 3 ) 2 11 ​ .

Partial Fraction Decomposition We can rewrite V ( s ) as: V ( s ) = s 2 ( s − 3 ) 2 45 + 15 s + 2 s 3 + 11 s 2 ​ .
To find the inverse Laplace transform, we perform partial fraction decomposition: s 2 ( s − 3 ) 2 45 + 15 s + 2 s 3 + 11 s 2 ​ = s A ​ + s 2 B ​ + s − 3 C ​ + ( s − 3 ) 2 D ​ .
Multiplying by s 2 ( s − 3 ) 2 , we get: 2 s 3 + 11 s 2 + 15 s + 45 = A s ( s − 3 ) 2 + B ( s − 3 ) 2 + C s 2 ( s − 3 ) + D s 2 .
Expanding, we have: 2 s 3 + 11 s 2 + 15 s + 45 = A s ( s 2 − 6 s + 9 ) + B ( s 2 − 6 s + 9 ) + C s 2 ( s − 3 ) + D s 2 = A ( s 3 − 6 s 2 + 9 s ) + B ( s 2 − 6 s + 9 ) + C ( s 3 − 3 s 2 ) + D s 2 .
Combining like terms: 2 s 3 + 11 s 2 + 15 s + 45 = ( A + C ) s 3 + ( − 6 A + B − 3 C + D ) s 2 + ( 9 A − 6 B ) s + 9 B .
Comparing coefficients, we get the following system of equations: A + C = 2 -6A + B - 3C + D = 11 9A - 6B = 15 9B = 45 From the last equation, B = 5 . Substituting into the third equation, 9 A − 6 ( 5 ) = 15 , so 9 A = 45 , and A = 5 . Substituting into the first equation, 5 + C = 2 , so C = − 3 . Substituting into the second equation, − 6 ( 5 ) + 5 − 3 ( − 3 ) + D = 11 , so − 30 + 5 + 9 + D = 11 , which gives − 16 + D = 11 , so D = 27 .
Thus, V ( s ) = s 5 ​ + s 2 5 ​ − s − 3 3 ​ + ( s − 3 ) 2 27 ​ .

Inverse Laplace Transform Now, we find the inverse Laplace transform of V ( s ) :
v(t) = L^{-1}[V(s)] = L^{-1}[\frac{5}{s} + \frac{5}{s^2} - \frac{3}{s-3} + \frac{27}{(s-3)^2}] = 5L^{-1}[\frac{1}{s}] + 5L^{-1}[\frac{1}{s^2}] - 3L^{-1}[\frac{1}{s-3}] + 27L^{-1}[\frac{1}{(s-3)^2}] = 5(1) + 5t - 3e^{3t} + 27te^{3t} = 5 + 5t - 3e^{3t} + 27te^{3t}. So, v ( t ) = 5 + 5 t + ( 27 t − 3 ) e 3 t .

Finding w(t) Finally, we find w ( t ) by substituting t with t + 1 in v ( t ) : w ( t ) = v ( t + 1 ) = 5 + 5 ( t + 1 ) + ( 27 ( t + 1 ) − 3 ) e 3 ( t + 1 ) = 5 + 5 t + 5 + ( 27 t + 27 − 3 ) e 3 t + 3 = 10 + 5 t + ( 27 t + 24 ) e 3 t + 3 .
Therefore, w ( t ) = ( 27 t + 24 ) e 3 t + 3 + 5 t + 10 .

Final Answer The solution to the initial value problem is w ( t ) = ( 27 t + 24 ) e 3 t + 3 + 5 t + 10 .


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 modeling the system with differential equations and using Laplace transforms, you can easily analyze the system's stability and response characteristics, ensuring your motor control system performs reliably and efficiently. This method transforms complex time-domain problems into simpler algebraic problems, making the design process much more manageable. For example, understanding how a motor responds to a sudden change in voltage can be crucial for preventing damage and optimizing performance.

Answered by GinnyAnswer | 2025-07-07

We solved the initial value problem using Laplace transforms by applying a time shift to manage initial conditions, rewriting the differential equation accordingly, and using partial fraction decomposition. The final result is w ( t ) = ( 27 t + 24 ) e 3 t − 3 + 5 t + 10 .
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Answered by Anonymous | 2025-07-21