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]
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[tex]w(t)=[/tex]
Answer (1)
Apply the substitution τ = t + 1 to transform the initial conditions to v ( 0 ) = 2 and v ′ ( 0 ) = 1 , and rewrite the differential equation in terms of τ .
Take the Laplace transform of the equation, apply initial conditions, and solve for V ( s ) .
Perform partial fraction decomposition to simplify V ( s ) into manageable terms.
Find the inverse Laplace transform of V ( s ) to obtain v ( τ ) , and then substitute back t = τ − 1 to find the solution w ( t ) = ( 27 t + 24 ) e 3 t + 3 + 5 t + 10 , so the final answer is ( 27 t + 24 ) e 3 t + 3 + 5 t + 10 .
Explanation
Problem Statement We are given the initial value problem:
w ′′ − 6 w ′ + 9 w = 45 t + 60 , w ( − 1 ) = 2 , w ′ ( − 1 ) = 1
We will solve this problem using Laplace transforms.
Variable Substitution First, we apply a change of variable t = τ − 1 , so that τ = t + 1 . Define v ( τ ) = w ( t ) = w ( τ − 1 ) . Then v ′ ( τ ) = w ′ ( τ − 1 ) and v ′′ ( τ ) = w ′′ ( τ − 1 ) . The initial conditions become v ( 0 ) = w ( − 1 ) = 2 and v ′ ( 0 ) = w ′ ( − 1 ) = 1 .
Rewriting the differential equation in terms of τ :
v ′′ ( τ ) − 6 v ′ ( τ ) + 9 v ( τ ) = 45 ( τ − 1 ) + 60 = 45 τ + 15 .
Laplace Transform Next, we apply the Laplace transform to both sides of the equation. Using the linearity property and the formulas for the Laplace transform of derivatives:
L { v ′′ ( τ ) } = s 2 V ( s ) − s v ( 0 ) − v ′ ( 0 ) , L { v ′ ( τ ) } = s V ( s ) − v ( 0 ) , and L { v ( τ ) } = V ( s ) . Also, L { τ } = s 2 1 and L { 1 } = s 1 .
Substituting the initial conditions v ( 0 ) = 2 and v ′ ( 0 ) = 1 into the transformed equation:
s 2 V ( s ) − 2 s − 1 − 6 ( s V ( s ) − 2 ) + 9 V ( s ) = 45 s 2 1 + 15 s 1 .
Solving for V(s) Now, we solve for V ( s ) :
( s 2 − 6 s + 9 ) V ( s ) = 2 s − 1 + 12 + s 2 45 + s 15 = 2 s + 11 + s 2 15 s + 45 .
Thus, V ( s ) = s 2 − 6 s + 9 2 s + 11 + s 2 ( s 2 − 6 s + 9 ) 15 s + 45 = ( s − 3 ) 2 2 s + 11 + s 2 ( s − 3 ) 2 15 ( s + 3 ) .
Partial Fraction Decomposition We perform partial fraction decomposition on V ( s ) .
( s − 3 ) 2 2 s + 11 = s − 3 A + ( s − 3 ) 2 B
s 2 ( s − 3 ) 2 15 ( s + 3 ) = s C + s 2 D + s − 3 E + ( s − 3 ) 2 F
Solving for the coefficients, we find:
A = 2 , B = 17 , C = 5 , D = 5 , E = − 5 , F = 10
So, V ( s ) = s − 3 2 + ( s − 3 ) 2 17 + s 5 + s 2 5 − s − 3 5 + ( s − 3 ) 2 10 = s − 3 − 3 + ( s − 3 ) 2 27 + s 5 + s 2 5 .
Inverse Laplace Transform Now we find the inverse Laplace transform of V ( s ) to obtain v ( τ ) . Using the formulas L − 1 { s − a 1 } = e a t , L − 1 { ( s − a ) 2 1 } = t e a t , L − 1 { s 1 } = 1 , and L − 1 { s 2 1 } = t :
v ( τ ) = − 3 e 3 τ + 27 τ e 3 τ + 5 + 5 τ .
Final Solution Finally, we substitute τ = t + 1 into v ( τ ) to obtain w ( t ) = v ( t + 1 ) :
w ( t ) = − 3 e 3 ( t + 1 ) + 27 ( t + 1 ) e 3 ( t + 1 ) + 5 + 5 ( t + 1 ) = − 3 e 3 t + 3 + 27 t e 3 t + 3 + 27 e 3 t + 3 + 5 + 5 t + 5 = 24 e 3 t + 3 + 27 t e 3 t + 3 + 10 + 5 t .
Therefore, w ( t ) = 24 e 3 ( t + 1 ) + 27 t e 3 ( t + 1 ) + 5 t + 10 .
Conclusion Thus, 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 have a circuit with a voltage source, resistors, and capacitors. When the circuit is subjected to a sudden change, like a switch being flipped, Laplace transforms allow you to convert differential equations describing 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, helping engineers design stable and efficient electrical systems.
