Solve for $Y(s)$, the Laplace transform of the solution $y(t)$ to the initial value problem below.
$y^{\prime \prime}-9 y^{\prime}+18 y=2 t e^{3 t}, y(0)=6, y^{\prime}(0)=-5$
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$Y(s)=\frac{2}{(s-3)^3(s-6)}+\frac{6 s+1}{(s-3)(s-6)}$
Answer (1)
Apply Laplace transform to the given differential equation.
Use Laplace transform properties to express derivatives in terms of Y ( s ) , y ( 0 ) , and y ′ ( 0 ) .
Substitute initial conditions y ( 0 ) = 6 and y ′ ( 0 ) = − 5 .
Solve for Y ( s ) and simplify: Y ( s ) = ( s − 3 ) 3 ( s − 6 ) 6 s 3 − 95 s 2 + 408 s − 529 .
Explanation
Problem Setup We are given the initial value problem y ′′ − 9 y ′ + 18 y = 2 t e 3 t with initial conditions y ( 0 ) = 6 and y ′ ( 0 ) = − 5 . Our goal is to find Y ( s ) , the Laplace transform of y ( t ) .
Applying Laplace Transform First, we apply the Laplace transform to both sides of the differential equation: L { y ′′ − 9 y ′ + 18 y } = L { 2 t e 3 t } Using the linearity property of the Laplace transform, we have: L { y ′′ } − 9 L { y ′ } + 18 L { y } = 2 L { t e 3 t }
Substituting Initial Conditions and Laplace Transforms Next, we use the properties of Laplace transforms for derivatives. Recall that L { y ′ ( t ) } = s Y ( s ) − y ( 0 ) and L { y ′′ ( t ) } = s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) . Also, from the Laplace transform table, we know that L { t e a t } = ( s − a ) 2 1 , so L { 2 t e 3 t } = ( s − 3 ) 2 2 . Substituting these into the equation, we get: ( s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 )) − 9 ( s Y ( s ) − y ( 0 )) + 18 Y ( s ) = ( s − 3 ) 2 2 Now, we plug in the initial conditions y ( 0 ) = 6 and y ′ ( 0 ) = − 5 :
( s 2 Y ( s ) − 6 s − ( − 5 )) − 9 ( s Y ( s ) − 6 ) + 18 Y ( s ) = ( s − 3 ) 2 2 s 2 Y ( s ) − 6 s + 5 − 9 s Y ( s ) + 54 + 18 Y ( s ) = ( s − 3 ) 2 2 ( s 2 − 9 s + 18 ) Y ( s ) − 6 s + 59 = ( s − 3 ) 2 2
Solving for Y(s) Now, we solve for Y ( s ) :
( s 2 − 9 s + 18 ) Y ( s ) = ( s − 3 ) 2 2 + 6 s − 59 Y ( s ) = s 2 − 9 s + 18 1 ( ( s − 3 ) 2 2 + 6 s − 59 ) Since s 2 − 9 s + 18 = ( s − 3 ) ( s − 6 ) , we have: Y ( s ) = ( s − 3 ) ( s − 6 ) 1 ( ( s − 3 ) 2 2 + 6 s − 59 ) Y ( s ) = ( s − 3 ) 3 ( s − 6 ) 2 + ( s − 3 ) ( s − 6 ) 6 s − 59
Simplifying Y(s) Combining the fractions, we get: Y ( s ) = ( s − 3 ) 3 ( s − 6 ) 2 + ( 6 s − 59 ) ( s − 3 ) 2 Y ( s ) = ( s − 3 ) 3 ( s − 6 ) 2 + ( 6 s − 59 ) ( s 2 − 6 s + 9 ) Y ( s ) = ( s − 3 ) 3 ( s − 6 ) 2 + 6 s 3 − 36 s 2 + 54 s − 59 s 2 + 354 s − 531 Y ( s ) = ( s − 3 ) 3 ( s − 6 ) 6 s 3 − 95 s 2 + 408 s − 529
Final Answer Therefore, the Laplace transform of the solution y ( t ) is: Y ( s ) = ( s − 3 ) 3 ( s − 6 ) 6 s 3 − 95 s 2 + 408 s − 529
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with a complex input signal, like a voltage source that varies with time. By using Laplace transforms, you can convert the differential equations that describe the circuit's behavior into algebraic equations in the s-domain. This makes it much easier to solve for the circuit's output, such as the current or voltage at a particular point. Once you have the solution in the s-domain, you can use the inverse Laplace transform to convert it back to the time domain and see how the circuit responds to the input signal over time. This is crucial for designing and troubleshooting electrical systems.
