Solve for $Y(s)$, the Laplace transform of the solution $y(t)$ to the initial value problem below.
$y^{\prime \prime}+6 y=6 t^2-7, y(0)=0, y^{\prime}(0)=-9$
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$Y(s)=$
$\square$
Answer (2)
Apply Laplace transform to the given differential equation: y ′′ + 6 y = 6 t 2 − 7 .
Use Laplace transform properties and initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = − 9 to get an algebraic equation in terms of Y ( s ) .
Solve the algebraic equation for Y ( s ) .
The Laplace transform of the solution is: s 3 ( s 2 + 6 ) − 9 s 3 − 7 s 2 + 12 .
Explanation
Problem Setup We are given the initial value problem:
y ′′ + 6 y = 6 t 2 − 7 , y ( 0 ) = 0 , y ′ ( 0 ) = − 9
Our goal is to find Y ( s ) , the Laplace transform of y ( t ) .
Applying Laplace Transform Taking the Laplace transform of both sides of the differential equation, we have:
L { y ′′ ( t ) } + 6 L { y ( t ) } = 6 L { t 2 } − 7 L { 1 }
Using the properties of Laplace transforms, we know that L { y ′′ ( t ) } = s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) , L { y ( t ) } = Y ( s ) , L { t 2 } = s 3 2 , and L { 1 } = s 1 .
Substituting Initial Conditions and Simplifying Substituting the initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = − 9 , we get:
s 2 Y ( s ) − s ( 0 ) − ( − 9 ) + 6 Y ( s ) = 6 ( s 3 2 ) − 7 ( s 1 )
Simplifying, we have:
s 2 Y ( s ) + 9 + 6 Y ( s ) = s 3 12 − s 7
Solving for Y(s) Now, we solve for Y ( s ) :
( s 2 + 6 ) Y ( s ) = s 3 12 − s 7 − 9
Y ( s ) = s 2 + 6 s 3 12 − s 7 − 9
Y ( s ) = s 3 ( s 2 + 6 ) 12 − 7 s 2 − 9 s 3
Y ( s ) = s 3 ( s 2 + 6 ) − 9 s 3 − 7 s 2 + 12
Final Answer Therefore, the Laplace transform of the solution y ( t ) is:
Y ( s ) = s 3 ( s 2 + 6 ) − 9 s 3 − 7 s 2 + 12
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with inductors, capacitors, and resistors, and you want to determine the voltage or current response to a particular input signal. By 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 engineers to quickly understand how a circuit will respond to different inputs and optimize its design for specific applications.
To find Y ( s ) , the Laplace transform of the solution to the initial value problem, we apply the Laplace transform to the differential equation, substitute initial conditions, and solve for Y ( s ) . The resulting expression is s 3 ( s 2 + 6 ) − 9 s 3 − 7 s 2 + 12 .
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