Use the Laplace transform to solve the given initial-value problem.
[tex]y^{\prime \prime}+y=\sqrt{2} \sin (\sqrt{2} t), y(0)=8, y^{\prime}(0)=0[/tex]
[tex]y(t)=[/tex]
Answer (2)
Apply Laplace transform to the differential equation: y ′′ + y = 2 sin ( 2 t ) .
Use initial conditions y ( 0 ) = 8 and y ′ ( 0 ) = 0 and Laplace transform properties to find Y ( s ) = s 2 + 1 8 s + s 2 + 1 2 − s 2 + 2 2 .
Apply inverse Laplace transform to find y ( t ) = 8 cos ( t ) + 2 sin ( t ) − 2 sin ( 2 t ) .
The solution to the initial-value problem is y ( t ) = 8 cos ( t ) + 2 sin ( t ) − 2 sin ( 2 t ) .
Explanation
Problem Setup We are given the differential equation y ′′ + y = 2 sin ( 2 t ) with initial conditions y ( 0 ) = 8 and y ′ ( 0 ) = 0 . Our goal is to solve this initial-value problem using the Laplace transform.
Applying Laplace Transform First, we apply the Laplace transform to both sides of the differential equation. Using the linearity property of the Laplace transform, we have
L { y ′′ + y } = L { 2 sin ( 2 t ) }
L { y ′′ } + L { y } = 2 L { sin ( 2 t ) }
Using Laplace Transform Properties and Initial Conditions We know that L { y ′′ } = s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) and L { y } = Y ( s ) , where Y ( s ) is the Laplace transform of y ( t ) . Also, L { sin ( a t ) } = s 2 + a 2 a . Therefore, L { sin ( 2 t ) } = s 2 + 2 2 . Substituting these into the equation, we get
s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) + Y ( s ) = 2 ⋅ s 2 + 2 2
Using the initial conditions y ( 0 ) = 8 and y ′ ( 0 ) = 0 , we have
s 2 Y ( s ) − 8 s − 0 + Y ( s ) = s 2 + 2 2
( s 2 + 1 ) Y ( s ) = 8 s + s 2 + 2 2
Solving for Y(s) and Partial Fraction Decomposition Now, we solve for Y ( s ) :
Y ( s ) = s 2 + 1 8 s + ( s 2 + 1 ) ( s 2 + 2 ) 2
To find the inverse Laplace transform, we need to perform partial fraction decomposition on the second term. Let
( s 2 + 1 ) ( s 2 + 2 ) 2 = s 2 + 1 A + s 2 + 2 B
2 = A ( s 2 + 2 ) + B ( s 2 + 1 )
2 = ( A + B ) s 2 + ( 2 A + B )
Comparing coefficients, we have A + B = 0 and 2 A + B = 2 . Subtracting the first equation from the second, we get A = 2 . Then, B = − A = − 2 . Thus,
( s 2 + 1 ) ( s 2 + 2 ) 2 = s 2 + 1 2 − s 2 + 2 2
Substituting Back into Y(s) Substituting this back into the expression for Y ( s ) , we get
Y ( s ) = s 2 + 1 8 s + s 2 + 1 2 − s 2 + 2 2
Y ( s ) = s 2 + 1 8 s + s 2 + 1 2 − 2 2 ⋅ s 2 + 2 2
Y ( s ) = s 2 + 1 8 s + s 2 + 1 2 − 2 ⋅ s 2 + 2 2
Applying Inverse Laplace Transform Now, we apply the inverse Laplace transform to each term. We know that L − 1 { s 2 + a 2 s } = cos ( a t ) and L − 1 { s 2 + a 2 a } = sin ( a t ) . Therefore,
y ( t ) = L − 1 { s 2 + 1 8 s } + L − 1 { s 2 + 1 2 } − L − 1 { 2 ⋅ s 2 + 2 2 }
y ( t ) = 8 cos ( t ) + 2 sin ( t ) − 2 sin ( 2 t )
Final Solution Thus, the solution to the initial-value problem is
y ( t ) = 8 cos ( t ) + 2 sin ( t ) − 2 sin ( 2 t )
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with a capacitor, resistor, and an inductor, and you want to find the current as a function of time when you apply a voltage source. Using Laplace transforms, you can convert the differential equations that describe the circuit into algebraic equations, solve for the current in the s-domain, and then use the inverse Laplace transform to find the current in the time domain. This makes analyzing complex circuits much easier!
To solve the given initial-value problem using Laplace transforms, we first transform the differential equation, apply the initial conditions, and solve for Y ( s ) . After partial fraction decomposition, we take the inverse Laplace transform to arrive at the solution. The final answer is y ( t ) = 8 cos ( t ) + 2 sin ( t ) − 2 sin ( 2 t ) .
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