Solve the initial value problem below using the method of Laplace transforms.
[tex]y^{\prime \prime}+16 y=16 t^2-32 t+66, y(0)=0, y^{\prime}(0)=18[/tex]
[tex]y(t)=[/tex]
Answer (2)
Apply Laplace transform to the differential equation.
Substitute initial conditions and solve for Y ( s ) .
Perform partial fraction decomposition on Y ( s ) .
Apply inverse Laplace transform to find y ( t ) : y ( t ) = t 2 − 2 t − 4 cos ( 4 t ) + 5 sin ( 4 t ) + 4
Explanation
Problem Setup We are given the initial value problem:
y ′′ + 16 y = 16 t 2 − 32 t + 66 , y ( 0 ) = 0 , y ′ ( 0 ) = 18
We will solve this problem using Laplace transforms.
Applying Laplace Transform First, we apply the Laplace transform to both sides of the differential equation:
L { y ′′ + 16 y } = L { 16 t 2 − 32 t + 66 }
Using the linearity property of the Laplace transform, we have:
L { y ′′ } + 16 L { y } = 16 L { t 2 } − 32 L { t } + 66 L { 1 }
Using Laplace Transform Formulas Now, we use the Laplace transform formulas:
L { y ′′ } = s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) L { y } = Y ( s ) L { t 2 } = s 3 2 L { t } = s 2 1 L { 1 } = s 1
Substituting these into the equation, we get:
s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) + 16 Y ( s ) = 16 ⋅ s 3 2 − 32 ⋅ s 2 1 + 66 ⋅ s 1
Substituting Initial Conditions Next, we substitute the initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = 18 :
s 2 Y ( s ) − s ( 0 ) − 18 + 16 Y ( s ) = s 3 32 − s 2 32 + s 66
( s 2 + 16 ) Y ( s ) − 18 = s 3 32 − s 2 32 + s 66
Solving for Y(s) Now, we solve for Y ( s ) :
( s 2 + 16 ) Y ( s ) = s 3 32 − s 2 32 + s 66 + 18
Y ( s ) = s 3 ( s 2 + 16 ) 32 − s 2 ( s 2 + 16 ) 32 + s ( s 2 + 16 ) 66 + s 2 + 16 18
Y ( s ) = s 3 ( s 2 + 16 ) 2 ( 9 s 3 + 33 s 2 − 16 s + 16 )
Partial Fraction Decomposition We perform partial fraction decomposition on Y ( s ) . From the tool, we have:
Y ( s ) = − s 2 + 16 4 ( s − 5 ) + s 4 − s 2 2 + s 3 2
Y ( s ) = − s 2 + 16 4 s + s 2 + 16 20 + s 4 − s 2 2 + s 3 2
Applying Inverse Laplace Transform Now, we apply the inverse Laplace transform to each term:
L − 1 { Y ( s ) } = L − 1 { − s 2 + 16 4 s } + L − 1 { s 2 + 16 20 } + L − 1 { s 4 } − L − 1 { s 2 2 } + L − 1 { s 3 2 }
Using the inverse Laplace transform formulas:
L − 1 { s 2 + a 2 s } = cos ( a t ) L − 1 { s 2 + a 2 a } = sin ( a t ) L − 1 { s 1 } = 1 L − 1 { s 2 1 } = t L − 1 { s 3 2 } = t 2
We get:
y ( t ) = − 4 cos ( 4 t ) + 5 sin ( 4 t ) + 4 − 2 t + t 2
Final Solution Therefore, the solution to the initial value problem is:
y ( t ) = t 2 − 2 t − 4 cos ( 4 t ) + 5 sin ( 4 t ) + 4
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with a voltage source, resistors, and capacitors. By using Laplace transforms, you can convert the differential equations that describe the circuit's behavior into algebraic equations, making it much easier to solve for the current and voltage at different points in the circuit. This helps engineers design and troubleshoot complex electrical systems efficiently.
We used Laplace transforms to solve the differential equation by first applying transforms, substituting initial conditions, and finally performing partial fraction decomposition. The solution is derived and expressed as y ( t ) = t 2 − 2 t − 4 cos ( 4 t ) + 5 sin ( 4 t ) + 4 . This comprehensive approach provides an analytical solution to the initial value problem.
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