Solve the initial value problem below using the method of Laplace transforms.
[tex]y^{\prime \prime}+y=t^2-6 t+7, y(0)=0, y^{\prime}(0)=-1[/tex]
[tex]y(t)=\square[/tex]
Answer (2)
Apply Laplace transform to the given differential equation y ′′ + y = t 2 − 6 t + 7 with initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = − 1 .
Use Laplace transform properties to convert the differential equation into an algebraic equation in terms of Y ( s ) .
Solve for Y ( s ) and perform partial fraction decomposition to simplify the expression.
Apply the inverse Laplace transform to each term to find the solution y ( t ) = 9 − 6 t − t 2 − 7 cos ( t ) + 3 sin ( t ) .
The solution to the initial value problem is y ( t ) = 9 − 6 t − t 2 − 7 cos ( t ) + 3 sin ( t )
Explanation
Problem Setup We are given the initial value problem:
y ′′ + y = t 2 − 6 t + 7 , y ( 0 ) = 0 , y ′ ( 0 ) = − 1
We will solve this using Laplace transforms.
Applying Laplace Transform Taking the Laplace transform of both sides of the differential equation, we get:
L { y ′′ + y } = L { t 2 − 6 t + 7 }
Using the linearity property of the Laplace transform:
L { y ′′ } + L { y } = L { t 2 } − 6 L { t } + 7 L { 1 }
Using Laplace Transform Properties Using the Laplace transform properties for derivatives and the given initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = − 1 , we have:
L { y ′′ } = s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) = s 2 Y ( s ) − s ( 0 ) − ( − 1 ) = s 2 Y ( s ) + 1
L { y } = Y ( s )
Also, using the Laplace transforms of t n and a constant:
L { t 2 } = s 3 2
L { t } = s 2 1
L { 1 } = s 1
Substituting into Transformed Equation Substituting these into the transformed equation, we get:
s 2 Y ( s ) + 1 + Y ( s ) = s 3 2 − s 2 6 + s 7
( s 2 + 1 ) Y ( s ) = s 3 2 − s 2 6 + s 7 − 1
Y ( s ) = s 3 ( s 2 + 1 ) 2 − s 2 ( s 2 + 1 ) 6 + s ( s 2 + 1 ) 7 − s 2 + 1 1
Partial Fraction Decomposition Now we need to find the inverse Laplace transform of Y ( s ) . We can use partial fraction decomposition to simplify the terms. First, we rewrite Y ( s ) as a single fraction:
Y ( s ) = s 3 ( s 2 + 1 ) 2 − 6 s + 7 s 2 − s 3 − s 5
However, it's easier to decompose each term separately:
s 3 ( s 2 + 1 ) 2 = s A + s 2 B + s 3 C + s 2 + 1 Ds + E
s 2 ( s 2 + 1 ) 6 = s A + s 2 B + s 2 + 1 C s + D
s ( s 2 + 1 ) 7 = s A + s 2 + 1 B s + C
s 2 + 1 1
After solving for the coefficients, we get:
s 3 ( s 2 + 1 ) 2 = s 2 − s 3 2 − s 2 + 1 2
s 2 ( s 2 + 1 ) 6 = s 2 6 − s 2 + 1 6
s ( s 2 + 1 ) 7 = s 7 − s 2 + 1 7 s
So,
Y ( s ) = ( s 2 − s 3 2 − s 2 + 1 2 ) − ( s 2 6 − s 2 + 1 6 ) + ( s 7 − s 2 + 1 7 s ) − s 2 + 1 1
Y ( s ) = s 9 − s 2 6 − s 3 2 − s 2 + 1 7 s + s 2 + 1 3
Inverse Laplace Transform Now, we take the inverse Laplace transform of each term:
L − 1 { s 9 } = 9
L − 1 { s 2 6 } = 6 t
L − 1 { s 3 2 } = t 2
L − 1 { s 2 + 1 7 s } = 7 cos ( t )
L − 1 { s 2 + 1 3 } = 3 sin ( t )
Final Solution Therefore, the solution is:
y ( t ) = 9 − 6 t − t 2 − 7 cos ( t ) + 3 sin ( t )
Examples
Laplace transforms are incredibly useful in electrical engineering for analyzing circuits. Imagine you have a circuit with resistors, capacitors, and inductors, and you want to find the current flowing through it when you apply a voltage source. The circuit's behavior can be described by a differential equation, and using Laplace transforms, you can convert this differential equation into an algebraic equation, making it easier to solve for the current as a function of time. This is also applicable in mechanical engineering for analyzing vibrations and control systems.
To solve the given initial value problem using Laplace transforms, we first transform the differential equation into an algebraic equation. We then find Y ( s ) , perform partial fraction decomposition, and finally apply the inverse Laplace transform to obtain the solution. The resulting solution is y ( t ) = 9 − 6 t − t 2 − 7 cos ( t ) + 3 sin ( t ) .
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