Solve the initial value problem below using the method of Laplace transforms.
[tex]y^{\prime \prime}+6 y^{\prime}+8 y=-8 \cos t-7 \sin t, y\left(\frac{\pi}{2}\right)=1, y^{\prime}\left(\frac{\pi}{2}\right)=0[/tex]
[tex]y(t)=[/tex]
Answer (1)
Apply Laplace transform to the given differential equation, converting it into an algebraic equation in terms of U ( s ) , considering the shifted initial conditions.
Solve for U ( s ) by applying partial fraction decomposition.
Apply the inverse Laplace transform to find u ( t ) .
Substitute t with t − 2 π to obtain the final solution: y ( t ) = − sin t − cos t + e π e − 2 t + e 2 π e − 4 t
Explanation
Problem Analysis We are given the second-order linear non-homogeneous differential equation y ′′ + 6 y ′ + 8 y = − 8 cos t − 7 sin t with initial conditions y ( 2 π ) = 1 and y ′ ( 2 π ) = 0 . Our goal is to solve this initial value problem using Laplace transforms.
Transforming the Differential Equation To handle the initial conditions given at t = 2 π instead of t = 0 , we introduce a new function u ( t ) = y ( t + 2 π ) . Then, y ( t ) = u ( t − 2 π ) , y ′ ( 2 π ) = u ′ ( 0 ) = 0 , and y ( 2 π ) = u ( 0 ) = 1 . We also have y ′ ( t ) = u ′ ( t − 2 π ) and y ′′ ( t ) = u ′′ ( t − 2 π ) . Substituting t with t + 2 π in the original equation, we get
u ′′ ( t ) + 6 u ′ ( t ) + 8 u ( t ) = − 8 cos ( t + 2 π ) − 7 sin ( t + 2 π )
Using trigonometric identities, cos ( t + 2 π ) = − sin t and sin ( t + 2 π ) = cos t . Thus, the equation becomes
u ′′ ( t ) + 6 u ′ ( t ) + 8 u ( t ) = 8 sin t − 7 cos t
with initial conditions u ( 0 ) = 1 and u ′ ( 0 ) = 0 .
Solving for U(s) Now, we apply the Laplace transform to both sides of the equation. Let U ( s ) = L [ u ( t )] . Using the properties of Laplace transforms, we have:
L [ u ′ ( t )] = s U ( s ) − u ( 0 ) = s U ( s ) − 1
L [ u ′′ ( t )] = s 2 U ( s ) − s u ( 0 ) − u ′ ( 0 ) = s 2 U ( s ) − s
L [ sin t ] = s 2 + 1 1
L [ cos t ] = s 2 + 1 s
So, the Laplace transform of the differential equation is:
( s 2 U ( s ) − s ) + 6 ( s U ( s ) − 1 ) + 8 U ( s ) = 8 s 2 + 1 1 − 7 s 2 + 1 s
( s 2 + 6 s + 8 ) U ( s ) − s − 6 = s 2 + 1 8 − 7 s
( s 2 + 6 s + 8 ) U ( s ) = s 2 + 1 8 − 7 s + s + 6
U ( s ) = ( s 2 + 1 ) ( s 2 + 6 s + 8 ) 8 − 7 s + s 2 + 6 s + 8 s + 6
U ( s ) = ( s 2 + 1 ) ( s + 2 ) ( s + 4 ) 8 − 7 s + ( s + 2 ) ( s + 4 ) s + 6
Partial Fraction Decomposition We can simplify U ( s ) by finding a common denominator:
U ( s ) = ( s 2 + 1 ) ( s + 2 ) ( s + 4 ) 8 − 7 s + ( s + 6 ) ( s 2 + 1 ) = ( s 2 + 1 ) ( s + 2 ) ( s + 4 ) 8 − 7 s + s 3 + 6 s 2 + s + 6 = ( s 2 + 1 ) ( s + 2 ) ( s + 4 ) s 3 + 6 s 2 − 6 s + 14
Now, we perform partial fraction decomposition on U ( s ) . We assume that
( s 2 + 1 ) ( s + 2 ) ( s + 4 ) s 3 + 6 s 2 − 6 s + 14 = s 2 + 1 A s + B + s + 2 C + s + 4 D
Multiplying both sides by ( s 2 + 1 ) ( s + 2 ) ( s + 4 ) , we get
s 3 + 6 s 2 − 6 s + 14 = ( A s + B ) ( s + 2 ) ( s + 4 ) + C ( s 2 + 1 ) ( s + 4 ) + D ( s 2 + 1 ) ( s + 2 )
s 3 + 6 s 2 − 6 s + 14 = ( A s + B ) ( s 2 + 6 s + 8 ) + C ( s 3 + 4 s 2 + s + 4 ) + D ( s 3 + 2 s 2 + s + 2 )
s 3 + 6 s 2 − 6 s + 14 = A s 3 + 6 A s 2 + 8 A s + B s 2 + 6 B s + 8 B + C s 3 + 4 C s 2 + C s + 4 C + D s 3 + 2 D s 2 + Ds + 2 D
Comparing coefficients, we have:
s 3 : A + C + D = 1
s 2 : 6 A + B + 4 C + 2 D = 6
s : 8 A + 6 B + C + D = − 6
1 : 8 B + 4 C + 2 D = 14
Solving this system of equations, we find A = − 1 , B = 1 , C = 1 , D = 1 . Therefore,
U ( s ) = s 2 + 1 − s + 1 + s + 2 1 + s + 4 1 = − s 2 + 1 s + s 2 + 1 1 + s + 2 1 + s + 4 1
Inverse Laplace Transform Now, we take the inverse Laplace transform of U ( s ) to find u ( t ) :
u ( t ) = L − 1 [ U ( s )] = L − 1 [ − s 2 + 1 s + s 2 + 1 1 + s + 2 1 + s + 4 1 ]
u ( t ) = − cos t + sin t + e − 2 t + e − 4 t
Since y ( t ) = u ( t − 2 π ) , we substitute t with t − 2 π :
y ( t ) = − cos ( t − 2 π ) + sin ( t − 2 π ) + e − 2 ( t − 2 π ) + e − 4 ( t − 2 π )
Using trigonometric identities, cos ( t − 2 π ) = sin t and sin ( t − 2 π ) = − cos t . Thus,
y ( t ) = − sin t − cos t + e − 2 t + π + e − 4 t + 2 π
y ( t ) = − sin t − cos t + e π e − 2 t + e 2 π e − 4 t
Final Solution Therefore, the solution to the initial value problem is:
y ( t ) = − sin t − cos t + e π e − 2 t + e 2 π e − 4 t
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. This makes it much easier to solve for the current and voltage in the circuit as a function of time, especially when dealing with complex circuits or time-varying sources. This technique simplifies the design and analysis of electrical systems, ensuring they perform as expected.
