Solve the initial value problem below using the method of Laplace transforms. [tex]y^{\prime \prime}-y^{\prime}-12 y=-2 \cos t-5 \sin t, y\left(\frac{\pi}{2}\right)=1, y^{\prime}\left(\frac{\pi}{2}\right)=0[/tex] [tex]y(t)=\square[/tex]
Answer (2)
Perform a change of variable t ′ = t − 2 π to simplify initial conditions.
Take the Laplace transform of the transformed differential equation.
Solve for Y ( s ) in the Laplace domain and perform partial fraction decomposition.
Find the inverse Laplace transform and substitute back to obtain the solution: y ( t ) = 70 23 e − 3 t + 2 3 π + 119 33 e 4 t − 2 π + 170 67 sin ( t ) + 170 21 cos ( t )
Explanation
Problem Setup We are given the second-order linear non-homogeneous differential equation y ′′ − y ′ − 12 y = − 2 cos t − 5 sin t with initial conditions y ( 2 π ) = 1 and y ′ ( 2 π ) = 0 . We will solve this using Laplace transforms.
Change of Variable Let's perform a change of variable t ′ = t − 2 π . Then t = t ′ + 2 π , and we have y ( 2 π ) = y ( 0 ) = 1 and y ′ ( 2 π ) = y ′ ( 0 ) = 0 . The differential equation becomes y ′′ ( t ′ ) − y ′ ( t ′ ) − 12 y ( t ′ ) = − 2 cos ( t ′ + 2 π ) − 5 sin ( t ′ + 2 π ) = 2 sin ( t ′ ) − 5 cos ( t ′ ) .
Laplace Transform Now, we take the Laplace transform of the transformed equation: L { y ′′ ( t ′ ) − y ′ ( t ′ ) − 12 y ( t ′ ) } = L { 2 sin ( t ′ ) − 5 cos ( t ′ ) } Using the properties of Laplace transforms, we get s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) − ( s Y ( s ) − y ( 0 )) − 12 Y ( s ) = s 2 + 1 2 − s 2 + 1 5 s .
Solving for Y(s) Substituting the initial conditions y ( 0 ) = 1 and y ′ ( 0 ) = 0 , we have s 2 Y ( s ) − s − ( s Y ( s ) − 1 ) − 12 Y ( s ) = s 2 + 1 2 − s 2 + 1 5 s .
Solving for Y ( s ) , we get Y ( s ) ( s 2 − s − 12 ) − s + 1 = s 2 + 1 2 − 5 s , so Y ( s ) = ( s 2 + 1 ) ( s 2 − s − 12 ) 2 − 5 s + s 2 − s − 12 s − 1 = ( s 2 + 1 ) ( s 2 − s − 12 ) 2 − 5 s + ( s − 1 ) ( s 2 + 1 ) = ( s 2 + 1 ) ( s − 4 ) ( s + 3 ) 2 − 5 s + s 3 − s 2 + s − 1 = ( s 2 + 1 ) ( s − 4 ) ( s + 3 ) s 3 − s 2 − 4 s + 1 .
Partial Fraction Decomposition We perform partial fraction decomposition on Y ( s ) :
Y ( s ) = s − 4 A + s + 3 B + s 2 + 1 C s + D .
Using a calculator, we find that Y ( s ) = 119 ( s − 4 ) 33 + 70 ( s + 3 ) 23 + 170 ( s 2 + 1 ) ( 67 s − 21 ) .
Inverse Laplace Transform Now we find the inverse Laplace transform of Y ( s ) :
y ( t ′ ) = L − 1 { 119 ( s − 4 ) 33 + 70 ( s + 3 ) 23 + 170 ( s 2 + 1 ) ( 67 s − 21 ) } = 119 33 e 4 t ′ + 70 23 e − 3 t ′ + 170 67 cos ( t ′ ) − 170 21 sin ( t ′ ) .
Substitute Back Finally, we substitute back t ′ = t − 2 π to get y ( t ) :
y ( t ) = 119 33 e 4 ( t − 2 π ) + 70 23 e − 3 ( t − 2 π ) + 170 67 cos ( t − 2 π ) − 170 21 sin ( t − 2 π ) = 119 33 e 4 t − 2 π + 70 23 e − 3 t + 2 3 π + 170 67 sin ( t ) + 170 21 cos ( t ) .
Final Answer Therefore, the solution to the initial value problem is: y ( t ) = 70 23 e − 3 t + 2 3 π + 119 33 e 4 t − 2 π + 170 67 sin ( t ) + 170 21 cos ( 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, making it much easier to solve for the current and voltage at different points in the circuit. This helps engineers design and troubleshoot circuits efficiently.
The solution to the initial value problem using Laplace transforms results in the function y ( t ) = 70 23 e − 3 t + 2 3 π + 119 33 e 4 t − 2 π + 170 67 sin ( t ) + 170 21 cos ( t ) . This approach involved changing variables, applying the Laplace transform, solving for Y ( s ) , and then performing the inverse transform to find the solution in the time domain.
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