Solving Initial Value Problems Using Laplace Transforms A Step-by-Step Guide

by Admin 77 views

The Laplace transform is a powerful mathematical tool for solving linear differential equations, particularly those with initial conditions. It transforms a differential equation in the time domain into an algebraic equation in the complex frequency domain (s-domain), which is often easier to solve. After finding the solution in the s-domain, the inverse Laplace transform is used to obtain the solution in the time domain. This method is especially useful for problems involving discontinuous or impulsive forcing functions. In this article, we will demonstrate how to solve a second-order linear differential equation with constant coefficients using the Laplace transform method. This technique is widely used in engineering, physics, and applied mathematics to analyze and design systems that evolve over time, such as electrical circuits, mechanical systems, and control systems. The Laplace transform provides a systematic way to handle initial conditions and forcing functions, making it a versatile tool for solving a wide range of problems. By mastering this method, you can gain a deeper understanding of how systems respond to various inputs and disturbances, and you can develop effective strategies for controlling and optimizing their behavior. The ability to solve differential equations is a fundamental skill for engineers and scientists, and the Laplace transform method is an essential part of their toolkit. With practice and experience, you can become proficient in using this technique to tackle complex problems and gain valuable insights into the dynamics of real-world systems. Let's delve into the step-by-step process of applying the Laplace transform to solve initial value problems. This approach not only provides a solution but also enhances the understanding of the underlying system's behavior and response.

We are given the following second-order linear differential equation with constant coefficients:

y′′−y′−12y=−2cos⁡t−5sin⁡ty^{\prime \prime}-y^{\prime}-12 y=-2 \cos t-5 \sin t

with initial conditions:

y(π2)=1,y′(π2)=0y(\frac{\pi}{2})=1, y^{\prime}(\frac{\pi}{2})=0

Our goal is to find the solution y(t)y(t) using the Laplace transform method. This problem exemplifies a common scenario in various engineering and physics applications, where understanding the system's response to external forces or disturbances is crucial. The presence of trigonometric functions on the right-hand side indicates a forced oscillation, and the initial conditions provide specific constraints on the solution at a particular time. The Laplace transform method is particularly well-suited for this type of problem because it can handle both the differential equation and the initial conditions in a systematic manner. By transforming the equation into the s-domain, we can convert the differentiation operations into algebraic manipulations, which simplifies the solution process. After solving for the Laplace transform of the solution, we can then use the inverse Laplace transform to obtain the solution in the time domain. This approach not only provides the solution but also offers insights into the system's behavior and stability. By analyzing the poles and zeros of the Laplace transform, we can determine the system's natural frequencies, damping characteristics, and response to different types of inputs. The Laplace transform method is a versatile tool for analyzing linear systems, and mastering it will enable you to tackle a wide range of engineering and scientific problems. Let's proceed with the step-by-step application of the Laplace transform to solve this initial value problem.

Apply the Laplace transform to both sides of the differential equation. Recall the following properties of the Laplace transform:

Ly′′(t)=s2Y(s)−sy(0)−y′(0)\mathcal{L}{y^{\prime \prime}(t)} = s^2Y(s) - sy(0) - y^{\prime}(0)

Ly′(t)=sY(s)−y(0)\mathcal{L}{y^{\prime}(t)} = sY(s) - y(0)

Ly(t)=Y(s)\mathcal{L}{y(t)} = Y(s)

Lcos⁡(at)=ss2+a2\mathcal{L}{\cos(at)} = \frac{s}{s^2 + a^2}

Lsin⁡(at)=as2+a2\mathcal{L}{\sin(at)} = \frac{a}{s^2 + a^2}

However, we have initial conditions at t=π2t = \frac{\pi}{2}, not t=0t = 0. To handle this, we make a change of variable. Let τ=t−π2\tau = t - \frac{\pi}{2}, so t=τ+π2t = \tau + \frac{\pi}{2}. Define a new function z(τ)=y(τ+π2)=y(t)z(\tau) = y(\tau + \frac{\pi}{2}) = y(t). Then,

z′(τ)=y′(t)z^{\prime}(\tau) = y^{\prime}(t)

z′′(τ)=y′′(t)z^{\prime \prime}(\tau) = y^{\prime \prime}(t)

and the initial conditions become:

z(0)=y(π2)=1z(0) = y(\frac{\pi}{2}) = 1

z′(0)=y′(π2)=0z^{\prime}(0) = y^{\prime}(\frac{\pi}{2}) = 0

The differential equation in terms of z(τ)z(\tau) is:

z′′−z′−12z=−2cos⁡(τ+π2)−5sin⁡(τ+π2)z^{\prime \prime} - z^{\prime} - 12z = -2 \cos(\tau + \frac{\pi}{2}) - 5 \sin(\tau + \frac{\pi}{2})

Using trigonometric identities:

cos⁡(τ+π2)=−sin⁡(τ)\cos(\tau + \frac{\pi}{2}) = -\sin(\tau)

sin⁡(τ+π2)=cos⁡(τ)\sin(\tau + \frac{\pi}{2}) = \cos(\tau)

So the equation becomes:

z′′−z′−12z=2sin⁡(τ)−5cos⁡(τ)z^{\prime \prime} - z^{\prime} - 12z = 2 \sin(\tau) - 5 \cos(\tau)

Now, apply the Laplace transform to this equation:

Lz′′−Lz′−12Lz=2Lsin⁡(τ)−5Lcos⁡(τ)\mathcal{L}{z^{\prime \prime}} - \mathcal{L}{z^{\prime}} - 12\mathcal{L}{z} = 2\mathcal{L}{\sin(\tau)} - 5\mathcal{L}{\cos(\tau)}

Using the Laplace transform properties and the initial conditions, we get:

[s2Z(s)−sz(0)−z′(0)]−[sZ(s)−z(0)]−12Z(s)=2(1s2+1)−5(ss2+1)[s^2Z(s) - sz(0) - z^{\prime}(0)] - [sZ(s) - z(0)] - 12Z(s) = 2(\frac{1}{s^2 + 1}) - 5(\frac{s}{s^2 + 1})

Substituting the initial conditions z(0)=1z(0) = 1 and z′(0)=0z^{\prime}(0) = 0:

s2Z(s)−s−sZ(s)+1−12Z(s)=2−5ss2+1s^2Z(s) - s - sZ(s) + 1 - 12Z(s) = \frac{2 - 5s}{s^2 + 1}

This step is crucial as it sets the stage for solving the differential equation in the s-domain. By applying the Laplace transform, we convert the differential equation into an algebraic equation, which is much easier to manipulate. The use of the trigonometric identities simplifies the forcing function, and the substitution of initial conditions provides specific values for the terms involving s. This process transforms the problem from a differential equation to an algebraic one, making it more manageable. The next step involves solving for Z(s), which represents the Laplace transform of the solution. This will involve algebraic manipulations and the combination of terms. The ability to perform these steps accurately is essential for successfully applying the Laplace transform method. Let's proceed to the next step and solve for Z(s).

Rearrange the equation to solve for Z(s)Z(s):

(s2−s−12)Z(s)=s−1+2−5ss2+1(s^2 - s - 12)Z(s) = s - 1 + \frac{2 - 5s}{s^2 + 1}

Factor the quadratic term:

(s−4)(s+3)Z(s)=(s−1)(s2+1)+2−5ss2+1(s - 4)(s + 3)Z(s) = \frac{(s - 1)(s^2 + 1) + 2 - 5s}{s^2 + 1}

Simplify the numerator on the right-hand side:

(s−4)(s+3)Z(s)=s3−s2+s−1+2−5ss2+1(s - 4)(s + 3)Z(s) = \frac{s^3 - s^2 + s - 1 + 2 - 5s}{s^2 + 1}

(s−4)(s+3)Z(s)=s3−s2−4s+1s2+1(s - 4)(s + 3)Z(s) = \frac{s^3 - s^2 - 4s + 1}{s^2 + 1}

Now, isolate Z(s)Z(s):

Z(s)=s3−s2−4s+1(s−4)(s+3)(s2+1)Z(s) = \frac{s^3 - s^2 - 4s + 1}{(s - 4)(s + 3)(s^2 + 1)}

This step involves algebraic manipulation to isolate Z(s), which represents the Laplace transform of the solution. The process includes factoring the quadratic term, simplifying the numerator, and dividing both sides by the factored quadratic. These algebraic steps are critical to obtaining the correct expression for Z(s). The complexity of the expression highlights the importance of careful and accurate algebraic manipulation. The result, Z(s), is a rational function in s, which is typical in Laplace transform problems. The denominator consists of factors corresponding to the roots of the characteristic equation and the forcing function. The numerator is a polynomial that arises from the initial conditions and the forcing function. The next step involves finding the inverse Laplace transform of Z(s) to obtain the solution in the time domain. This often requires partial fraction decomposition, which breaks down the rational function into simpler terms that can be easily inverted using Laplace transform tables. The expression for Z(s) provides valuable information about the system's behavior, including its stability and response characteristics. The poles of Z(s) (the roots of the denominator) determine the system's natural frequencies and damping. Let's proceed to the next step and find the inverse Laplace transform of Z(s).

To find the inverse Laplace transform, we need to perform partial fraction decomposition on Z(s)Z(s). Let:

s3−s2−4s+1(s−4)(s+3)(s2+1)=As−4+Bs+3+Cs+Ds2+1\frac{s^3 - s^2 - 4s + 1}{(s - 4)(s + 3)(s^2 + 1)} = \frac{A}{s - 4} + \frac{B}{s + 3} + \frac{Cs + D}{s^2 + 1}

Multiply both sides by (s−4)(s+3)(s2+1)(s - 4)(s + 3)(s^2 + 1):

s3−s2−4s+1=A(s+3)(s2+1)+B(s−4)(s2+1)+(Cs+D)(s−4)(s+3)s^3 - s^2 - 4s + 1 = A(s + 3)(s^2 + 1) + B(s - 4)(s^2 + 1) + (Cs + D)(s - 4)(s + 3)

To find A, let s=4s = 4:

43−42−4(4)+1=A(4+3)(42+1)4^3 - 4^2 - 4(4) + 1 = A(4 + 3)(4^2 + 1)

64−16−16+1=A(7)(17)64 - 16 - 16 + 1 = A(7)(17)

33=119A33 = 119A

A=33119A = \frac{33}{119}

To find B, let s=−3s = -3:

(−3)3−(−3)2−4(−3)+1=B(−3−4)((−3)2+1)(-3)^3 - (-3)^2 - 4(-3) + 1 = B(-3 - 4)((-3)^2 + 1)

−27−9+12+1=B(−7)(10)-27 - 9 + 12 + 1 = B(-7)(10)

−23=−70B-23 = -70B

B=2370B = \frac{23}{70}

To find C and D, we can expand the equation and equate coefficients. Expanding the right side:

s3−s2−4s+1=A(s3+3s2+s+3)+B(s3−4s2+s−4)+(Cs+D)(s2−s−12)s^3 - s^2 - 4s + 1 = A(s^3 + 3s^2 + s + 3) + B(s^3 - 4s^2 + s - 4) + (Cs + D)(s^2 - s - 12)

s3−s2−4s+1=A(s3+3s2+s+3)+B(s3−4s2+s−4)+Cs3−Cs2−12Cs+Ds2−Ds−12Ds^3 - s^2 - 4s + 1 = A(s^3 + 3s^2 + s + 3) + B(s^3 - 4s^2 + s - 4) + Cs^3 - Cs^2 - 12Cs + Ds^2 - Ds - 12D

Equating coefficients of s3s^3:

1=A+B+C1 = A + B + C

C=1−A−B=1−33119−2370=1−33119−2370=8330−2310−32818330=27398330=391190C = 1 - A - B = 1 - \frac{33}{119} - \frac{23}{70} = 1 - \frac{33}{119} - \frac{23}{70} = \frac{8330 - 2310 - 3281}{8330} = \frac{2739}{8330} = \frac{39}{1190}

Equating constant terms:

1=3A−4B−12D1 = 3A - 4B - 12D

12D=3A−4B−1=3(33119)−4(2370)−1=99119−4635−1=495−1564−595595=−166459512D = 3A - 4B - 1 = 3(\frac{33}{119}) - 4(\frac{23}{70}) - 1 = \frac{99}{119} - \frac{46}{35} - 1 = \frac{495 - 1564 - 595}{595} = \frac{-1664}{595}

D=−166412∗595=−4161785D = \frac{-1664}{12 * 595} = \frac{-416}{1785}

So we have:

A=33119A = \frac{33}{119}

B=2370B = \frac{23}{70}

C=391190C = \frac{39}{1190}

D=−4161785D = \frac{-416}{1785}

This step is a critical process in solving differential equations using Laplace transforms. Partial fraction decomposition allows us to break down a complex rational function into simpler fractions, making it easier to find the inverse Laplace transform. The method involves expressing the rational function as a sum of simpler fractions with denominators corresponding to the factors of the original denominator. The numerators of these simpler fractions are constants or linear expressions that need to be determined. The process typically involves setting up an equation with unknown coefficients, clearing denominators, and then solving for the coefficients by substituting specific values of s or equating coefficients of like powers of s. The accuracy of this step is crucial, as any errors in the partial fraction decomposition will propagate through the rest of the solution. The resulting simpler fractions can then be individually inverted using standard Laplace transform tables or properties. This technique is widely used in engineering and physics to solve linear systems, and mastering it is essential for anyone working with Laplace transforms. Let's proceed to the next step, where we will use these results to find the inverse Laplace transform and obtain the solution in the time domain.

Now we have:

Z(s)=33119(s−4)+2370(s+3)+391190s−4161785s2+1Z(s) = \frac{33}{119(s - 4)} + \frac{23}{70(s + 3)} + \frac{\frac{39}{1190}s - \frac{416}{1785}}{s^2 + 1}

We can rewrite the last term as:

391190ss2+1−4161785s2+1\frac{\frac{39}{1190}s}{s^2 + 1} - \frac{\frac{416}{1785}}{s^2 + 1}

Now, take the inverse Laplace transform:

z(τ)=33119e4τ+2370e−3τ+391190cos⁡(τ)−4161785sin⁡(τ)z(\tau) = \frac{33}{119}e^{4\tau} + \frac{23}{70}e^{-3\tau} + \frac{39}{1190}\cos(\tau) - \frac{416}{1785}\sin(\tau)

Substitute back τ=t−π2\tau = t - \frac{\pi}{2}:

y(t)=33119e4(t−π2)+2370e−3(t−π2)+391190cos⁡(t−π2)−4161785sin⁡(t−π2)y(t) = \frac{33}{119}e^{4(t - \frac{\pi}{2})} + \frac{23}{70}e^{-3(t - \frac{\pi}{2})} + \frac{39}{1190}\cos(t - \frac{\pi}{2}) - \frac{416}{1785}\sin(t - \frac{\pi}{2})

Using trigonometric identities:

cos⁡(t−π2)=sin⁡(t)\cos(t - \frac{\pi}{2}) = \sin(t)

sin⁡(t−π2)=−cos⁡(t)\sin(t - \frac{\pi}{2}) = -\cos(t)

So:

y(t)=33119e4(t−π2)+2370e−3(t−π2)+391190sin⁡(t)+4161785cos⁡(t)y(t) = \frac{33}{119}e^{4(t - \frac{\pi}{2})} + \frac{23}{70}e^{-3(t - \frac{\pi}{2})} + \frac{39}{1190}\sin(t) + \frac{416}{1785}\cos(t)

This step involves applying the inverse Laplace transform to each term obtained from the partial fraction decomposition. The inverse Laplace transform converts the function from the s-domain back to the time domain, giving us the solution to the original differential equation. The standard Laplace transform pairs are used to invert each term, such as the exponential terms and the trigonometric terms. The constants obtained from the partial fraction decomposition are multiplied by the corresponding inverse Laplace transforms. After obtaining the solution in terms of the shifted time variable τ, we substitute back to the original time variable t. This substitution is necessary because the initial conditions were given at a non-zero time. Finally, trigonometric identities are used to simplify the solution and express it in a more standard form. The resulting solution, y(t), describes the behavior of the system over time and satisfies both the differential equation and the initial conditions. The solution consists of exponential terms and trigonometric terms, which indicate the system's natural response and its forced response to the input. The exponential terms decay over time if the system is stable, while the trigonometric terms oscillate with specific frequencies. Let's review the entire process and summarize the final solution.

The solution to the initial value problem is:

y(t)=33119e4(t−π2)+2370e−3(t−π2)+391190sin⁡(t)+4161785cos⁡(t)y(t) = \frac{33}{119}e^{4(t - \frac{\pi}{2})} + \frac{23}{70}e^{-3(t - \frac{\pi}{2})} + \frac{39}{1190}\sin(t) + \frac{416}{1785}\cos(t)

This final solution represents the complete response of the system to the given initial conditions and forcing function. The solution consists of two exponential terms and two trigonometric terms, each with specific coefficients. The exponential terms, e4(t−π2)e^{4(t - \frac{\pi}{2})} and e−3(t−π2)e^{-3(t - \frac{\pi}{2})}, represent the system's natural response, which is determined by the homogeneous part of the differential equation. The term e4(t−π2)e^{4(t - \frac{\pi}{2})} grows exponentially as t increases, indicating an unstable mode, while the term e−3(t−π2)e^{-3(t - \frac{\pi}{2})} decays exponentially, indicating a stable mode. The trigonometric terms, sin⁡(t)\sin(t) and cos⁡(t)\cos(t), represent the system's forced response, which is caused by the external forcing function. These terms oscillate with a frequency of 1, which corresponds to the frequency of the sine and cosine terms in the forcing function. The coefficients of these terms determine the amplitude and phase of the oscillations. The coefficients 33119\frac{33}{119} and 2370\frac{23}{70} are associated with the exponential terms, while the coefficients 391190\frac{39}{1190} and 4161785\frac{416}{1785} are associated with the trigonometric terms. These coefficients are determined by the initial conditions and the parameters of the differential equation. The solution y(t) satisfies both the differential equation and the initial conditions, which can be verified by substituting y(t) and its derivatives into the equation and checking that the initial conditions are met. The Laplace transform method provides a systematic way to solve linear differential equations with constant coefficients, and this example demonstrates the key steps involved in the process. By understanding the Laplace transform and its properties, you can effectively solve a wide range of engineering and scientific problems.

In this article, we have successfully solved the given initial value problem using the method of Laplace transforms. The key steps involved applying the Laplace transform to the differential equation, handling the initial conditions at t=π2t = \frac{\pi}{2} by using a change of variable, solving for Z(s)Z(s) in the s-domain, performing partial fraction decomposition, and finally, taking the inverse Laplace transform to obtain the solution in the time domain. This method is a powerful tool for solving linear differential equations, especially those with initial conditions and non-homogeneous terms. The Laplace transform transforms the differential equation into an algebraic equation, which is often easier to solve. The use of partial fraction decomposition allows us to break down complex rational functions into simpler terms, making it easier to find the inverse Laplace transform. The final solution provides the behavior of the system over time, considering both the initial conditions and the external forces. The Laplace transform method is widely used in various fields, including engineering, physics, and applied mathematics, to analyze and design dynamic systems. By mastering this method, you can gain a deeper understanding of system behavior and develop effective strategies for solving complex problems. The ability to solve differential equations is a fundamental skill for engineers and scientists, and the Laplace transform method is an essential part of their toolkit. With practice and experience, you can become proficient in using this technique to tackle complex problems and gain valuable insights into the dynamics of real-world systems. This method not only provides a solution but also enhances the understanding of the underlying system's behavior and response, making it a valuable tool for analyzing and designing dynamic systems.