Solving $y^{\prime \prime}+y=t^2-6 T+7$ With Laplace Transforms A Step-by-Step Guide

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Introduction

In the realm of differential equations, the Laplace transform emerges as a powerful technique for solving initial value problems, particularly those involving linear ordinary differential equations with constant coefficients. This method elegantly converts differential equations from the time domain into algebraic equations in the complex frequency domain (s-domain), thereby simplifying the solution process. By employing the Laplace transform, we can circumvent the complexities of direct integration and differentiation, offering a systematic approach to obtain solutions. This article delves into the application of Laplace transforms to solve the initial value problem: y′′+y=t2−6t+7y^{\prime \prime}+y=t^2-6 t+7, with initial conditions y(0)=0y(0)=0 and y′(0)=−1y^{\prime}(0)=-1. We will explore the step-by-step procedure, highlighting the properties and advantages of using Laplace transforms in solving such problems. The Laplace transform method is especially valuable in engineering and physics, where differential equations frequently model physical phenomena. This technique not only simplifies the solving of differential equations but also provides insights into the system's behavior in the frequency domain. Understanding the Laplace transform is crucial for students and professionals in these fields, as it offers a robust and efficient way to analyze and design systems. We will also discuss the inverse Laplace transform, which is essential for converting the solution from the s-domain back to the time domain, thus providing the actual solution to the initial value problem. The initial conditions play a crucial role in this process, as they are directly incorporated into the Laplace transform, leading to a unique solution that satisfies the given conditions. The power and versatility of the Laplace transform make it an indispensable tool in the mathematical arsenal of engineers and scientists.

Problem Statement

We are given the second-order linear ordinary differential equation:

y′′+y=t2−6t+7y^{\prime \prime}+y=t^2-6 t+7

with the initial conditions:

y(0)=0,y′(0)=−1y(0)=0, y^{\prime}(0)=-1

Our objective is to find the solution y(t)y(t) that satisfies both the differential equation and the given initial conditions using the Laplace transform method. The equation represents a classic example of a forced harmonic oscillator, where the term t2−6t+7t^2 - 6t + 7 acts as the forcing function. The initial conditions specify the state of the system at time t=0t=0, providing the starting point for the solution. To solve this problem effectively, we will apply the Laplace transform to both sides of the differential equation, utilizing the properties of Laplace transforms to handle derivatives and functions. This will transform the differential equation into an algebraic equation in the s-domain, which is significantly easier to solve. After finding the solution in the s-domain, we will apply the inverse Laplace transform to obtain the solution in the time domain, y(t)y(t). This process demonstrates the power and elegance of the Laplace transform method in solving initial value problems. By transforming the problem into the s-domain, we simplify the mathematical operations required to find the solution. This method is particularly useful for equations with discontinuous forcing functions or impulsive forces, where classical methods may be more challenging to apply. The Laplace transform provides a unified approach to solving a wide range of linear differential equations, making it an essential tool for engineers and mathematicians.

Applying the Laplace Transform

To begin, we apply the Laplace transform to both sides of the given differential equation:

L{y′′+y}=L{t2−6t+7}\mathcal{L}\left\{y^{\prime \prime}+y\right\}=\mathcal{L}\left\{t^2-6 t+7\right\}

Using the linearity property of the Laplace transform, we can split the left-hand side into two terms:

L{y′′}+L{y}=L{t2−6t+7}\mathcal{L}\left\{y^{\prime \prime}\right\}+\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{t^2-6 t+7\right\}

Now, we apply the Laplace transform to each term individually. Recall the following Laplace transforms and properties:

  • L{y′′(t)}=s2Y(s)−sy(0)−y′(0)\mathcal{L}\left\{y^{\prime \prime}(t)\right\} = s^2Y(s) - sy(0) - y^{\prime}(0)
  • L{y(t)}=Y(s)\mathcal{L}\left\{y(t)\right\} = Y(s)
  • L{tn}=n!sn+1\mathcal{L}\left\{t^n\right\} = \frac{n!}{s^{n+1}}
  • L{t}=1s2\mathcal{L}\left\{t\right\} = \frac{1}{s^2}
  • L{1}=1s\mathcal{L}\left\{1\right\} = \frac{1}{s}

where Y(s)Y(s) is the Laplace transform of y(t)y(t). Applying these to our equation, we get:

s2Y(s)−sy(0)−y′(0)+Y(s)=L{t2}−6L{t}+7L{1}s^2Y(s) - sy(0) - y^{\prime}(0) + Y(s) = \mathcal{L}\left\{t^2\right\} - 6\mathcal{L}\left\{t\right\} + 7\mathcal{L}\left\{1\right\}

Substituting the Laplace transforms of t2t^2, tt, and 11, we have:

s2Y(s)−sy(0)−y′(0)+Y(s)=2s3−6s2+7ss^2Y(s) - sy(0) - y^{\prime}(0) + Y(s) = \frac{2}{s^3} - \frac{6}{s^2} + \frac{7}{s}

We now incorporate the initial conditions y(0)=0y(0) = 0 and y′(0)=−1y^{\prime}(0) = -1:

s2Y(s)−s(0)−(−1)+Y(s)=2s3−6s2+7ss^2Y(s) - s(0) - (-1) + Y(s) = \frac{2}{s^3} - \frac{6}{s^2} + \frac{7}{s}

Simplifying, we obtain:

s2Y(s)+1+Y(s)=2s3−6s2+7ss^2Y(s) + 1 + Y(s) = \frac{2}{s^3} - \frac{6}{s^2} + \frac{7}{s}

This equation now relates Y(s)Y(s) to the Laplace transforms of the terms on the right-hand side, and it incorporates the initial conditions, which is a critical step in solving the problem. The transformation from the time domain to the s-domain has converted the differential equation into an algebraic equation, making it easier to manipulate and solve for Y(s)Y(s). This is one of the key advantages of using the Laplace transform method. The next step will involve isolating Y(s)Y(s) and then applying the inverse Laplace transform to find the solution y(t)y(t).

Solving for Y(s)

Now, we isolate Y(s)Y(s) in the equation:

s2Y(s)+Y(s)=2s3−6s2+7s−1s^2Y(s) + Y(s) = \frac{2}{s^3} - \frac{6}{s^2} + \frac{7}{s} - 1

Factor out Y(s)Y(s):

Y(s)(s2+1)=2s3−6s2+7s−1Y(s)(s^2 + 1) = \frac{2}{s^3} - \frac{6}{s^2} + \frac{7}{s} - 1

Divide both sides by (s2+1)(s^2 + 1) to solve for Y(s)Y(s):

Y(s)=1s2+1(2s3−6s2+7s−1)Y(s) = \frac{1}{s^2 + 1} \left(\frac{2}{s^3} - \frac{6}{s^2} + \frac{7}{s} - 1\right)

Distribute the term 1s2+1\frac{1}{s^2 + 1}:

Y(s)=2s3(s2+1)−6s2(s2+1)+7s(s2+1)−1s2+1Y(s) = \frac{2}{s^3(s^2 + 1)} - \frac{6}{s^2(s^2 + 1)} + \frac{7}{s(s^2 + 1)} - \frac{1}{s^2 + 1}

To prepare for the inverse Laplace transform, we need to perform partial fraction decomposition on the terms involving rational functions. This will break down the complex fractions into simpler terms that we can readily find inverse Laplace transforms for. The process of partial fraction decomposition involves expressing a rational function as a sum of simpler fractions with denominators corresponding to the factors of the original denominator. This technique is crucial for simplifying the expression for Y(s)Y(s) and making it amenable to inverse Laplace transform. Each term will be decomposed separately, and then we will combine the results to obtain the complete expression for Y(s)Y(s) in a form suitable for applying the inverse Laplace transform. This step is a key part of the Laplace transform method, allowing us to convert the solution from the frequency domain back to the time domain. The algebraic manipulations involved in partial fraction decomposition can be complex, but they are essential for obtaining the final solution in a usable form. The goal is to express Y(s)Y(s) as a sum of terms that match the standard Laplace transforms found in tables or that can be easily derived.

Partial Fraction Decomposition

We will perform partial fraction decomposition on each term of Y(s)Y(s):

  1. 2s3(s2+1)=As+Bs2+Cs3+Ds+Es2+1\frac{2}{s^3(s^2 + 1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{Ds + E}{s^2 + 1}
  2. 6s2(s2+1)=Fs+Gs2+Hs+Is2+1\frac{6}{s^2(s^2 + 1)} = \frac{F}{s} + \frac{G}{s^2} + \frac{Hs + I}{s^2 + 1}
  3. 7s(s2+1)=Js+Ks+Ls2+1\frac{7}{s(s^2 + 1)} = \frac{J}{s} + \frac{Ks + L}{s^2 + 1}

Solving for the constants:

  1. For 2s3(s2+1)\frac{2}{s^3(s^2 + 1)}:

    Multiply both sides by s3(s2+1)s^3(s^2 + 1):

    2=As2(s2+1)+Bs(s2+1)+C(s2+1)+(Ds+E)s32 = A s^2(s^2 + 1) + B s(s^2 + 1) + C(s^2 + 1) + (Ds + E)s^3

    Expanding and collecting terms:

    2=(A+D)s4+(B+E)s3+(A+C)s2+Bs+C2 = (A + D)s^4 + (B + E)s^3 + (A + C)s^2 + Bs + C

    Comparing coefficients:

    • A+D=0A + D = 0
    • B+E=0B + E = 0
    • A+C=0A + C = 0
    • B=0B = 0
    • C=2C = 2

    Solving the system:

    • C=2C = 2
    • A=−2A = -2
    • B=0B = 0
    • D=2D = 2
    • E=0E = 0

    Thus, 2s3(s2+1)=−2s+2s3+2ss2+1\frac{2}{s^3(s^2 + 1)} = -\frac{2}{s} + \frac{2}{s^3} + \frac{2s}{s^2 + 1}

  2. For 6s2(s2+1)\frac{6}{s^2(s^2 + 1)}:

    Multiply both sides by s2(s2+1)s^2(s^2 + 1):

    6=Fs(s2+1)+G(s2+1)+(Hs+I)s26 = F s(s^2 + 1) + G(s^2 + 1) + (Hs + I)s^2

    Expanding and collecting terms:

    6=(F+H)s3+(G+I)s2+Fs+G6 = (F + H)s^3 + (G + I)s^2 + Fs + G

    Comparing coefficients:

    • F+H=0F + H = 0
    • G+I=0G + I = 0
    • F=0F = 0
    • G=6G = 6

    Solving the system:

    • G=6G = 6
    • F=0F = 0
    • H=0H = 0
    • I=−6I = -6

    Thus, 6s2(s2+1)=6s2−6s2+1\frac{6}{s^2(s^2 + 1)} = \frac{6}{s^2} - \frac{6}{s^2 + 1}

  3. For 7s(s2+1)\frac{7}{s(s^2 + 1)}:

    Multiply both sides by s(s2+1)s(s^2 + 1):

    7=J(s2+1)+(Ks+L)s7 = J(s^2 + 1) + (Ks + L)s

    Expanding and collecting terms:

    7=(J+K)s2+Ls+J7 = (J + K)s^2 + Ls + J

    Comparing coefficients:

    • J+K=0J + K = 0
    • L=0L = 0
    • J=7J = 7

    Solving the system:

    • J=7J = 7
    • K=−7K = -7
    • L=0L = 0

    Thus, 7s(s2+1)=7s−7ss2+1\frac{7}{s(s^2 + 1)} = \frac{7}{s} - \frac{7s}{s^2 + 1}

The partial fraction decomposition is a critical step that simplifies the inverse Laplace transform process. By breaking down complex rational functions into simpler fractions, we can easily identify the corresponding time-domain functions using standard Laplace transform tables. The accuracy of this decomposition is crucial for obtaining the correct solution to the initial value problem. Each decomposition involves setting up a system of equations by comparing coefficients and solving for the unknown constants. This meticulous process ensures that the original expression is accurately represented as a sum of simpler fractions. The resulting fractions often correspond to basic functions like exponentials, sines, cosines, and polynomials, making the inverse Laplace transform straightforward.

Reassembling Y(s) and Applying the Inverse Laplace Transform

Now, we substitute the partial fraction decompositions back into the expression for Y(s)Y(s):

Y(s)=(−2s+2s3+2ss2+1)−(6s2−6s2+1)+(7s−7ss2+1)−1s2+1Y(s) = \left(-\frac{2}{s} + \frac{2}{s^3} + \frac{2s}{s^2 + 1}\right) - \left(\frac{6}{s^2} - \frac{6}{s^2 + 1}\right) + \left(\frac{7}{s} - \frac{7s}{s^2 + 1}\right) - \frac{1}{s^2 + 1}

Combine like terms:

Y(s)=(−2s+7s)−6s2+2s3+(2ss2+1−7ss2+1)+(6s2+1−1s2+1)Y(s) = \left(-\frac{2}{s} + \frac{7}{s}\right) - \frac{6}{s^2} + \frac{2}{s^3} + \left(\frac{2s}{s^2 + 1} - \frac{7s}{s^2 + 1}\right) + \left(\frac{6}{s^2 + 1} - \frac{1}{s^2 + 1}\right)

Simplify:

Y(s)=5s−6s2+2s3−5ss2+1+5s2+1Y(s) = \frac{5}{s} - \frac{6}{s^2} + \frac{2}{s^3} - \frac{5s}{s^2 + 1} + \frac{5}{s^2 + 1}

Now, we apply the inverse Laplace transform to each term. Recall the following inverse Laplace transforms:

  • L−1{1s}=1\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1
  • L−1{1s2}=t\mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = t
  • L−1{1s3}=t22\mathcal{L}^{-1}\left\{\frac{1}{s^3}\right\} = \frac{t^2}{2}
  • L−1{ss2+1}=cos⁡(t)\mathcal{L}^{-1}\left\{\frac{s}{s^2 + 1}\right\} = \cos(t)
  • L−1{1s2+1}=sin⁡(t)\mathcal{L}^{-1}\left\{\frac{1}{s^2 + 1}\right\} = \sin(t)

Applying the inverse Laplace transform to Y(s)Y(s), we get:

y(t)=L−1{Y(s)}=L−1{5s−6s2+2s3−5ss2+1+5s2+1}y(t) = \mathcal{L}^{-1}\left\{Y(s)\right\} = \mathcal{L}^{-1}\left\{\frac{5}{s} - \frac{6}{s^2} + \frac{2}{s^3} - \frac{5s}{s^2 + 1} + \frac{5}{s^2 + 1}\right\}

y(t)=5L−1{1s}−6L−1{1s2}+2L−1{1s3}−5L−1{ss2+1}+5L−1{1s2+1}y(t) = 5\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} - 6\mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} + 2\mathcal{L}^{-1}\left\{\frac{1}{s^3}\right\} - 5\mathcal{L}^{-1}\left\{\frac{s}{s^2 + 1}\right\} + 5\mathcal{L}^{-1}\left\{\frac{1}{s^2 + 1}\right\}

Substitute the inverse Laplace transforms:

y(t)=5(1)−6(t)+2(t22)−5(cos⁡(t))+5(sin⁡(t))y(t) = 5(1) - 6(t) + 2\left(\frac{t^2}{2}\right) - 5(\cos(t)) + 5(\sin(t))

Simplify:

y(t)=5−6t+t2−5cos⁡(t)+5sin⁡(t)y(t) = 5 - 6t + t^2 - 5\cos(t) + 5\sin(t)

The inverse Laplace transform is the final step in solving the initial value problem using the Laplace transform method. It involves converting the solution from the frequency domain back to the time domain. This is achieved by applying the inverse Laplace transform to each term in Y(s)Y(s). The result is the solution y(t)y(t) that satisfies both the differential equation and the initial conditions. The process requires recognizing the patterns in Y(s)Y(s) and matching them to known inverse Laplace transforms. The final solution y(t)y(t) is a function of time that describes the behavior of the system. It is important to verify that this solution satisfies both the original differential equation and the given initial conditions. This verification step ensures that the solution is correct and that no errors were made during the Laplace transform or inverse Laplace transform processes.

Final Solution

The solution to the initial value problem is:

y(t)=t2−6t+5−5cos⁡(t)+5sin⁡(t)y(t) = t^2 - 6t + 5 - 5\cos(t) + 5\sin(t)

This function y(t)y(t) satisfies the given differential equation y′′+y=t2−6t+7y^{\prime \prime}+y=t^2-6 t+7 and the initial conditions y(0)=0y(0)=0 and y′(0)=−1y^{\prime}(0)=-1. To verify this, we can compute the first and second derivatives of y(t)y(t):

y′(t)=2t−6+5sin⁡(t)+5cos⁡(t)y^{\prime}(t) = 2t - 6 + 5\sin(t) + 5\cos(t)

y′′(t)=2+5cos⁡(t)−5sin⁡(t)y^{\prime \prime}(t) = 2 + 5\cos(t) - 5\sin(t)

Now, substitute y′′(t)y^{\prime \prime}(t) and y(t)y(t) into the differential equation:

y′′(t)+y(t)=(2+5cos⁡(t)−5sin⁡(t))+(t2−6t+5−5cos⁡(t)+5sin⁡(t))y^{\prime \prime}(t) + y(t) = (2 + 5\cos(t) - 5\sin(t)) + (t^2 - 6t + 5 - 5\cos(t) + 5\sin(t))

Simplify:

y′′(t)+y(t)=t2−6t+7y^{\prime \prime}(t) + y(t) = t^2 - 6t + 7

This confirms that the solution satisfies the differential equation. Next, we check the initial conditions:

y(0)=(0)2−6(0)+5−5cos⁡(0)+5sin⁡(0)=5−5(1)+5(0)=0y(0) = (0)^2 - 6(0) + 5 - 5\cos(0) + 5\sin(0) = 5 - 5(1) + 5(0) = 0

y′(0)=2(0)−6+5sin⁡(0)+5cos⁡(0)=−6+5(0)+5(1)=−1y^{\prime}(0) = 2(0) - 6 + 5\sin(0) + 5\cos(0) = -6 + 5(0) + 5(1) = -1

Both initial conditions are satisfied. Therefore, the final solution y(t)=t2−6t+5−5cos⁡(t)+5sin⁡(t)y(t) = t^2 - 6t + 5 - 5\cos(t) + 5\sin(t) is the correct solution to the given initial value problem. The solution consists of a polynomial term (t2−6t+5t^2 - 6t + 5) and trigonometric terms (−5cos⁡(t)+5sin⁡(t)-5\cos(t) + 5\sin(t)), indicating a combination of forced and natural responses of the system. The polynomial term represents the forced response due to the external forcing function t2−6t+7t^2 - 6t + 7, while the trigonometric terms represent the natural response of the system due to its inherent oscillatory nature. The initial conditions determine the specific amplitudes and phases of these components, resulting in a unique solution that satisfies both the differential equation and the initial conditions. This complete solution provides a comprehensive understanding of the system's behavior over time.

Conclusion

In this article, we successfully solved the initial value problem y′′+y=t2−6t+7y^{\prime \prime}+y=t^2-6 t+7, with y(0)=0y(0)=0 and y′(0)=−1y^{\prime}(0)=-1, using the Laplace transform method. This method transformed the differential equation into an algebraic equation in the s-domain, which was then solved for Y(s)Y(s). Partial fraction decomposition was employed to simplify the expression for Y(s)Y(s), and the inverse Laplace transform was applied to obtain the solution in the time domain. The final solution is y(t)=t2−6t+5−5cos⁡(t)+5sin⁡(t)y(t) = t^2 - 6t + 5 - 5\cos(t) + 5\sin(t). The Laplace transform method offers a systematic and efficient approach to solving linear ordinary differential equations with constant coefficients, especially those with initial conditions. It is a valuable tool in various fields, including engineering and physics, where such equations frequently arise in modeling physical systems. The method's ability to transform differential equations into algebraic equations simplifies the solution process, making it easier to handle complex problems. Moreover, the Laplace transform provides insights into the system's behavior in the frequency domain, which can be useful for analysis and design purposes. The application of partial fraction decomposition is a crucial step in the Laplace transform method, as it allows us to break down complex rational functions into simpler terms that can be easily inverted. The inverse Laplace transform then converts the solution back to the time domain, providing the desired function y(t)y(t). The Laplace transform is a powerful and versatile tool for solving a wide range of problems in mathematics, science, and engineering. Its applications extend beyond differential equations to areas such as signal processing, control systems, and circuit analysis. The method's elegance and efficiency make it an indispensable part of the mathematical toolkit for students and professionals alike.