Solving Initial Value Problems Using Laplace Transforms A Detailed Guide

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Introduction

In the realm of differential equations, the quest to find solutions that satisfy specific initial conditions is a common challenge. One powerful technique for tackling such initial value problems is the Laplace transform method. This approach elegantly converts differential equations into algebraic equations, which are often easier to solve. Once the solution is obtained in the Laplace domain, an inverse transform is applied to bring the solution back to the original time domain. In this article, we delve into the application of Laplace transforms to solve the following initial value problem:

wβ€²β€²βˆ’12wβ€²+36w=180t+696,w(βˆ’4)=βˆ’5,wβ€²(βˆ’4)=βˆ’25w^{\prime \prime} - 12w^{\prime} + 36w = 180t + 696, \quad w(-4) = -5, \quad w^{\prime}(-4) = -25

We will systematically walk through the steps involved, providing a detailed explanation of each stage. The Laplace transform method is particularly advantageous for linear differential equations with constant coefficients, as it streamlines the solution process and offers a structured way to handle initial conditions. This method is widely used in engineering and physics to analyze systems that evolve over time, such as electrical circuits, mechanical systems, and control systems.

Understanding Laplace transforms and their application to solving differential equations is a fundamental skill for engineers and scientists. This article aims to provide a comprehensive guide to this technique, enabling readers to confidently tackle a wide range of initial value problems. We will leverage the properties of Laplace transforms and their inverses, along with a table of common transforms, to efficiently solve the given problem. By the end of this article, you will have a solid understanding of how to apply Laplace transforms to solve second-order linear differential equations with initial conditions.

Laplace Transforms

The Laplace transform is a mathematical tool that converts a function of time, denoted as f(t)f(t), into a function of a complex variable ss, denoted as F(s)F(s). This transformation is defined by the integral:

F(s)=Lf(t)=∫0∞eβˆ’stf(t)dtF(s) = \mathcal{L}{f(t)} = \int_{0}^{\infty} e^{-st} f(t) dt

where ss is a complex number. The Laplace transform is particularly useful for solving linear differential equations with constant coefficients because it transforms differential equations into algebraic equations, which are generally easier to solve. The key to effectively using Laplace transforms lies in understanding their properties and having a table of common transforms readily available.

One of the most important properties of Laplace transforms is their linearity. This means that the Laplace transform of a linear combination of functions is the linear combination of their individual Laplace transforms: Laf(t)+bg(t)=aLf(t)+bLg(t)\mathcal{L}{af(t) + bg(t)} = a\mathcal{L}{f(t)} + b\mathcal{L}{g(t)}, where aa and bb are constants. Another crucial property is the transform of derivatives. The Laplace transform of the first derivative fβ€²(t)f'(t) is given by sF(s)βˆ’f(0)sF(s) - f(0), and the Laplace transform of the second derivative fβ€²β€²(t)f''(t) is given by s2F(s)βˆ’sf(0)βˆ’fβ€²(0)s^2F(s) - sf(0) - f'(0). These properties are instrumental in converting differential equations into algebraic equations in the Laplace domain.

The Laplace transform of some common functions includes:

  • L1=1s\mathcal{L}{1} = \frac{1}{s}
  • Lt=1s2\mathcal{L}{t} = \frac{1}{s^2}
  • Leat=1sβˆ’a\mathcal{L}{e^{at}} = \frac{1}{s-a}
  • Lsin⁑(at)=as2+a2\mathcal{L}{\sin(at)} = \frac{a}{s^2 + a^2}
  • Lcos⁑(at)=ss2+a2\mathcal{L}{\cos(at)} = \frac{s}{s^2 + a^2}

These basic transforms, along with the properties of Laplace transforms, allow us to handle a wide variety of functions and differential equations. The process of solving an initial value problem using Laplace transforms involves three main steps: taking the Laplace transform of the differential equation, solving the resulting algebraic equation for the Laplace transform of the solution, and then applying the inverse Laplace transform to obtain the solution in the time domain. The inverse Laplace transform, denoted as Lβˆ’1\mathcal{L}^{-1}, undoes the Laplace transform and brings us back to the original function of time.

Step 1: Apply the Laplace Transform

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

Lwβ€²β€²βˆ’12wβ€²+36w=L180t+696\mathcal{L}{w^{\prime \prime} - 12w^{\prime} + 36w} = \mathcal{L}{180t + 696}

Using the linearity property of the Laplace transform, we can rewrite this as:

Lwβ€²β€²βˆ’12Lwβ€²+36Lw=180Lt+696L1\mathcal{L}{w^{\prime \prime}} - 12\mathcal{L}{w^{\prime}} + 36\mathcal{L}{w} = 180\mathcal{L}{t} + 696\mathcal{L}{1}

Now, we use the Laplace transforms of derivatives and the given initial conditions. Let W(s)=Lw(t)W(s) = \mathcal{L}{w(t)}. We have:

  • Lwβ€²(t)=sW(s)βˆ’w(0)\mathcal{L}{w^{\prime}(t)} = sW(s) - w(0)
  • Lwβ€²β€²(t)=s2W(s)βˆ’sw(0)βˆ’wβ€²(0)\mathcal{L}{w^{\prime \prime}(t)} = s^2W(s) - sw(0) - w^{\prime}(0)

However, our initial conditions are given at t=βˆ’4t = -4, not t=0t = 0. To handle this, we introduce a change of variable: let Ο„=t+4\tau = t + 4. Then t=Ο„βˆ’4t = \tau - 4, and the differential equation becomes:

d2wdΟ„2βˆ’12dwdΟ„+36w=180(Ο„βˆ’4)+696\frac{d^2w}{d\tau^2} - 12\frac{dw}{d\tau} + 36w = 180(\tau - 4) + 696

Simplifying the right-hand side, we get:

d2wdΟ„2βˆ’12dwdΟ„+36w=180Ο„βˆ’720+696=180Ο„βˆ’24\frac{d^2w}{d\tau^2} - 12\frac{dw}{d\tau} + 36w = 180\tau - 720 + 696 = 180\tau - 24

Now, the initial conditions are w(0)=βˆ’5w(0) = -5 and wβ€²(0)=βˆ’25w^{\prime}(0) = -25. Applying the Laplace transform to the transformed equation, we get:

Ld2wdΟ„2βˆ’12LdwdΟ„+36Lw=180LΟ„βˆ’24L1\mathcal{L}{\frac{d^2w}{d\tau^2}} - 12\mathcal{L}{\frac{dw}{d\tau}} + 36\mathcal{L}{w} = 180\mathcal{L}{\tau} - 24\mathcal{L}{1}

Substituting the Laplace transforms of the derivatives and using W(s)=Lw(Ο„)W(s) = \mathcal{L}{w(\tau)}, we have:

[s2W(s)βˆ’sw(0)βˆ’wβ€²(0)]βˆ’12[sW(s)βˆ’w(0)]+36W(s)=180(1s2)βˆ’24(1s)[s^2W(s) - sw(0) - w^{\prime}(0)] - 12[sW(s) - w(0)] + 36W(s) = 180\left(\frac{1}{s^2}\right) - 24\left(\frac{1}{s}\right)

Plugging in the initial conditions w(0)=βˆ’5w(0) = -5 and wβ€²(0)=βˆ’25w^{\prime}(0) = -25, we get:

s2W(s)βˆ’s(βˆ’5)βˆ’(βˆ’25)βˆ’12[sW(s)βˆ’(βˆ’5)]+36W(s)=180s2βˆ’24ss^2W(s) - s(-5) - (-25) - 12[sW(s) - (-5)] + 36W(s) = \frac{180}{s^2} - \frac{24}{s}

Simplifying, we obtain:

s2W(s)+5s+25βˆ’12sW(s)βˆ’60+36W(s)=180s2βˆ’24ss^2W(s) + 5s + 25 - 12sW(s) - 60 + 36W(s) = \frac{180}{s^2} - \frac{24}{s}

(s2βˆ’12s+36)W(s)=180s2βˆ’24sβˆ’5s+35(s^2 - 12s + 36)W(s) = \frac{180}{s^2} - \frac{24}{s} - 5s + 35

Step 2: Solve for W(s)

Now, we solve for W(s)W(s). Notice that s2βˆ’12s+36=(sβˆ’6)2s^2 - 12s + 36 = (s - 6)^2, so we have:

(sβˆ’6)2W(s)=180s2βˆ’24sβˆ’5s+35(s - 6)^2W(s) = \frac{180}{s^2} - \frac{24}{s} - 5s + 35

Divide both sides by (sβˆ’6)2(s - 6)^2 to isolate W(s)W(s):

W(s)=1(sβˆ’6)2(180s2βˆ’24sβˆ’5s+35)W(s) = \frac{1}{(s - 6)^2}\left(\frac{180}{s^2} - \frac{24}{s} - 5s + 35\right)

Distribute and combine terms:

W(s)=180s2(sβˆ’6)2βˆ’24s(sβˆ’6)2βˆ’5s(sβˆ’6)2+35(sβˆ’6)2W(s) = \frac{180}{s^2(s - 6)^2} - \frac{24}{s(s - 6)^2} - \frac{5s}{(s - 6)^2} + \frac{35}{(s - 6)^2}

Step 3: Partial Fraction Decomposition

To find the inverse Laplace transform, we need to perform partial fraction decomposition on each term. Let's decompose the first term, 180s2(sβˆ’6)2\frac{180}{s^2(s - 6)^2}:

180s2(sβˆ’6)2=As+Bs2+Csβˆ’6+D(sβˆ’6)2\frac{180}{s^2(s - 6)^2} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s - 6} + \frac{D}{(s - 6)^2}

Multiplying both sides by s2(sβˆ’6)2s^2(s - 6)^2, we get:

180=As(sβˆ’6)2+B(sβˆ’6)2+Cs2(sβˆ’6)+Ds2180 = As(s - 6)^2 + B(s - 6)^2 + Cs^2(s - 6) + Ds^2

Let s=0s = 0: 180=36Bβ€…β€ŠβŸΉβ€…β€ŠB=5180 = 36B \implies B = 5

Let s=6s = 6: 180=36Dβ€…β€ŠβŸΉβ€…β€ŠD=5180 = 36D \implies D = 5

Expanding the equation, we have:

180=A(s3βˆ’12s2+36s)+B(s2βˆ’12s+36)+C(s3βˆ’6s2)+Ds2180 = A(s^3 - 12s^2 + 36s) + B(s^2 - 12s + 36) + C(s^3 - 6s^2) + Ds^2

180=As3βˆ’12As2+36As+Bs2βˆ’12Bs+36B+Cs3βˆ’6Cs2+Ds2180 = As^3 - 12As^2 + 36As + Bs^2 - 12Bs + 36B + Cs^3 - 6Cs^2 + Ds^2

Comparing coefficients:

  • s3:0=A+Cs^3: 0 = A + C
  • s2:0=βˆ’12A+Bβˆ’6C+Ds^2: 0 = -12A + B - 6C + D
  • s:0=36Aβˆ’12Bs: 0 = 36A - 12B
  • Constant: 180=36B180 = 36B

From 0=36Aβˆ’12B0 = 36A - 12B, we have 36A=12B=12(5)β€…β€ŠβŸΉβ€…β€ŠA=5336A = 12B = 12(5) \implies A = \frac{5}{3}. Since 0=A+C0 = A + C, C=βˆ’53C = -\frac{5}{3}.

So, 180s2(sβˆ’6)2=53s+5s2βˆ’53(sβˆ’6)+5(sβˆ’6)2\frac{180}{s^2(s - 6)^2} = \frac{5}{3s} + \frac{5}{s^2} - \frac{5}{3(s - 6)} + \frac{5}{(s - 6)^2}.

Next, decompose βˆ’24s(sβˆ’6)2\frac{-24}{s(s - 6)^2}:

βˆ’24s(sβˆ’6)2=Es+Fsβˆ’6+G(sβˆ’6)2\frac{-24}{s(s - 6)^2} = \frac{E}{s} + \frac{F}{s - 6} + \frac{G}{(s - 6)^2}

βˆ’24=E(sβˆ’6)2+Fs(sβˆ’6)+Gs-24 = E(s - 6)^2 + Fs(s - 6) + Gs

Let s=0s = 0: βˆ’24=36Eβ€…β€ŠβŸΉβ€…β€ŠE=βˆ’23-24 = 36E \implies E = -\frac{2}{3}

Let s=6s = 6: βˆ’24=6Gβ€…β€ŠβŸΉβ€…β€ŠG=βˆ’4-24 = 6G \implies G = -4

Expanding and comparing coefficients:

βˆ’24=E(s2βˆ’12s+36)+F(s2βˆ’6s)+Gs-24 = E(s^2 - 12s + 36) + F(s^2 - 6s) + Gs

βˆ’24=Es2βˆ’12Es+36E+Fs2βˆ’6Fs+Gs-24 = Es^2 - 12Es + 36E + Fs^2 - 6Fs + Gs

  • s2:0=E+Fβ€…β€ŠβŸΉβ€…β€ŠF=βˆ’E=23s^2: 0 = E + F \implies F = -E = \frac{2}{3}

So, βˆ’24s(sβˆ’6)2=βˆ’23s+23(sβˆ’6)βˆ’4(sβˆ’6)2\frac{-24}{s(s - 6)^2} = -\frac{2}{3s} + \frac{2}{3(s - 6)} - \frac{4}{(s - 6)^2}.

Next, decompose βˆ’5s(sβˆ’6)2\frac{-5s}{(s - 6)^2}:

βˆ’5s(sβˆ’6)2=Hsβˆ’6+I(sβˆ’6)2\frac{-5s}{(s - 6)^2} = \frac{H}{s - 6} + \frac{I}{(s - 6)^2}

βˆ’5s=H(sβˆ’6)+I-5s = H(s - 6) + I

Let s=6s = 6: βˆ’30=I-30 = I

Comparing coefficients of ss: βˆ’5=H-5 = H

So, βˆ’5s(sβˆ’6)2=βˆ’5sβˆ’6βˆ’30(sβˆ’6)2\frac{-5s}{(s - 6)^2} = \frac{-5}{s - 6} - \frac{30}{(s - 6)^2}.

The last term is 35(sβˆ’6)2\frac{35}{(s - 6)^2}.

Step 4: Combine Terms and Apply Inverse Laplace Transform

Combining all the partial fractions, we have:

W(s)=(53s+5s2βˆ’53(sβˆ’6)+5(sβˆ’6)2)+(βˆ’23s+23(sβˆ’6)βˆ’4(sβˆ’6)2)+(βˆ’5sβˆ’6βˆ’30(sβˆ’6)2)+35(sβˆ’6)2W(s) = \left(\frac{5}{3s} + \frac{5}{s^2} - \frac{5}{3(s - 6)} + \frac{5}{(s - 6)^2}\right) + \left(-\frac{2}{3s} + \frac{2}{3(s - 6)} - \frac{4}{(s - 6)^2}\right) + \left(\frac{-5}{s - 6} - \frac{30}{(s - 6)^2}\right) + \frac{35}{(s - 6)^2}

W(s)=1s+5s2+53(sβˆ’6)βˆ’29(sβˆ’6)2W(s) = \frac{1}{s} + \frac{5}{s^2} + \frac{5}{3(s - 6)} - \frac{29}{(s - 6)^2}

Now, we apply the inverse Laplace transform:

w(Ο„)=Lβˆ’1W(s)=Lβˆ’1(1s+5s2+53(sβˆ’6)βˆ’29(sβˆ’6)2)w(\tau) = \mathcal{L}^{-1}{W(s)} = \mathcal{L}^{-1}{\left(\frac{1}{s} + \frac{5}{s^2} + \frac{5}{3(s - 6)} - \frac{29}{(s - 6)^2}\right)}

Using the inverse Laplace transform table:

w(Ο„)=1+5Ο„+53e6Ο„βˆ’29Ο„e6Ο„w(\tau) = 1 + 5\tau + \frac{5}{3}e^{6\tau} - 29\tau e^{6\tau}

Finally, substitute back Ο„=t+4\tau = t + 4:

w(t)=1+5(t+4)+53e6(t+4)βˆ’29(t+4)e6(t+4)w(t) = 1 + 5(t + 4) + \frac{5}{3}e^{6(t + 4)} - 29(t + 4)e^{6(t + 4)}

w(t)=1+5t+20+53e6t+24βˆ’29te6t+24βˆ’116e6t+24w(t) = 1 + 5t + 20 + \frac{5}{3}e^{6t + 24} - 29te^{6t + 24} - 116e^{6t + 24}

w(t)=5t+21+(53βˆ’116)e6t+24βˆ’29te6t+24w(t) = 5t + 21 + \left(\frac{5}{3} - 116\right)e^{6t + 24} - 29te^{6t + 24}

w(t)=5t+21βˆ’3433e6(t+4)βˆ’29(t+4)e6(t+4)w(t) = 5t + 21 - \frac{343}{3}e^{6(t + 4)} - 29(t + 4)e^{6(t + 4)}

Conclusion

In this article, we have demonstrated how to solve an initial value problem using the Laplace transform method. We started by applying the Laplace transform to the differential equation, incorporating the initial conditions given at t=βˆ’4t = -4 by using a change of variable. We then solved for W(s)W(s) in the Laplace domain, performed partial fraction decomposition to simplify the expression, and finally applied the inverse Laplace transform to obtain the solution in the time domain. This method provides a systematic approach to solving linear differential equations with constant coefficients, particularly when dealing with non-zero initial conditions. The Laplace transform is a powerful tool in engineering and mathematics, and mastering its application is essential for solving a wide range of problems. By understanding the steps involved and practicing with various examples, one can effectively utilize this technique to tackle complex differential equations and initial value problems.