Solving Initial Value Problems Using Laplace Transforms

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In this article, we will delve into the application of Laplace transforms to solve initial value problems, focusing on the differential equation:

w′′−12w′+36w=180t+696w^{\prime \prime} - 12w^{\prime} + 36w = 180t + 696, with initial conditions w(−4)=−5w(-4) = -5 and w′(−4)=−25w^{\prime}(-4) = -25.

Laplace transforms provide a powerful tool for solving linear differential equations, particularly those with constant coefficients. They convert differential equations into algebraic equations, which are generally easier to solve. Once the solution is obtained in the Laplace domain, the inverse Laplace transform is applied to return to the original time domain. This method is especially effective for problems with initial conditions, as these conditions are naturally incorporated into the transformation process.

Understanding Laplace Transforms

The Laplace transform of a function f(t)f(t), denoted by F(s)F(s) or Lf(t)\mathcal{L}{f(t)}, is defined as:

F(s)=Lf(t)=∫0∞e−stf(t)dtF(s) = \mathcal{L}{f(t)} = \int_{0}^{\infty} e^{-st} f(t) dt

where ss is a complex variable. The Laplace transform essentially converts a function of time, f(t)f(t), into a function of complex frequency, F(s)F(s). This transformation is particularly useful for solving differential equations because it turns differentiation into multiplication and integration into division in the ss-domain.

The inverse Laplace transform, denoted by f(t)=L−1F(s)f(t) = \mathcal{L}^{-1}{F(s)}, reverses this process, converting a function in the ss-domain back to a function in the time domain. While the integral definition for the inverse Laplace transform exists, in practice, we often use tables of known Laplace transforms and partial fraction decomposition to find inverse transforms.

Key Properties and Transforms

To effectively use Laplace transforms, it's crucial to understand some key properties and transforms. Here are a few essential ones:

  • Linearity: The Laplace transform is a linear operator, meaning that for constants aa and bb, and functions f(t)f(t) and g(t)g(t): Laf(t)+bg(t)=aLf(t)+bLg(t)=aF(s)+bG(s)\mathcal{L}{af(t) + bg(t)} = a\mathcal{L}{f(t)} + b\mathcal{L}{g(t)} = aF(s) + bG(s)
  • Transform of Derivatives: This property is particularly useful for solving differential equations. The Laplace transform of the first and second derivatives are:
    • Lf′(t)=sF(s)−f(0)\mathcal{L}{f^{\prime}(t)} = sF(s) - f(0)
    • Lf′′(t)=s2F(s)−sf(0)−f′(0)\mathcal{L}{f^{\prime \prime}(t)} = s^2F(s) - sf(0) - f^{\prime}(0)
  • Time Shifting (Translation): If Lf(t)=F(s)\mathcal{L}{f(t)} = F(s), then: Lf(t−a)u(t−a)=e−asF(s)\mathcal{L}{f(t-a)u(t-a)} = e^{-as}F(s) where u(t−a)u(t-a) is the Heaviside step function.
  • Transforms of Common Functions:
    • L1=1s\mathcal{L}{1} = \frac{1}{s}
    • Lt=1s2\mathcal{L}{t} = \frac{1}{s^2}
    • Ltn=n!sn+1\mathcal{L}{t^n} = \frac{n!}{s^{n+1}}
    • Leat=1s−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 properties and transforms are the building blocks for solving differential equations using Laplace transforms. By applying these rules, we can convert a differential equation into an algebraic equation in the ss-domain, solve for the transformed solution, and then use the inverse Laplace transform to find the solution in the time domain.

Step-by-Step Solution Using Laplace Transforms

Now, let's apply the Laplace transform method to solve the given initial value problem:

w′′−12w′+36w=180t+696w^{\prime \prime} - 12w^{\prime} + 36w = 180t + 696, with w(−4)=−5w(-4) = -5 and w′(−4)=−25w^{\prime}(-4) = -25.

1. Time Shift the Problem

Since the initial conditions are given at t=−4t = -4, it's convenient to shift the time variable to t=0t = 0. Let τ=t+4\tau = t + 4, so t=τ−4t = \tau - 4. Define a new function v(τ)=w(t)=w(τ−4)v(\tau) = w(t) = w(\tau - 4). Then,

  • w′(t)=v′(τ)w^{\prime}(t) = v^{\prime}(\tau)
  • w′′(t)=v′′(τ)w^{\prime \prime}(t) = v^{\prime \prime}(\tau)

The differential equation becomes:

v′′(τ)−12v′(τ)+36v(τ)=180(τ−4)+696v^{\prime \prime}(\tau) - 12v^{\prime}(\tau) + 36v(\tau) = 180(\tau - 4) + 696

Simplifying the right-hand side:

180(τ−4)+696=180τ−720+696=180τ−24180(\tau - 4) + 696 = 180\tau - 720 + 696 = 180\tau - 24

So the shifted differential equation is:

v′′(τ)−12v′(τ)+36v(τ)=180τ−24v^{\prime \prime}(\tau) - 12v^{\prime}(\tau) + 36v(\tau) = 180\tau - 24

The initial conditions transform to:

  • v(0)=w(−4)=−5v(0) = w(-4) = -5
  • v′(0)=w′(−4)=−25v^{\prime}(0) = w^{\prime}(-4) = -25

2. Apply the Laplace Transform

Now, we apply the Laplace transform to both sides of the shifted differential equation. Let V(s)=Lv(τ)V(s) = \mathcal{L}{v(\tau)}. Using the properties of Laplace transforms, we have:

Lv′′(τ)=s2V(s)−sv(0)−v′(0)=s2V(s)+5s+25\mathcal{L}{v^{\prime \prime}(\tau)} = s^2V(s) - sv(0) - v^{\prime}(0) = s^2V(s) + 5s + 25

Lv′(τ)=sV(s)−v(0)=sV(s)+5\mathcal{L}{v^{\prime}(\tau)} = sV(s) - v(0) = sV(s) + 5

Lv(τ)=V(s)\mathcal{L}{v(\tau)} = V(s)

L180τ−24=180Lτ−24L1=1801s2−241s\mathcal{L}{180\tau - 24} = 180\mathcal{L}{\tau} - 24\mathcal{L}{1} = 180\frac{1}{s^2} - 24\frac{1}{s}

Applying the Laplace transform to the entire equation:

(s2V(s)+5s+25)−12(sV(s)+5)+36V(s)=180s2−24s(s^2V(s) + 5s + 25) - 12(sV(s) + 5) + 36V(s) = \frac{180}{s^2} - \frac{24}{s}

3. Solve for V(s)

Collect terms involving V(s)V(s):

(s2−12s+36)V(s)+5s+25−60=180s2−24s(s^2 - 12s + 36)V(s) + 5s + 25 - 60 = \frac{180}{s^2} - \frac{24}{s}

(s2−12s+36)V(s)=180s2−24s−5s+35(s^2 - 12s + 36)V(s) = \frac{180}{s^2} - \frac{24}{s} - 5s + 35

Notice that s2−12s+36=(s−6)2s^2 - 12s + 36 = (s - 6)^2, so

(s−6)2V(s)=180s2−24s−5s+35(s - 6)^2V(s) = \frac{180}{s^2} - \frac{24}{s} - 5s + 35

Now, solve for V(s)V(s):

V(s)=1(s−6)2(180s2−24s−5s+35)V(s) = \frac{1}{(s - 6)^2}\left(\frac{180}{s^2} - \frac{24}{s} - 5s + 35\right)

V(s)=180s2(s−6)2−24s(s−6)2−5s(s−6)2+35(s−6)2V(s) = \frac{180}{s^2(s - 6)^2} - \frac{24}{s(s - 6)^2} - \frac{5s}{(s - 6)^2} + \frac{35}{(s - 6)^2}

4. Partial Fraction Decomposition

To find the inverse Laplace transform, we need to perform partial fraction decomposition on each term. Let's decompose each term separately:

Term 1: 180s2(s−6)2\frac{180}{s^2(s - 6)^2}

180s2(s−6)2=As+Bs2+Cs−6+D(s−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 through by s2(s−6)2s^2(s - 6)^2:

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

  • Set s=0s = 0: 180=36B  ⟹  B=5180 = 36B \implies B = 5
  • Set s=6s = 6: 180=36D  ⟹  D=5180 = 36D \implies D = 5
  • Expand and collect terms: 180=As(s2−12s+36)+5(s2−12s+36)+Cs2(s−6)+5s2180 = As(s^2 - 12s + 36) + 5(s^2 - 12s + 36) + Cs^2(s - 6) + 5s^2 180=As3−12As2+36As+5s2−60s+180+Cs3−6Cs2+5s2180 = As^3 - 12As^2 + 36As + 5s^2 - 60s + 180 + Cs^3 - 6Cs^2 + 5s^2
  • Equate coefficients:
    • s3:0=A+Cs^3: 0 = A + C
    • s2:0=−12A+5−6C+5  ⟹  0=−12A−6C+10s^2: 0 = -12A + 5 - 6C + 5 \implies 0 = -12A - 6C + 10
    • s:0=36A−60s: 0 = 36A - 60

From 36A=6036A = 60, we get A=53A = \frac{5}{3}. Then, C=−53C = -\frac{5}{3}.

So, 180s2(s−6)2=5/3s+5s2−5/3s−6+5(s−6)2\frac{180}{s^2(s - 6)^2} = \frac{5/3}{s} + \frac{5}{s^2} - \frac{5/3}{s - 6} + \frac{5}{(s - 6)^2}

Term 2: −24s(s−6)2\frac{ -24}{s(s - 6)^2}

−24s(s−6)2=Es+Fs−6+G(s−6)2\frac{-24}{s(s - 6)^2} = \frac{E}{s} + \frac{F}{s - 6} + \frac{G}{(s - 6)^2}

Multiplying through by s(s−6)2s(s - 6)^2:

−24=E(s−6)2+Fs(s−6)+Gs-24 = E(s - 6)^2 + Fs(s - 6) + Gs

  • Set s=0s = 0: −24=36E  ⟹  E=−23-24 = 36E \implies E = -\frac{2}{3}
  • Set s=6s = 6: −24=6G  ⟹  G=−4-24 = 6G \implies G = -4
  • Expand and equate coefficients: −24=E(s2−12s+36)+Fs2−6Fs+Gs-24 = E(s^2 - 12s + 36) + Fs^2 - 6Fs + Gs −24=−23(s2−12s+36)+Fs2−6Fs−4s-24 = -\frac{2}{3}(s^2 - 12s + 36) + Fs^2 - 6Fs - 4s
  • s2:0=−23+F  ⟹  F=23s^2: 0 = -\frac{2}{3} + F \implies F = \frac{2}{3}

So, −24s(s−6)2=−2/3s+2/3s−6−4(s−6)2\frac{-24}{s(s - 6)^2} = -\frac{2/3}{s} + \frac{2/3}{s - 6} - \frac{4}{(s - 6)^2}

Term 3: −5s(s−6)2\frac{-5s}{(s - 6)^2}

−5s(s−6)2=Hs−6+I(s−6)2\frac{-5s}{(s - 6)^2} = \frac{H}{s - 6} + \frac{I}{(s - 6)^2}

−5s=H(s−6)+I-5s = H(s - 6) + I

  • Set s=6s = 6: −30=I-30 = I
  • Set s=0s = 0: 0=−6H−30  ⟹  H=−50 = -6H - 30 \implies H = -5

So, −5s(s−6)2=−5s−6−30(s−6)2\frac{-5s}{(s - 6)^2} = -\frac{5}{s - 6} - \frac{30}{(s - 6)^2}

Term 4: 35(s−6)2\frac{35}{(s - 6)^2}

This term is already in a suitable form.

5. Combine Partial Fractions

Now, let's combine all the partial fractions:

V(s)=(5/3s+5s2−5/3s−6+5(s−6)2)+(−2/3s+2/3s−6−4(s−6)2)+(−5s−6−30(s−6)2)+35(s−6)2V(s) = \left(\frac{5/3}{s} + \frac{5}{s^2} - \frac{5/3}{s - 6} + \frac{5}{(s - 6)^2}\right) + \left(-\frac{2/3}{s} + \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}

Combine like terms:

V(s)=(5/3−2/3s)+5s2+(−53+23−5)1s−6+(5−4−30+35)1(s−6)2V(s) = \left(\frac{5/3 - 2/3}{s}\right) + \frac{5}{s^2} + \left(-\frac{5}{3} + \frac{2}{3} - 5\right)\frac{1}{s - 6} + \left(5 - 4 - 30 + 35\right)\frac{1}{(s - 6)^2}

V(s)=1s+5s2−18−531s−6+61(s−6)2V(s) = \frac{1}{s} + \frac{5}{s^2} - \frac{18 -5}{3}\frac{1}{s - 6} + 6\frac{1}{(s - 6)^2}

V(s)=1s+5s2−61s−6+61(s−6)2V(s) = \frac{1}{s} + \frac{5}{s^2} - 6\frac{1}{s - 6} + 6\frac{1}{(s - 6)^2}

6. Inverse Laplace Transform

Apply the inverse Laplace transform to each term:

v(τ)=L−1V(s)=L−1(1s)+5L−1(1s2)−6L−1(1s−6)+6L−1(1(s−6)2)v(\tau) = \mathcal{L}^{-1}{V(s)} = \mathcal{L}^{-1}{\left(\frac{1}{s}\right)} + 5\mathcal{L}^{-1}{\left(\frac{1}{s^2}\right)} - 6\mathcal{L}^{-1}{\left(\frac{1}{s - 6}\right)} + 6\mathcal{L}^{-1}{\left(\frac{1}{(s - 6)^2}\right)}

Using the standard Laplace transforms:

v(τ)=1+5τ−6e6τ+6τe6τv(\tau) = 1 + 5\tau - 6e^{6\tau} + 6\tau e^{6\tau}

7. Shift Back to Original Time Variable

Finally, substitute back t=τ−4t = \tau - 4 to get the solution in terms of tt:

w(t)=v(t+4)=1+5(t+4)−6e6(t+4)+6(t+4)e6(t+4)w(t) = v(t + 4) = 1 + 5(t + 4) - 6e^{6(t + 4)} + 6(t + 4)e^{6(t + 4)}

Simplifying:

w(t)=1+5t+20−6e6t+24+6te6t+24+24e6t+24w(t) = 1 + 5t + 20 - 6e^{6t + 24} + 6te^{6t + 24} + 24e^{6t + 24}

w(t)=5t+21+(6t+18)e6t+24w(t) = 5t + 21 + (6t + 18)e^{6t + 24}

Conclusion

We have successfully solved the initial value problem using Laplace transforms. This method involves transforming the differential equation into an algebraic equation, solving for the transformed solution, and then applying the inverse Laplace transform to obtain the solution in the original time domain. The key steps include shifting the time variable to handle initial conditions at non-zero time, applying the Laplace transform, performing partial fraction decomposition, and finally, applying the inverse Laplace transform. This approach is particularly powerful for linear differential equations with constant coefficients and is widely used in engineering and physics.

By understanding and applying Laplace transforms, engineers and scientists can efficiently solve a wide range of initial value problems. The method's ability to convert differential equations into algebraic ones simplifies the solution process, making it an invaluable tool in various fields. This detailed walkthrough illustrates the step-by-step process, ensuring clarity and comprehension in solving such problems.