Solving $w'' - 6w' + 9w = 45t + 60$ With Laplace Transforms

Introduction

In this article, we will explore how to solve initial value problems (IVPs) using the powerful method of Laplace transforms. Laplace transforms provide a systematic way to convert differential equations into algebraic equations, which are often easier to solve. Once we find the solution in the Laplace domain, we can use inverse Laplace transforms to obtain the solution in the original time domain. This method is particularly useful for linear differential equations with constant coefficients, especially those with discontinuous forcing functions or impulsive forces. The specific initial value problem we will tackle is:

$w^{\prime \prime} - 6w^{\prime} + 9w = 45t + 60$, with initial conditions $w(-1) = 2$ and $w^{\prime}(-1) = 1$.

This second-order linear differential equation with constant coefficients is a classic example where Laplace transforms can simplify the solution process. We will go through each step in detail, from transforming the differential equation to finding the inverse Laplace transform. We'll also discuss the properties of Laplace transforms that make this method effective and highlight common pitfalls to avoid. By the end of this article, you should have a solid understanding of how to apply Laplace transforms to solve similar initial value problems.

Background on Laplace Transforms

The Laplace transform is an integral transform that converts a function of time, $f(t)$, into a function of a complex variable, $s$. It is defined as:

$F(s) = \mathcal{L}{f(t)} = \int_{0}^{\infty} e^{-st} f(t) dt$

where $F(s)$ is the Laplace transform of $f(t)$. The Laplace transform is particularly useful for solving linear differential equations because it transforms differentiation into multiplication. Specifically, the Laplace transform of the first derivative $f'(t)$ is given by:

$\mathcal{L}{f'(t)} = sF(s) - f(0)$

and the Laplace transform of the second derivative $f''(t)$ is:

$\mathcal{L}{f''(t)} = s^2F(s) - sf(0) - f'(0)$

These properties allow us to convert a differential equation into an algebraic equation in the $s$-domain. Solving this algebraic equation gives us the Laplace transform of the solution, which we can then invert to find the solution in the time domain. Inverse Laplace transforms are typically found using tables of Laplace transforms or by employing techniques such as partial fraction decomposition.

Key Properties and Transforms

Several key properties and transforms are frequently used when solving differential equations with Laplace transforms. Some of the most important include:

• Linearity: $\mathcal{L}{af(t) + bg(t)} = a\mathcal{L}{f(t)} + b\mathcal{L}{g(t)}$
• Time Invariance: $\mathcal{L}{f(t-a)u(t-a)} = e^{-as}F(s)$, where $u(t-a)$ is the Heaviside step function.
• Transform of Derivatives: As mentioned earlier, these are crucial for converting differential equations into algebraic equations.
• Transforms of Basic Functions:
  • $\mathcal{L}{1} = \frac{1}{s}$
  • $\mathcal{L}{t} = \frac{1}{s^2}$
  • $\mathcal{L}{e^{at}} = \frac{1}{s-a}$
  • $\mathcal{L}{\sin(at)} = \frac{a}{s^2 + a^2}$
  • $\mathcal{L}{\cos(at)} = \frac{s}{s^2 + a^2}$

Shifting Theorems

The shifting theorems are particularly useful when dealing with initial conditions that are not given at $t=0$. The first shifting theorem states that if $\mathcal{L}{f(t)} = F(s)$, then:

$\mathcal{L}{e^{at}f(t)} = F(s-a)$

The second shifting theorem, also known as the time-shifting theorem, states that:

$\mathcal{L}{f(t-a)u(t-a)} = e^{-as}F(s)$

where $u(t-a)$ is the Heaviside step function. These theorems are essential for handling the initial conditions given at $t=-1$ in our problem.

Step-by-Step Solution Using Laplace Transforms

1. Time Shifting to $t=0$

Our initial conditions are given at $t = -1$, but Laplace transforms are most easily applied when initial conditions are at $t = 0$. Therefore, we introduce a time shift. Let $\tau = t + 1$, so $t = \tau - 1$. Define a new function $v(\tau) = w(\tau - 1) = w(t)$. Then, the derivatives of $v$ with respect to $\tau$ are related to the derivatives of $w$ with respect to $t$ as follows:

• $v'(\tau) = w'(t)$
• $v''(\tau) = w''(t)$

The differential equation becomes:

$v''(\tau) - 6v'(\tau) + 9v(\tau) = 45(\tau - 1) + 60 = 45\tau + 15$

The initial conditions transform to:

• $v(0) = w(-1) = 2$
• $v'(0) = w'(-1) = 1$

2. Apply the Laplace Transform

Now we apply the Laplace transform to both sides of the shifted differential equation. Using the linearity property and the transforms of derivatives, we have:

$\mathcal{L}{v''(\tau) - 6v'(\tau) + 9v(\tau)} = \mathcal{L}{45\tau + 15}$

$s^2V(s) - sv(0) - v'(0) - 6[sV(s) - v(0)] + 9V(s) = 45\mathcal{L}{\tau} + 15\mathcal{L}{1}$

Substituting the initial conditions $v(0) = 2$ and $v'(0) = 1$, and using the Laplace transforms $\mathcal{L}{\tau} = \frac{1}{s^2}$ and $\mathcal{L}{1} = \frac{1}{s}$, we get:

$s^2V(s) - 2s - 1 - 6sV(s) + 12 + 9V(s) = \frac{45}{s^2} + \frac{15}{s}$

3. Solve for $V(s)$

Rearrange the equation to solve for $V(s)$:

$(s^2 - 6s + 9)V(s) = \frac{45}{s^2} + \frac{15}{s} + 2s - 11$

$V(s) = \frac{45}{s^2(s-3)^2} + \frac{15}{s(s-3)^2} + \frac{2s - 11}{(s-3)^2}$

4. Partial Fraction Decomposition

To find the inverse Laplace transform, we need to decompose $V(s)$ into partial fractions. For the first term, we have:

$\frac{45}{s^2(s-3)^2} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s-3} + \frac{D}{(s-3)^2}$

Solving for the coefficients, we find $A = \frac{10}{3}$, $B = 5$, $C = -\frac{10}{3}$, and $D = \frac{5}{3}$.

For the second term:

$\frac{15}{s(s-3)^2} = \frac{E}{s} + \frac{F}{s-3} + \frac{G}{(s-3)^2}$

Solving for the coefficients, we find $E = \frac{5}{3}$, $F = -\frac{5}{3}$, and $G = \frac{5}{3}$.

For the third term:

$\frac{2s - 11}{(s-3)^2} = \frac{H}{s-3} + \frac{I}{(s-3)^2}$

Solving for the coefficients, we find $H = 2$ and $I = -5$.

So, $V(s)$ becomes:

$V(s) = \frac{10}{3s} + \frac{5}{s^2} - \frac{10}{3(s-3)} + \frac{5}{3(s-3)^2} + \frac{5}{3s} - \frac{5}{3(s-3)} + \frac{5}{3(s-3)^2} + \frac{2}{s-3} - \frac{5}{(s-3)^2}$

Combining like terms:

$V(s) = \frac{5}{s} + \frac{5}{s^2} - \frac{5}{s-3} - \frac{5}{3(s-3)^2}$

5. Inverse Laplace Transform

Now we take the inverse Laplace transform of $V(s)$ to find $v(\tau)$:

$v(\tau) = \mathcal{L}^{-1}{V(s)} = \mathcal{L}^{-1}{\frac{5}{s}} + \mathcal{L}^{-1}{\frac{5}{s^2}} - \mathcal{L}^{-1}{\frac{5}{s-3}} - \mathcal{L}^{-1}{\frac{5}{3(s-3)^2}}$

Using the inverse Laplace transforms of basic functions, we get:

$v(\tau) = 5 + 5\tau - 5e^{3\tau} - \frac{5}{3}\tau e^{3\tau}$

6. Time Shift Back to $t$

Finally, we substitute back $t = \tau - 1$ to find $w(t)$:

$w(t) = v(t + 1) = 5 + 5(t + 1) - 5e^{3(t + 1)} - \frac{5}{3}(t + 1)e^{3(t + 1)}$

Simplifying, we get:

$w(t) = 5t + 10 - 5e^{3(t + 1)} - \frac{5}{3}(t + 1)e^{3(t + 1)}$

Conclusion

In this article, we successfully solved the initial value problem $w'' - 6w' + 9w = 45t + 60$ with initial conditions $w(-1) = 2$ and $w'(-1) = 1$ using Laplace transforms. The key steps involved shifting the time variable to apply the Laplace transform effectively, transforming the differential equation into an algebraic equation, solving for $V(s)$, performing partial fraction decomposition, finding the inverse Laplace transform, and shifting back to the original time variable. This method demonstrates the power and versatility of Laplace transforms in solving linear differential equations, particularly those with initial conditions given at times other than $t = 0$. The final solution is:

$w(t) = 5t + 10 - 5e^{3(t + 1)} - \frac{5}{3}(t + 1)e^{3(t + 1)}$

By understanding and applying the properties and techniques discussed, you can tackle a wide range of initial value problems using Laplace transforms.