Solving Laplace Transform Of IVP: Dy/dt - Y = 2cos(kt)

Let's dive into the fascinating world of Laplace transforms! This article will guide you through the process of solving an initial value problem (IVP) using Laplace transforms. Specifically, we'll tackle the differential equation $\frac{dy}{dt} - y = 2\cos(kt)$ with the initial condition $y(0) = -6$. Get ready to transform your understanding of differential equations!

Understanding Laplace Transforms

Before we jump into the problem, let's quickly recap what Laplace transforms are all about. The Laplace transform is a powerful mathematical tool that transforms a function of time, $f(t)$, into a function of complex frequency, $F(s)$. Think of it as a way to convert differential equations into algebraic equations, which are often easier to solve. The Laplace transform 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 $f(t)$.
• $s$ is a complex frequency variable.
• $\mathcal{L}$ is the Laplace transform operator.

Key properties of Laplace transforms that we'll use in solving the IVP include:

• Linearity: $\mathcal{L}{af(t) + bg(t)} = a\mathcal{L}{f(t)} + b\mathcal{L}{g(t)}$, where $a$ and $b$ are constants.
• Derivative: $\mathcal{L}{\frac{dy}{dt}} = sY(s) - y(0)$, where $Y(s)$ is the Laplace transform of $y(t)$ and $y(0)$ is the initial condition.
• Cosine Function: $\mathcal{L}{\cos(kt)} = \frac{s}{s^2 + k^2}$

With these tools in our arsenal, we're ready to tackle the problem at hand!

Applying the Laplace Transform

Now, let's apply the Laplace transform to our given initial value problem (IVP): $\frac{dy}{dt} - y = 2\cos(kt)$, with $y(0) = -6$. The initial value problem is a cornerstone in differential equations, as it provides a specific solution path, and the Laplace transform method offers an elegant way to navigate this path. We will transform each term in the equation, leveraging the properties mentioned earlier, to transition from the time domain to the complex frequency domain. This step is crucial because it converts the differential equation into an algebraic equation, which is significantly easier to manipulate and solve. By carefully applying the Laplace transform, we set the stage for finding the solution to the IVP.

First, we apply the Laplace transform to both sides of the equation:

$\mathcal{L}{\frac{dy}{dt} - y} = \mathcal{L}{2\cos(kt)}$

Using the linearity property, we can split the left side:

$\mathcal{L}{\frac{dy}{dt}} - \mathcal{L}{y} = 2\mathcal{L}{\cos(kt)}$

Let $Y(s) = \mathcal{L}{y(t)}$. Applying the derivative property and the Laplace transform of the cosine function, we get:

$[sY(s) - y(0)] - Y(s) = 2\frac{s}{s^2 + k^2}$

Now, we substitute the initial condition $y(0) = -6$:

$[sY(s) - (-6)] - Y(s) = \frac{2s}{s^2 + k^2}$

This simplifies to:

$sY(s) + 6 - Y(s) = \frac{2s}{s^2 + k^2}$

Solving for Y(s)

Our next goal is to isolate $Y(s)$, which represents the Laplace transform of the solution $y(t)$. This step is vital because once we have $Y(s)$, we can apply the inverse Laplace transform to find $y(t)$. Isolating $Y(s)$ involves algebraic manipulation of the equation we derived in the previous step. This includes combining like terms, moving terms around, and factoring out $Y(s)$. The process is similar to solving for a variable in any algebraic equation, but the context here is the complex frequency domain, where $Y(s)$ lives. By carefully performing these algebraic steps, we can express $Y(s)$ in a form that is amenable to the inverse Laplace transform, thus bringing us closer to the solution of the original differential equation.

We need to solve for $Y(s)$. Let's rearrange the equation:

$sY(s) - Y(s) = \frac{2s}{s^2 + k^2} - 6$

Factor out $Y(s)$:

$Y(s)(s - 1) = \frac{2s}{s^2 + k^2} - 6$

Now, divide both sides by $(s - 1)$:

$Y(s) = \frac{1}{s - 1} \left( \frac{2s}{s^2 + k^2} - 6 \right)$

Distribute the $\frac{1}{s - 1}$ term:

$Y(s) = \frac{2s}{(s - 1)(s^2 + k^2)} - \frac{6}{s - 1}$

Partial Fraction Decomposition

The term $\frac{2s}{(s - 1)(s^2 + k^2)}$ looks complicated, doesn't it? This is where partial fraction decomposition comes to the rescue. Partial fraction decomposition is a technique that allows us to break down a complex rational function (a fraction where both the numerator and denominator are polynomials) into simpler fractions. This is incredibly useful in the context of Laplace transforms because the simpler fractions often have known inverse Laplace transforms. By decomposing $\frac{2s}{(s - 1)(s^2 + k^2)}$, we can express it as a sum of fractions with denominators $(s - 1)$ and $(s^2 + k^2)$, making it easier to find the inverse Laplace transform of each term. This technique is a cornerstone in solving differential equations using Laplace transforms, as it simplifies the process of moving from the complex frequency domain back to the time domain.

To simplify this, we'll use partial fraction decomposition on the first term. We want to express $\frac{2s}{(s - 1)(s^2 + k^2)}$ in the form:

$\frac{2s}{(s - 1)(s^2 + k^2)} = \frac{A}{s - 1} + \frac{Bs + C}{s^2 + k^2}$

Multiply both sides by $(s - 1)(s^2 + k^2)$ to clear the denominators:

$2s = A(s^2 + k^2) + (Bs + C)(s - 1)$

Expand the right side:

$2s = As^2 + Ak^2 + Bs^2 - Bs + Cs - C$

Now, group the terms by powers of $s$:

$2s = (A + B)s^2 + (-B + C)s + (Ak^2 - C)$

Equate the coefficients of the powers of $s$:

• $s^2$ terms: $A + B = 0$
• $s$ terms: $-B + C = 2$
• Constant terms: $Ak^2 - C = 0$

From the first equation, $B = -A$. Substituting into the second equation, we get $A + C = 2$, so $C = 2 - A$. Substituting into the third equation, we have $Ak^2 - (2 - A) = 0$, which gives $Ak^2 + A = 2$, or $A(k^2 + 1) = 2$. Thus:

$A = \frac{2}{k^2 + 1}$

Now we can find $B$ and $C$:

$B = -A = -\frac{2}{k^2 + 1}$

$C = 2 - A = 2 - \frac{2}{k^2 + 1} = \frac{2k^2}{k^2 + 1}$

So, our partial fraction decomposition is:

$\frac{2s}{(s - 1)(s^2 + k^2)} = \frac{2}{(k^2 + 1)(s - 1)} + \frac{-\frac{2}{k^2 + 1}s + \frac{2k^2}{k^2 + 1}}{s^2 + k^2}$

Substitute this back into the expression for $Y(s)$:

$Y(s) = \frac{2}{(k^2 + 1)(s - 1)} + \frac{-\frac{2}{k^2 + 1}s + \frac{2k^2}{k^2 + 1}}{s^2 + k^2} - \frac{6}{s - 1}$

Inverse Laplace Transform

We've made significant progress in transforming the original differential equation into an algebraic equation in the Laplace transform domain, and then solving for $Y(s)$. Now, the crucial final step is to apply the inverse Laplace transform to $Y(s)$ to obtain $y(t)$, the solution to our initial value problem. The inverse Laplace transform is the reverse operation of the Laplace transform; it brings us back from the complex frequency domain to the time domain. Each term in $Y(s)$ corresponds to a term in $y(t)$, and by applying the inverse transform to each part, we construct the overall solution. This step showcases the elegance of the Laplace transform method, where we solve a differential equation by transforming it, manipulating it algebraically, and then transforming the solution back to the original domain.

Now, we need to find the inverse Laplace transform of $Y(s)$. Let's rewrite $Y(s)$ to make it easier to work with:

$Y(s) = \frac{2}{(k^2 + 1)(s - 1)} - \frac{6}{s - 1} - \frac{2s}{(k^2 + 1)(s^2 + k^2)} + \frac{2k^2}{(k^2 + 1)(s^2 + k^2)}$

Combine the terms with $(s - 1)$ in the denominator:

$Y(s) = \frac{2 - 6(k^2 + 1)}{(k^2 + 1)(s - 1)} - \frac{2s}{(k^2 + 1)(s^2 + k^2)} + \frac{2k^2}{(k^2 + 1)(s^2 + k^2)}$

Simplify the first term:

$Y(s) = \frac{-6k^2 - 4}{(k^2 + 1)(s - 1)} - \frac{2s}{(k^2 + 1)(s^2 + k^2)} + \frac{2k^2}{(k^2 + 1)(s^2 + k^2)}$

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

• $\mathcal{L}^{-1}{\frac{1}{s - a}} = e^{at}$
• $\mathcal{L}^{-1}{\frac{s}{s^2 + k^2}} = \cos(kt)$
• $\mathcal{L}^{-1}{\frac{k}{s^2 + k^2}} = \sin(kt)$

Applying these, we get:

$y(t) = \mathcal{L}^{-1}{Y(s)} = \frac{-6k^2 - 4}{k^2 + 1}e^t - \frac{2}{k^2 + 1}\cos(kt) + \frac{2k}{k^2 + 1}\sin(kt)$

The Solution

We've reached the final destination! The expression we've derived, $y(t) = \frac{-6k^2 - 4}{k^2 + 1}e^t - \frac{2}{k^2 + 1}\cos(kt) + \frac{2k}{k^2 + 1}\sin(kt)$, is the solution to the initial value problem $\frac{dy}{dt} - y = 2\cos(kt)$ with the initial condition $y(0) = -6$. This journey, starting with the Laplace transform and culminating in $y(t)$, showcases the power of this method in solving differential equations. Each step, from transforming the equation to performing partial fraction decomposition and finally applying the inverse transform, has contributed to unveiling the specific function $y(t)$ that satisfies both the differential equation and the initial condition. This solution provides a complete description of the system's behavior over time, fulfilling the goal we set out to achieve.

Therefore, the solution to the IVP is:

$y(t) = \frac{-6k^2 - 4}{k^2 + 1}e^t - \frac{2}{k^2 + 1}\cos(kt) + \frac{2k}{k^2 + 1}\sin(kt)$

Conclusion

In this article, we've successfully navigated the process of solving an initial value problem using the Laplace transform. We started by transforming the differential equation into an algebraic equation, solved for $Y(s)$, performed partial fraction decomposition, and finally, applied the inverse Laplace transform to obtain the solution $y(t)$. This powerful technique provides a systematic way to solve linear differential equations, especially those with initial conditions. We hope this journey through Laplace transforms has been enlightening and that you now feel more confident in tackling similar problems!

For further exploration and deeper understanding of Laplace transforms, you can visit Wolfram MathWorld's Laplace Transform page. This resource provides a comprehensive overview of the topic, including properties, examples, and applications. Happy transforming!