Solving Y''-4y=δ(t-1) With Laplace Transforms And Initial Conditions

$$

Introduction

In this article, we will delve into solving the second-order linear ordinary differential equation $y'' - 4y = \delta(t-1)$ with the initial conditions $y(0) = 0$ and $y'(0) = 1$ using the powerful technique of Laplace transforms. This type of problem frequently arises in various fields of engineering and physics, where we need to analyze systems subjected to impulsive forces or disturbances modeled by the Dirac delta function. The presence of the Dirac delta function, denoted by $\delta(t-1)$, makes this problem particularly interesting and requires a careful application of Laplace transform properties. Our objective is to find the solution $y(t)$ that satisfies both the differential equation and the given initial conditions. We will first take the Laplace transform of the entire equation, incorporating the initial conditions, and then solve for the Laplace transform of the solution, denoted by $Y(s)$. The resulting expression for $Y(s)$ will then be manipulated using partial fraction decomposition and inverse Laplace transforms to obtain the final solution $y(t)$ in the time domain. This method allows us to systematically handle the Dirac delta function and initial conditions, providing a clear and efficient pathway to the solution. Understanding the properties of Laplace transforms and their applications in solving differential equations is crucial for engineers and scientists dealing with dynamic systems and control theory. This article provides a comprehensive step-by-step guide to tackle such problems, enhancing problem-solving skills and fostering a deeper understanding of the interplay between differential equations and transform methods.

Laplace Transform Approach

To solve the given differential equation $y'' - 4y = \delta(t-1)$ with initial conditions $y(0) = 0$ and $y'(0) = 1$, we will utilize the Laplace transform method. The Laplace transform is a valuable tool for converting differential equations into algebraic equations, making them easier to solve. This technique is particularly effective when dealing with linear differential equations with constant coefficients and discontinuous forcing functions, such as the Dirac delta function. The Laplace transform of a function $y(t)$ is defined as $Y(s) = \mathcal{L}{y(t)} = \int_0^\infty e^{-st}y(t) dt$, where $s$ is a complex variable. Applying the Laplace transform to the given differential equation requires us to transform each term individually, using the properties of Laplace transforms for derivatives and the Dirac delta function. The Laplace transform of the second derivative $y''(t)$ is given by $s^2Y(s) - sy(0) - y'(0)$, and the Laplace transform of $y(t)$ is simply $Y(s)$. The Laplace transform of the Dirac delta function $\delta(t-1)$ is $e^{-s}$, which reflects the impulsive nature of the function at $t=1$. By substituting these transforms into the original equation, we obtain an algebraic equation in terms of $Y(s)$, which we can then solve for $Y(s)$. This step is crucial as it simplifies the problem from a differential equation to an algebraic one, allowing us to manipulate the equation using algebraic techniques. The initial conditions $y(0) = 0$ and $y'(0) = 1$ play a vital role in this process, as they are directly incorporated into the transformed equation, providing specific values that help determine the unique solution. The transformed equation is then rearranged to isolate $Y(s)$, representing the Laplace transform of the solution we seek. The next step involves finding the inverse Laplace transform of $Y(s)$ to obtain the solution $y(t)$ in the time domain. This process often involves techniques such as partial fraction decomposition and consulting Laplace transform tables to identify the corresponding time-domain functions. The Laplace transform method not only simplifies the solution process but also provides a systematic approach to handling initial conditions and discontinuous functions, making it an indispensable tool in solving differential equations encountered in various applications.

Applying the Laplace Transform

Applying the Laplace transform to the differential equation $y'' - 4y = \delta(t-1)$ involves using the properties of Laplace transforms for derivatives and the Dirac delta function. Recall that the Laplace transform of the second derivative $y''(t)$ is given by $s^2Y(s) - sy(0) - y'(0)$, where $Y(s)$ is the Laplace transform of $y(t)$. The Laplace transform of $y(t)$ is simply $Y(s)$, and the Laplace transform of the Dirac delta function $\delta(t-1)$ is $e^{-s}$. Substituting these into the given equation, we have:

$$\mathcal{L}{y''} - 4\mathcal{L}{y} = \mathcal{L}{\delta(t-1)}$$

$$[s^2Y(s) - sy(0) - y'(0)] - 4Y(s) = e^{-s}$$

Using the initial conditions $y(0) = 0$ and $y'(0) = 1$, we substitute these values into the equation:

$$s^2Y(s) - s(0) - 1 - 4Y(s) = e^{-s}$$

$$s^2Y(s) - 1 - 4Y(s) = e^{-s}$$

Now, we rearrange the equation to isolate $Y(s)$:

$$Y(s)(s^2 - 4) = 1 + e^{-s}$$

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

This equation represents the Laplace transform of the solution $y(t)$. It combines the effects of the differential equation and the initial conditions into a single algebraic expression. The next step involves finding the inverse Laplace transform of $Y(s)$ to obtain the solution $y(t)$ in the time domain. The expression for $Y(s)$ contains a term with $e^{-s}$, which corresponds to a time-shifted function in the time domain. To find the inverse Laplace transform, we will use techniques such as partial fraction decomposition to break down the rational function into simpler terms that can be easily inverted using Laplace transform tables. This process allows us to systematically convert the frequency-domain representation $Y(s)$ back into the time-domain solution $y(t)$. The resulting solution will reflect the behavior of the system in response to the Dirac delta function input and the given initial conditions.

Solving for $Y(s)$

In the previous section, we applied the Laplace transform to the differential equation and incorporated the initial conditions to arrive at the expression for $Y(s)$:

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

This equation represents the Laplace transform of the solution $y(t)$. To find the solution in the time domain, we need to find the inverse Laplace transform of $Y(s)$. The expression for $Y(s)$ can be separated into two parts:

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

This separation allows us to deal with each term individually when finding the inverse Laplace transform. The first term, $\frac{1}{s^2 - 4}$, represents the response of the system to the initial conditions and the homogeneous part of the differential equation. The second term, $\frac{e^{-s}}{s^2 - 4}$, represents the response of the system to the Dirac delta function input, which occurs at $t=1$. The factor $e^{-s}$ indicates a time shift of 1 unit in the time domain. To find the inverse Laplace transform of each term, we first perform partial fraction decomposition on $\frac{1}{s^2 - 4}$. We can write:

$$\frac{1}{s^2 - 4} = \frac{1}{(s - 2)(s + 2)} = \frac{A}{s - 2} + \frac{B}{s + 2}$$

Multiplying both sides by $(s - 2)(s + 2)$ gives:

$$1 = A(s + 2) + B(s - 2)$$

To solve for $A$ and $B$, we can use the method of substitution. Setting $s = 2$ gives:

$$1 = A(2 + 2) + B(2 - 2) \implies 1 = 4A \implies A = \frac{1}{4}$$

Setting $s = -2$ gives:

$$1 = A(-2 + 2) + B(-2 - 2) \implies 1 = -4B \implies B = -\frac{1}{4}$$

Thus, we have:

$$\frac{1}{s^2 - 4} = \frac{1/4}{s - 2} - \frac{1/4}{s + 2}$$

Now we can rewrite $Y(s)$ as:

$$Y(s) = \frac{1/4}{s - 2} - \frac{1/4}{s + 2} + e^{-s}\left(\frac{1/4}{s - 2} - \frac{1/4}{s + 2}\right)$$

This expression for $Y(s)$ is now in a form that allows us to easily find the inverse Laplace transform using standard Laplace transform pairs. The inverse Laplace transform of each term will give us the solution $y(t)$ in the time domain.

Inverse Laplace Transform

Having obtained $Y(s)$ in the form:

$$Y(s) = \frac{1/4}{s - 2} - \frac{1/4}{s + 2} + e^{-s}\left(\frac{1/4}{s - 2} - \frac{1/4}{s + 2}\right)$$

we now proceed to find the inverse Laplace transform to obtain $y(t)$. We will use the following Laplace transform pairs:

\mathcal{L}^{-1}\left{\frac{1}{s - a}\right} = e^{at}

$$\mathcal{L}^{-1}{e^{-as}F(s)} = f(t - a)u(t - a)$$

where $u(t - a)$ is the Heaviside step function, defined as:

$$u(t - a) = \begin{cases} 0, & t < a \\ 1, & t \geq a \end{cases}$$

Applying the inverse Laplace transform to each term in $Y(s)$, we have:

\mathcal{L}^{-1}\left{\frac{1/4}{s - 2}\right} = \frac{1}{4}e^{2t}

\mathcal{L}^{-1}\left{-\frac{1/4}{s + 2}\right} = -\frac{1}{4}e^{-2t}

For the terms involving $e^{-s}$, we apply the time-shifting property:

\mathcal{L}^{-1}\left{e^{-s}\frac{1/4}{s - 2}\right} = \frac{1}{4}e^{2(t - 1)}u(t - 1)

\mathcal{L}^{-1}\left{e^{-s}\left(-\frac{1/4}{s + 2}\right)\right} = -\frac{1}{4}e^{-2(t - 1)}u(t - 1)

Combining these results, we get the solution $y(t)$:

$$y(t) = \frac{1}{4}e^{2t} - \frac{1}{4}e^{-2t} + \frac{1}{4}e^{2(t - 1)}u(t - 1) - \frac{1}{4}e^{-2(t - 1)}u(t - 1)$$

We can simplify the first two terms using the definition of the hyperbolic sine function, $\sinh(x) = \frac{e^x - e^{-x}}{2}$:

$$\frac{1}{4}e^{2t} - \frac{1}{4}e^{-2t} = \frac{1}{2}\left(\frac{e^{2t} - e^{-2t}}{2}\right) = \frac{1}{2}\sinh(2t)$$

Similarly, for the terms involving the Heaviside function:

$$\frac{1}{4}e^{2(t - 1)}u(t - 1) - \frac{1}{4}e^{-2(t - 1)}u(t - 1) = \frac{1}{2}\left(\frac{e^{2(t - 1)} - e^{-2(t - 1)}}{2}\right)u(t - 1) = \frac{1}{2}\sinh(2(t - 1))u(t - 1)$$

Therefore, the final solution is:

$$y(t) = \frac{1}{2}\sinh(2t) + \frac{1}{2}\sinh(2(t - 1))u(t - 1)$$

This solution represents the response of the system to both the initial conditions and the Dirac delta function input at $t = 1$. The term $\frac{1}{2}\sinh(2t)$ is the homogeneous solution due to the initial conditions, and the term $\frac{1}{2}\sinh(2(t - 1))u(t - 1)$ is the particular solution due to the Dirac delta function input, shifted in time by 1 unit.

Verification of the Solution

To verify the solution obtained using Laplace transforms, we need to ensure that it satisfies both the original differential equation and the initial conditions. The solution we found is:

$$y(t) = \frac{1}{2}\sinh(2t) + \frac{1}{2}\sinh(2(t - 1))u(t - 1)$$

First, let's check the initial conditions $y(0) = 0$ and $y'(0) = 1$:

For $t = 0$:

$$y(0) = \frac{1}{2}\sinh(0) + \frac{1}{2}\sinh(2(0 - 1))u(0 - 1) = \frac{1}{2}(0) + \frac{1}{2}\sinh(-2)(0) = 0$$

So, $y(0) = 0$ is satisfied.

Next, we need to find the derivative $y'(t)$. The derivative of $\sinh(2t)$ is $2\cosh(2t)$. To differentiate the second term, we use the product rule and the fact that the derivative of $u(t - 1)$ is $\delta(t - 1)$:

$$y'(t) = \cosh(2t) + \frac{1}{2}[2\cosh(2(t - 1))u(t - 1) + \sinh(2(t - 1))\delta(t - 1)]$$

$$y'(t) = \cosh(2t) + \cosh(2(t - 1))u(t - 1) + \frac{1}{2}\sinh(2(t - 1))\delta(t - 1)$$

At $t = 0$:

$$y'(0) = \cosh(0) + \cosh(-2)u(-1) + \frac{1}{2}\sinh(-2)\delta(-1) = 1 + 0 + 0 = 1$$

So, $y'(0) = 1$ is also satisfied.

Now, let's check if the solution satisfies the differential equation $y'' - 4y = \delta(t - 1)$. We need to find the second derivative $y''(t)$. Differentiating $y'(t)$, we have:

$$y''(t) = 2\sinh(2t) + 2\sinh(2(t - 1))u(t - 1) + \cosh(2(t - 1))\delta(t - 1) + \frac{1}{2}\sinh(2(t - 1))\delta'(t - 1)$$

Now, we substitute $y(t)$ and $y''(t)$ into the differential equation:

$$y''(t) - 4y(t) = \left[2\sinh(2t) + 2\sinh(2(t - 1))u(t - 1) + \cosh(2(t - 1))\delta(t - 1) + \frac{1}{2}\sinh(2(t - 1))\delta'(t - 1)\right] - 4\left[\frac{1}{2}\sinh(2t) + \frac{1}{2}\sinh(2(t - 1))u(t - 1)\right]$$

$$y''(t) - 4y(t) = 2\sinh(2t) + 2\sinh(2(t - 1))u(t - 1) + \cosh(2(t - 1))\delta(t - 1) - 2\sinh(2t) - 2\sinh(2(t - 1))u(t - 1)$$

$$y''(t) - 4y(t) = \cosh(2(t - 1))\delta(t - 1)$$

Since $\cosh(0) = 1$, the equation simplifies to:

$$y''(t) - 4y(t) = \delta(t - 1)$$

Thus, the solution satisfies the differential equation. The solution $y(t) = \frac{1}{2}\sinh(2t) + \frac{1}{2}\sinh(2(t - 1))u(t - 1)$ satisfies both the initial conditions and the differential equation, confirming its correctness.

Conclusion

In this article, we have successfully solved the differential equation $y'' - 4y = \delta(t-1)$ with initial conditions $y(0) = 0$ and $y'(0) = 1$ using the Laplace transform method. The solution obtained is:

$$y(t) = \frac{1}{2}\sinh(2t) + \frac{1}{2}\sinh(2(t - 1))u(t - 1)$$

We began by applying the Laplace transform to the differential equation, incorporating the initial conditions and the property of the Dirac delta function. This transformed the differential equation into an algebraic equation in the $s$-domain, which was then solved for $Y(s)$, the Laplace transform of the solution. The resulting expression for $Y(s)$ was then decomposed using partial fraction decomposition to facilitate the inverse Laplace transform. The inverse Laplace transform was applied to each term, utilizing standard Laplace transform pairs and the time-shifting property to obtain the solution $y(t)$ in the time domain. The solution consists of two parts: the homogeneous solution due to the initial conditions and the particular solution due to the Dirac delta function input. The Heaviside step function $u(t - 1)$ in the second term indicates the time shift caused by the Dirac delta function, which acts as an impulse at $t = 1$. Finally, we verified that the obtained solution satisfies both the initial conditions and the original differential equation, ensuring its correctness. The Laplace transform method provides a powerful and systematic approach to solving linear differential equations with constant coefficients, particularly those involving discontinuous forcing functions like the Dirac delta function. This technique is widely used in engineering and physics to analyze the behavior of dynamic systems under various inputs and initial conditions. The step-by-step process outlined in this article provides a clear understanding of the method and its application, enhancing problem-solving skills in differential equations and transform methods.