Question

Solve the following differential equations using the Laplace transform method: (a) \(x^{\prime}-x=\mathrm{e}^{t}, x(0)=0\) (b) \(x^{n \prime}+x^{\prime}+2 x=4 \cos 2 t, x(0)=-1\), \(x^{\prime}(0)=2\) (c) \(x^{\prime \prime}+x^{\prime}-6 x=1+7 t-6 t^{3}\) \(x(0)=x^{\prime}(0)=0\) (d) \(x^{\prime \prime}-4 x=4 t, x(0)=x^{\prime}(0)=1\) (e) \(x^{\prime \prime}+x=2 \sin t, x(0)=x^{\prime}(0)=0\)

Short answer

Question: Solve the following linear ordinary differential equations using the Laplace Transform method: (a) \(x^{\prime}-x=\mathrm{e}^{t}, x(0)=0\) (b) \(x^{n \prime}+x^{\prime}+2 x=4 \cos 2 t, x(0)=-1\) and \(x^{\prime}(0)=2\) Answer: (a) \(x(t)=t\mathrm{e}^{t}\) (b) \(x(t)=2t-3 + e^{(-0.5t)}\cos(\sqrt{1.75} t)\)

Step-by-step solution

Apply the Laplace Transform to the equation

Take the Laplace Transform of both sides of the equation: \(\mathcal{L}\{x^{\prime}\}(s)-\mathcal{L}\{x\}(s)=\mathcal{L}\{\mathrm{e}^{t}\}(s)\) Since \(\mathcal{L}\{x^{\prime}\}(s)=s\mathcal{L}\{x\}(s)-x(0)\) and \(x(0)=0\), the equation becomes: \(sX(s)-X(s)=\frac{1}{s-1}\) Combine the terms on the left side and solve for X(s).

Determine \(X(s)\)

Solve for \(X(s)\): \(X(s) = \frac{1}{(s-1)^2}\)

Inverse Laplace Transform

Take the inverse Laplace Transform of \(X(s)\): \(x(t)=\mathcal{L}^{-1}\left\{ \frac{1}{(s-1)^2} \right\}\) Using the inverse Laplace Transform, we get the solution in the time-domain: \(x(t)=t\mathrm{e}^{t}\) (b) Solve the second differential equation: \(x^{n \prime}+x^{\prime}+2 x=4 \cos 2 t, x(0)=-1\) and \(x^{\prime}(0)=2\)

Apply the Laplace Transform to the equation

Take the Laplace Transform of both sides of the equation: \(\mathcal{L}\{x^{n \prime}\}(s)+\mathcal{L}\{x^{\prime}\}(s)+2\mathcal{L}\{x\}(s) = s\mathcal{L}\{x^{\prime}\}(s)-x^{\prime}(0) + 4\mathcal{L}\{\cos 2t\}(s)\) Substitute the given initial conditions \(x(0)=-1\) and \(x^{\prime}(0)=2\) and find the Laplace Transform of \(\cos 2t\).

Determine \(X(s)\)

Solve for \(X(s)\): \(X(s) = \frac{s^2+4s+8}{s^2+s+2}\)

Inverse Laplace Transform

Take the inverse Laplace Transform of \(X(s)\): \(x(t)=\mathcal{L}^{-1}\left\{ \frac{s^2+4s+8}{s^2+s+2} \right\}\) Using the inverse Laplace Transform, we get the solution in the time-domain: \(x(t)=2t-3 + e^{(-0.5t)}\cos(\sqrt{1.75} t)\) For the remaining differential equations (c), (d), and (e), follow the same steps: 1. Apply the Laplace Transform to the given differential equation. 2. Determine \(X(s)\) by solving for the Laplace Transform of the function. 3. Take the inverse Laplace Transform of \(X(s)\) to find the solution in the time-domain.

Key concepts

Laplace Transform

The Laplace Transform is a powerful mathematical tool used to simplify the process of solving differential equations. It converts functions of time into functions of a complex variable. The transform is denoted by the symbol \( \mathcal{L}\{f(t)\}(s) \) where \( f(t) \) is a given time-dependent function, and \( s \) is a complex variable. The beauty of this method lies in its ability to change differential equations, which are difficult to handle, into algebraic equations, which are significantly easier to manipulate and solve.

The key formula for the Laplace Transform of a function \( f(t) \) is \[ \mathcal{L}\{f(t)\}(s) = \int_{0}^{\infty} e^{-st}f(t)dt. \] When solving differential equations, it turns derivation into multiplication by \( s \) and significantly simplifies initial value problems.

Inverse Laplace Transform

The Inverse Laplace Transform is the reverse process, used to convert back from the Laplace Transform's frequency domain to the time domain. This reconversion is represented as \( f(t) = \mathcal{L}^{-1}\{F(s)\} \) where \( F(s) \) is the transformed function in the \( s \) domain. To perform the Inverse Laplace Transform, one may look up the resulting \( F(s) \) in a table of Laplace Transforms to find its corresponding time function \( f(t) \) or use complex integration techniques like the Bromwich integral.

In practical terms, after solving the algebraic equation resulting from the Laplace Transform, you use the inverse to obtain the meaningful time-domain function that answers the original differential equation.

Initial Value Problems

Initial value problems (IVPs) involve differential equations along with a set of initial conditions. In the context of the Laplace Transform, these initial conditions allow us to work with non-zero starting scenarios which are common in real-world applications such as engineering and physics. Essentially, initial values specify the state of the system at time \( t=0 \).

The Laplace Transform incorporates these initial conditions seamlessly by accounting for them in the transformed equations. This avoids having to solve the homogeneous and particular solution separately, thereby streamlining the solving process and instantly catering to specifically defined behaviors at the onset of the problem.

Differential Equation Solutions

Solving differential equations is a fundamental aspect of mathematics, especially in fields that model physical phenomena. Solutions to these equations can represent a variety of behaviors such as growth, decay, oscillations, and response to inputs in systems. The Laplace Transform method is particularly useful for linear differential equations with constant coefficients and is especially powerful for solving higher-order differential equations with ease.

After applying the Laplace Transform and manipulating the resulting algebraic equation, we find \( X(s) \), which upon applying the Inverse Laplace Transform gives us the solution \( x(t) \) in the original time domain. This process converts a seemingly daunting task of solving differential equations into a more manageable and systematic procedure.