Problem 4 - Advanced Engineering Mathematics

Question

In Problems \(1-20\), solve the given system of differential equations by systematic elimination. $$ \begin{aligned} &\frac{d x}{d t}-4 y=1 \\ &\frac{d y}{d t}+x=2 \end{aligned} $$

Step-by-Step Solution

Answer

Find general solutions: \( x(t) = C_1 \cos(2t) + C_2 \sin(2t) + 2 \) and \( y(t) = -\frac{C_1}{2} \sin(2t) + \frac{C_2}{2} \cos(2t) + C_3 \).

Step 1: Differentiate the First Equation with Respect to t

Start with the equation \( \frac{dx}{dt} - 4y = 1 \). Differentiate both sides with respect to \( t \). This yields \( \frac{d^2x}{dt^2} - 4\frac{dy}{dt} = 0 \).

Step 2: Solve for \( \frac{dy}{dt} \) from the Second Equation

The second equation given is \( \frac{dy}{dt} + x = 2 \). From this, solve for \( \frac{dy}{dt} \) to obtain \( \frac{dy}{dt} = 2 - x \).

Step 3: Substitute \( \frac{dy}{dt} \) into the Differentiated Equation

Substitute \( \frac{dy}{dt} = 2 - x \) into the differentiated equation \( \frac{d^2x}{dt^2} - 4\frac{dy}{dt} = 0 \). This transformation becomes \( \frac{d^2x}{dt^2} - 4(2 - x) = 0 \).

Step 4: Simplify and Solve for \( x \)

Simplify the equation from Step 3: \( \frac{d^2x}{dt^2} - 8 + 4x = 0 \), which rearranges to \( \frac{d^2x}{dt^2} + 4x = 8 \). This is a non-homogeneous second-order linear differential equation.

Step 5: Solve the Homogeneous Equation

Solve the associated homogeneous equation \( \frac{d^2x}{dt^2} + 4x = 0 \). The characteristic equation is \( r^2 + 4 = 0 \) with solutions \( r = \,\pm 2i \). Thus, the general solution is \( x_h(t) = C_1 \cos(2t) + C_2 \sin(2t) \).

Step 6: Find a Particular Solution for the Non-Homogeneous Equation

Guess a particular solution of the form \( x_p(t) = A \). Substitute it into \( \frac{d^2x}{dt^2} + 4x = 8 \). This simplifies to \( 4A = 8 \), so \( A = 2 \). The particular solution is \( x_p(t) = 2 \).

Step 7: Form the General Solution for \( x \)

Combine the homogeneous and particular solutions to get the general solution \( x(t) = C_1 \cos(2t) + C_2 \sin(2t) + 2 \).

Step 8: Solve for \( y \) Using \( x \) and Initial Equations

Substitute \( x(t) = C_1 \cos(2t) + C_2 \sin(2t) + 2 \) into \( \frac{dy}{dt} + x = 2 \). Rearranging gives \( \frac{dy}{dt} = -C_1 \cos(2t) - C_2 \sin(2t) \).

Step 9: Integrate to Find \( y \)

Integrate \( \frac{dy}{dt} = -C_1 \cos(2t) - C_2 \sin(2t) \) to find \( y(t) = -\frac{C_1}{2} \sin(2t) + \frac{C_2}{2} \cos(2t) + C_3 \).

Step 10: Form the General Solution for \( y(t) \)

The complete solution for \( y(t) \) is \( y(t) = -\frac{C_1}{2} \sin(2t) + \frac{C_2}{2} \cos(2t) + C_3 \).

Step 11: Combine the Solutions

The general solutions are \( x(t) = C_1 \cos(2t) + C_2 \sin(2t) + 2 \) and \( y(t) = -\frac{C_1}{2} \sin(2t) + \frac{C_2}{2} \cos(2t) + C_3 \). These functions satisfy the original system of differential equations.

Key Concepts

systematic elimination

Systematic elimination is a method used to solve a system of differential equations by eliminating variables one by one until a single equation in one variable remains.
In the given exercise, we have two linked differential equations involving functions of time, specifically for variables \( x(t) \) and \( y(t) \). By differentiating and substituting, we reduce the problem size.
This process simplifies the stately way of handling simultaneous equations, enabling clearer paths to the solutions.

• The first step is differentiating one of the equations with respect to time. This results in a higher-order derivative.
• Next, solve for the derivative of one variable from the second equation.
• Finally, substitute this expression back into the differentiated equation, resulting in an equation only in terms of one variable.

These steps showcase systematic elimination efficiently directs us toward solving one differential equation at a time.

non-homogeneous differential equation

A non-homogeneous differential equation contains a variable that does not depend on the function being solved for. This means there is an additional function on the right-hand side of the equation that disrupts its 'pure' form.
In our case, the equation \( \frac{d^2x}{dt^2} + 4x = 8 \) is non-homogeneous because of the constant term on the right side. Such equations require more than just the solution to its homogeneous counterpart.

• Recognizing a non-homogeneous differential equation is crucial as it affects the approach to finding the overall solution.
• While there's a systematic way, often involving guessing a form for the particular solution, to handle it, it is essential to find a specific function, called the particular solution, that satisfies the equation with the non-homogeneous part present.

This makes dealing with real-world problems meaningful as most equations we encounter aren't homogeneously ideal.

characteristic equation

The characteristic equation is a vital tool in solving linear differential equations. It is derived from the homogeneous part of a differential equation by assuming a solution of exponential form.
For the homogeneous equation \( \frac{d^2x}{dt^2} + 4x = 0 \), assuming a solution of the form \( x(t) = e^{rt} \) transforms it into a characteristic equation \( r^2 + 4 = 0 \).
This equation is crucial because:

• It helps us determine the nature of the solutions—real or complex—depending on the roots it produces.
• For complex roots, like \( r = \pm 2i \), the solutions will be oscillatory, involving sine and cosine functions.

For this particular example, it indicates the solution includes trigonometric functions, establishing a foundation for the general solution.

particular solution

The particular solution to a differential equation represents a concrete solution that satisfies the non-homogeneous equation given the external influences or forcing functions.
In the ongoing example, finding a particular solution to \( \frac{d^2x}{dt^2} + 4x = 8 \) involves a guess method. Assuming \( x_p(t) = A \), and substituting, gives \( 4A = 8 \). Solving this, we find \( A = 2 \).

• The particular solution aims to complement the homogeneous solution to satisfy the complete equation.
• This provides insight into how external influences mold the solution to include specific behaviors beyond the natural tendencies of the system.

When combined with the general homogeneous solution, it completes the general solution to the non-homogeneous equation.