Problem 37 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. $$ \frac{d^{2} x}{d t^{2}}+4 x=-5 \sin 2 t+3 \cos 2 t, x(0)=-1, x^{\prime}(0)=1 $$

Step-by-Step Solution

Answer

The solution is \( x(t) = -\frac{1}{4} \cos(2t) + \frac{1}{2} \sin(2t) \).

Step 1: Solve the Homogeneous Equation

The homogeneous equation is \( \frac{d^2 x}{dt^2} + 4x = 0 \). We solve this by finding the characteristic equation, which is \( r^2 + 4 = 0 \). Solving for \( r \), we get \( r = \pm 2i \). Thus, the solution to the homogeneous equation is \( x_h(t) = c_1 \cos(2t) + c_2 \sin(2t) \).

Step 2: Find the Particular Solution

To find a particular solution, \( x_p(t) \), we use the method of undetermined coefficients. Assume \( x_p(t) = A \sin(2t) + B \cos(2t) \). Substitute \( x_p(t) \) into the non-homogeneous equation:\[ \frac{d^2}{dt^2}(A \sin(2t) + B \cos(2t)) + 4(A \sin(2t) + B \cos(2t)) = -5 \sin(2t) + 3 \cos(2t)\]This results in \[ -(4A \sin(2t) + 4B \cos(2t)) + 4(A \sin(2t) + B \cos(2t)) = -5 \sin(2t) + 3 \cos(2t)\]Simplifying gives \(-4A = -5\) and \(-4B = 3\). Solving these gives \(A = \frac{5}{4}\) and \(B = -\frac{3}{4}\). Thus, \( x_p(t) = \frac{5}{4} \sin(2t) - \frac{3}{4} \cos(2t) \).

Step 3: Form the General Solution

The general solution is the sum of the homogeneous and particular solutions: \[ x(t) = c_1 \cos(2t) + c_2 \sin(2t) + \frac{5}{4} \sin(2t) - \frac{3}{4} \cos(2t) \].

Step 4: Apply Initial Conditions

Given: \( x(0) = -1 \) and \( x'(0) = 1 \). First, apply \( x(0) = -1 \): \[ x(0) = c_1 \cdot 1 + c_2 \cdot 0 - \frac{3}{4} = -1 \]Solving for \( c_1 \):\[ c_1 - \frac{3}{4} = -1 \rightarrow c_1 = -\frac{1}{4} \].For \( x'(0) = 1 \), differentiate the general solution:\[ x'(t) = -2c_1 \sin(2t) + 2c_2 \cos(2t) + \frac{5}{2} \cos(2t) + \frac{3}{2} \sin(2t) \].Evaluate at \( t = 0 \): \[ x'(0) = 0 + 2c_2 + \frac{5}{2} \cdot 1 + 0 = 1 \]Solve for \( c_2 \): \[ 2c_2 + \frac{5}{2} = 1 \rightarrow 2c_2 = -\frac{3}{2} \rightarrow c_2 = -\frac{3}{4} \].

Step 5: Write the Final Solution

Include the determined values of \( c_1 \) and \( c_2 \) from the initial conditions into the general solution:\[ x(t) = -\frac{1}{4} \cos(2t) - \frac{3}{4} \sin(2t) + \frac{5}{4} \sin(2t) - \frac{3}{4} \cos(2t) \].Simplify this to: \[ x(t) = -\frac{1}{4} \cos(2t) + \frac{1}{2} \sin(2t) \].

Key Concepts

Initial-Value Problems

An initial-value problem in differential equations is a type of boundary value problem where the solution is required to satisfy specified values at the start of the interval. In our case, these values include conditions such as \( x(0) = -1 \) and \( x'(0) = 1 \). These conditions tell us the value of the function and its derivative at \( t = 0 \), which together help us determine the constants in the general solution.

When faced with an initial-value problem, you will first find the general solution of the differential equation. Then, apply the initial conditions to this solution to find the specific constants. This results in a unique solution satisfying both the equation and the initial conditions.

• Step 1: Find the general solution of the differential equation.
• Step 2: Use the initial values to find any unknown constants.

This process ensures that the solution is not just one of the many possible solutions, but the exact one matching the given problem.

Homogeneous Equation

The homogeneous equation is a crucial aspect of solving linear differential equations. It is derived from the non-homogeneous equation by stripping away the non-homogeneous part (the terms not involving the function or its derivatives).

For example, given the differential equation \( \frac{d^2 x}{dt^2} + 4x = 0 \), you remove any forcing function (such as external driving forces in physical systems).

Solving this involves finding the characteristic equation, which is typically a quadratic equation that reveals the nature of the solution via its roots.

• If the roots are real and distinct, the solution involves exponential functions.
• If the roots are complex, as in our problem (\( r = \pm 2i \)), the solution is trigonometric, involving sine and cosine.

The homogeneous solution, once obtained using the roots of the characteristic equation, forms the complementary solution part of the full response to the differential equation.

Non-Homogeneous Equation

When a differential equation has a term not associated with the derivatives of the function in question, it is termed non-homogeneous. The solution to a non-homogeneous equation consists of two parts: the homogeneous solution we discussed earlier, and a particular solution.

In our exercise, the equation \( \frac{d^2 x}{dt^2} + 4x = -5 \sin 2t + 3 \cos 2t \) is non-homogeneous due to the presence of \(-5 \sin 2t + 3 \cos 2t\). These terms require us to find a particular solution that satisfies the entire equation.

Understanding the non-homogeneous part:

• This part may represent external forces or drives acting on the system, such as oscillations, forces, or other influences.
• Finding a particular solution matches this part, fulfilling what wouldn't be resolved by the homogeneous solution alone.

It's the combination of these two solutions, homogeneous and particular, that gives the general solution for the non-homogeneous differential equation.

Method of Undetermined Coefficients

The method of undetermined coefficients is a popular approach for finding the particular solution of non-homogeneous linear differential equations with constant coefficients.

This method assumes a form for the particular solution based on the non-homogeneous term itself. You propose a solution structure with unknown coefficients which you then adjust to balance the equation.

In our problem, we started with the guess \( x_p(t) = A \sin(2t) + B \cos(2t) \) because the non-homogeneous part involves \( \sin \) and \( \cos \) terms.
Steps to use this method effectively:

• Identify the form of the non-homogeneous term (e.g., sinusoidal, polynomial).
• Propose a solution with undetermined coefficients (A, B, etc.).
• Substitute this guessed solution into the differential equation.
• Solve for these coefficients.

This method works well when the non-homogeneous term is simple and matches the nature of the assumed solution form. If the equation had terms that repeat the structure of the homogeneous solution, you would multiply your trial solution by \( t \) to a suitable degree to ensure solutions are independent.