Problem 10 - Advanced Engineering Mathematics

Question

In Problems \(7-10, x=c_{1} \cos t+c_{2} \sin t\) is a two-parameter family of solutions of the second-order DE \(x^{\prime \prime}+x=0\). Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. $$ x(\pi / 4)=\sqrt{2}, \quad x^{\prime}(\pi / 4)=2 \sqrt{2} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Differential Equation

The given differential equation is a second-order homogeneous linear differential equation: \[ x'' + x = 0 \]The general solution for this equation is provided as a two-parameter family of solutions:\[ x(t) = c_1 \cos t + c_2 \sin t \] where \( c_1 \) and \( c_2 \) are constants to be determined using initial conditions.

Step 2: Finding the Derivative

Differentiate the general solution \( x(t) = c_1 \cos t + c_2 \sin t \) to obtain the first derivative:\[ x'(t) = -c_1 \sin t + c_2 \cos t \].This derivative will be used to apply the initial condition on the derivative.

Step 3: Applying Initial Condition for x(t)

Substitute the first initial condition \( x(\pi/4) = \sqrt{2} \) into the general solution:\[ c_1 \cos(\pi/4) + c_2 \sin(\pi/4) = \sqrt{2} \]Simplifying, we find:\[ \frac{c_1}{\sqrt{2}} + \frac{c_2}{\sqrt{2}} = \sqrt{2} \]Multiplying through by \(\sqrt{2}\) results in:\[ c_1 + c_2 = 2 \sqrt{2} \].

Step 4: Applying Initial Condition for x'(t)

Substitute the second initial condition \( x'(\pi/4) = 2\sqrt{2} \) into \( x'(t) \):\[ -c_1 \sin(\pi/4) + c_2 \cos(\pi/4) = 2\sqrt{2} \]Simplifying, we find:\[ -\frac{c_1}{\sqrt{2}} + \frac{c_2}{\sqrt{2}} = 2 \sqrt{2} \]Multiplying through by \(\sqrt{2}\) results in:\[ -c_1 + c_2 = 4 \].

Step 5: Solving the System of Equations

We now have two equations to solve for \( c_1 \) and \( c_2 \):1. \( c_1 + c_2 = 2\sqrt{2} \)2. \( -c_1 + c_2 = 4 \)Add these equations:\[ (c_1 + c_2) + (-c_1 + c_2) = 2\sqrt{2} + 4 \]\[ 2c_2 = 2\sqrt{2} + 4 \]\[ c_2 = \sqrt{2} + 2 \]Substitute \( c_2 \) back into the first equation:\[ c_1 + (\sqrt{2} + 2) = 2\sqrt{2} \]\[ c_1 = 2\sqrt{2} - \sqrt{2} - 2 \]\[ c_1 = \sqrt{2} - 2 \].

Step 6: Writing the Final Solution

Substitute the values of \( c_1 \) and \( c_2 \) into the general solution:\[ x(t) = (\sqrt{2} - 2) \cos t + (\sqrt{2} + 2) \sin t \]This is the particular solution to the initial value problem satisfying both the differential equation and the initial conditions.

Key Concepts

Initial value problem

An initial value problem is a type of differential equation paired with specified values, called initial conditions, which the solution must satisfy at a certain point. In the context of a second-order differential equation like \(x'' + x = 0\), the initial conditions are the values of the function and its first derivative at a specific point. Here, we have initial conditions \(x(\pi/4) = \sqrt{2}\) and \(x'(\pi/4) = 2\sqrt{2}\).

To solve an initial value problem, we need to find a specific solution to the differential equation that fits the given initial conditions, unlike generic solutions which involve arbitrary constants. First, determine the general solution of the given differential equation, and then use the initial conditions to identify the constants so the solution is particular to these given conditions.

Homogeneous linear differential equation

A homogeneous linear differential equation is a special kind of differential equation where all terms depend on the function and its derivatives, and there are no free-standing terms. In our equation \(x'' + x = 0\), both terms involve the function \(x\).

These types of equations usually have solutions in the form of linear combinations of functions. For our given example, the general solution is \(x(t) = c_1 \cos t + c_2 \sin t\), where \(c_1\) and \(c_2\) are constants. The term 'homogeneous' indicates the equation has no additional terms; it simply balances itself out to zero for all solutions.

• The superposition principle applies, which means any linear combination of solutions of a homogeneous equation is also a solution.
• This characteristic aids in the understanding of the solution’s behavior and complexity.

Applying initial conditions helps simplify this linear combination to one specific solution that fits them, known as a particular solution.

Particular solution

Finding a particular solution involves substituting initial conditions into the general solution to determine the constants (e.g., \(c_1\) and \(c_2\)) that define the specific solution which satisfies these conditions. In the initial value problem outlined, we utilize \(x(t) = c_1 \cos t + c_2 \sin t\) as the general solution. The given initial conditions are then used to solve for \(c_1\) and \(c_2\).

We first substitute \(x(\pi/4) = \sqrt{2}\) to find a relationship between \(c_1\) and \(c_2\), giving us an equation. Then, using \(x'(\pi/4) = 2\sqrt{2}\), we derive a second equation. Solving these simultaneously gives us the specific values of \(c_1\) and \(c_2\).

• Solving such conditions leads us to the exact particular solution: \(x(t) = (\sqrt{2} - 2) \cos t + (\sqrt{2} + 2) \sin t\).
• This solution represents one unique solution from the family of solutions provided by the general formula.

This approach shows how particular solutions adapt to fit specific initial conditions, giving us insights into the behavior of dynamic systems defined by differential equations.