Finding the Differential Equation by Eliminating Arbitrary Constants

The problem asks us to find the differential equation of minimum order by eliminating the arbitrary constants A and C from the given equation:

\(\qquad y = A [\sin (x + C) + \cos (x + C)]\)

The given equation contains two arbitrary constants, A and C. To eliminate these two constants, we need to differentiate the equation at least two times. The minimum order of the resulting differential equation will be 2, which is equal to the number of arbitrary constants.

Simplifying the Given Equation

Let's simplify the expression inside the square brackets using a trigonometric identity. We know that for any angle \(\theta\), \(\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \frac{\pi}{4})\).

Applying this to our equation with \(\theta = x + C\):

\(\qquad \sin (x + C) + \cos (x + C) = \sqrt{2} \sin((x + C) + \frac{\pi}{4}) = \sqrt{2} \sin(x + C + \frac{\pi}{4})\)

So the original equation becomes:

\(\qquad y = A \left[ \sqrt{2} \sin(x + C + \frac{\pi}{4}) \right]\)

\(\qquad y = (A \sqrt{2}) \sin(x + C + \frac{\pi}{4})\)

Let's introduce new constants. Let \(B = A \sqrt{2}\) and \(D = C + \frac{\pi}{4}\). The equation simplifies to:

\(\qquad y = B \sin(x + D)\)

This form \(y = B \sin(x + D)\) is easier to differentiate. It still contains two arbitrary constants, B (which depends on A) and D (which depends on C). The process of eliminating A and C is equivalent to eliminating B and D from this simplified form.

Differentiating the Equation

We will differentiate \(y = B \sin(x + D)\) with respect to \(x\) twice.

First Derivative (\(y'\)):

Differentiate \(y = B \sin(x + D)\) with respect to \(x\):

\(\qquad y' = \frac{d}{dx} (B \sin(x + D))\)

\(\qquad y' = B \cos(x + D) \cdot \frac{d}{dx}(x + D)\)

\(\qquad y' = B \cos(x + D) \cdot (1 + 0)\)

\(\qquad y' = B \cos(x + D)\)

Second Derivative (\(y''\)):

Differentiate \(y' = B \cos(x + D)\) with respect to \(x\):

\(\qquad y'' = \frac{d}{dx} (B \cos(x + D))\)

\(\qquad y'' = B (-\sin(x + D)) \cdot \frac{d}{dx}(x + D)\)

\(\qquad y'' = -B \sin(x + D) \cdot (1 + 0)\)

\(\qquad y'' = -B \sin(x + D)\)

Eliminating Arbitrary Constants B and D

We now have the following three equations:

1. \(y = B \sin(x + D)\)
2. \(y' = B \cos(x + D)\)
3. \(y'' = -B \sin(x + D)\)

Observe equations (1) and (3). Equation (1) is \(y = B \sin(x + D)\) and equation (3) is \(y'' = -B \sin(x + D)\). We can see that \(y'' = -(B \sin(x + D))\). Since \(y = B \sin(x + D)\), we can substitute \(y\) into the equation for \(y''\):

\(\qquad y'' = -(y)\)

Rearranging this equation, we get:

\(\qquad y'' + y = 0\)

This is a differential equation that does not contain the arbitrary constants A, C (or B, D). Since we differentiated twice to eliminate the two constants, this is the differential equation of minimum order.

Comparing with Options

Let's compare our derived differential equation with the given options:

• Option 1: \(y'' + (\sin x + \cos x) y' = 1\) - Does not match \(y'' + y = 0\).
• Option 2: \(y'' = (\sin x + \cos x) y'\) - Does not match \(y'' + y = 0\).
• Option 3: \(y'' = (y')^2 + \sin x \cos x\) - Does not match \(y'' + y = 0\).
• Option 4: \(y'' + y = 0\) - Matches our derived differential equation.

Thus, the differential equation of minimum order obtained by eliminating the arbitrary constants A and C is \(y'' + y = 0\).

Revision Table: Eliminating Constants

Additional Information: Forming Differential Equations

Forming a differential equation from a given family of curves involves eliminating the arbitrary constants present in the equation representing the family. The order of the resulting differential equation is equal to the number of independent arbitrary constants in the original equation.

Steps involved in forming a differential equation:

1. Identify the number of arbitrary constants in the given equation.
2. Differentiate the equation as many times as the number of arbitrary constants.
3. Eliminate the arbitrary constants from the original equation and its derivatives. This often involves substitution or solving a system of equations.
4. The resulting equation is the differential equation of the family of curves.

In this specific problem, the trigonometric identity \(\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \frac{\pi}{4})\) or \(\sqrt{2} \cos(\theta - \frac{\pi}{4})\) was helpful in simplifying the original expression \(A [\sin (x + C) + \cos (x + C)]\) into a form \(B \sin(x + D)\) or \(B \cos(x + D)\), which made the differentiation and elimination steps more straightforward. While \(y = B \sin(x+D)\) represents a family of sinusoidal functions, the original form \(y = A [\sin (x + C) + \cos (x + C)]\) also represents a family of sinusoidal functions. The sum of two sinusoids of the same frequency is also a sinusoid of that frequency.