The differential equation of minimum order by eliminating the arbitrary constants A and C in the equation y = A [sin (x + C) + cos (x + C)] is :

Options

A. `y'' +( sin x + cos x) y' = 1`
B. `y'' = (sin x + cos x ) y'`
C. `y'' = (y')^(2)) sin x cos x`
D. `y'' + y =0`

Answer

To find the differential equation of minimum order by eliminating the arbitrary constants \( A \) and \( C \) in the equation

\[ y = A \left[ \sin(x + C) + \cos(x + C) \right], \]

we will follow these steps:

### Step 1: Differentiate the equation once
First, we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = A \left[ \cos(x + C) + (-\sin(x + C)) \right] = A \left[ \cos(x + C) - \sin(x + C) \right]. \]

### Step 2: Differentiate the equation again
Next, we differentiate \( \frac{dy}{dx} \) to find the second derivative:

\[ \frac{d^2y}{dx^2} = A \left[ -\sin(x + C) - \cos(x + C) \right]. \]

### Step 3: Express the second derivative in terms of \( y \)
Now, we can express the second derivative in terms of \( y \). From the original equation, we know:

\[ y = A \left[ \sin(x + C) + \cos(x + C) \right]. \]

Thus, we can rearrange this to find \( A \):

\[ A = \frac{y}{\sin(x + C) + \cos(x + C)}. \]

However, we can also express \( \sin(x + C) + \cos(x + C) \) in terms of \( y \):

From the second derivative, we have:

\[ \frac{d^2y}{dx^2} = -\left( \sin(x + C) + \cos(x + C) \right) A. \]

Substituting \( A \) into this equation gives us:

\[ \frac{d^2y}{dx^2} = -\frac{y}{\sin(x + C) + \cos(x + C)} \left( \sin(x + C) + \cos(x + C) \right) = -y. \]

### Step 4: Form the differential equation
Thus, we can write the differential equation as:

\[ \frac{d^2y}{dx^2} + y = 0. \]

### Final Answer
The differential equation of minimum order by eliminating the arbitrary constants \( A \) and \( C \) is:

\[ \frac{d^2y}{dx^2} + y = 0. \]