The differential equation of the family of curves y=a cos (x + b) is

Options

A. `(d^2y)/(dx^2)-y=0`
B. `(d^2y)/(dx^2)+y=0`
C. `(d^2y)/(dx^2)+2y=0`
D. `(d^2y)/(dx^2)-2y =0`

Answer

To find the differential equation of the family of curves given by \( y = a \cos(x + b) \), we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation:
\[ y = a \cos(x + b) \]
To eliminate the constants \( a \) and \( b \), we differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = -a \sin(x + b) \]

### Step 2: Express \( a \) in terms of \( y \)
From the original equation, we can express \( a \) as:
\[ a = \frac{y}{\cos(x + b)} \]

### Step 3: Substitute \( a \) into the derivative
Now we substitute \( a \) into the derivative we found in Step 1:
\[ \frac{dy}{dx} = -\frac{y}{\cos(x + b)} \sin(x + b) \]
This simplifies to:
\[ \frac{dy}{dx} = -y \tan(x + b) \]

### Step 4: Eliminate \( b \)
To eliminate \( b \), we can differentiate \( \frac{dy}{dx} \) again:
\[ \frac{d^2y}{dx^2} = -\frac{dy}{dx} \tan(x + b) - y \sec^2(x + b) \]
However, we can also express \( \tan(x + b) \) in terms of \( y \) and \( \frac{dy}{dx} \).

### Step 5: Form the final differential equation
We can rearrange the equation to form a second-order differential equation:
\[ \frac{dy}{dx} + y \tan(x + b) = 0 \]

Since \( b \) is still present, we can express it in terms of \( y \) and \( \frac{dy}{dx} \) to eliminate it completely, leading us to the final form of the differential equation.

### Final Differential Equation
The final differential equation representing the family of curves \( y = a \cos(x + b) \) is:
\[ \frac{dy}{dx} + \frac{y}{\sqrt{1 - \left(\frac{y}{a}\right)^2}} = 0 \]