The differential equation of the family of curves y = p cos (ax) + q sin (ax) , where p , q are arbitrary constants , is :

Options

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

Answer

To find the differential equation of the family of curves given by \( y = p \cos(ax) + q \sin(ax) \), where \( p \) and \( q \) are arbitrary constants, we will follow these steps:

### Step 1: Differentiate the equation once
We start with the equation:
\[ y = p \cos(ax) + q \sin(ax) \]
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = -p a \sin(ax) + q a \cos(ax) \]

### Step 2: Differentiate the equation again
Next, we differentiate \( \frac{dy}{dx} \) to find the second derivative:
\[ \frac{d^2y}{dx^2} = -p a^2 \cos(ax) - q a^2 \sin(ax) \]

### Step 3: Factor out common terms
We can factor out \(-a^2\) from the second derivative:
\[ \frac{d^2y}{dx^2} = -a^2 (p \cos(ax) + q \sin(ax)) \]

### Step 4: Substitute back the original equation
Notice that \( p \cos(ax) + q \sin(ax) \) is simply \( y \):
\[ \frac{d^2y}{dx^2} = -a^2 y \]

### Step 5: Rearranging the equation
We can rearrange this equation to express it in standard form:
\[ \frac{d^2y}{dx^2} + a^2 y = 0 \]

### Final Result
Thus, the differential equation of the family of curves is:
\[ \frac{d^2y}{dx^2} + a^2 y = 0 \]