The differential equation which represents the family of curves `y=e^(cx)` is :

Options

A. `y'=cy`
B. `xy'-logy = 0`
C. `x log y =y y'`
D. `y log y = xy'`

Answer

To find the differential equation that represents the family of curves given by \( y = e^{cx} \), where \( c \) is an arbitrary constant, we will follow these steps:

### Step 1: Differentiate the given equation
Start with the equation:
\[ y = e^{cx} \]
Differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{cx}) = c e^{cx} \]

### Step 2: Substitute \( e^{cx} \) in terms of \( y \)
From the original equation, we know that:
\[ e^{cx} = y \]
Substituting this into the derivative gives:
\[ \frac{dy}{dx} = c y \]

### Step 3: Eliminate the arbitrary constant \( c \)
Now, we need to express \( c \) in terms of \( y \) and \( \frac{dy}{dx} \). Rearranging the equation from Step 2 gives:
\[ c = \frac{1}{y} \frac{dy}{dx} \]

### Step 4: Substitute \( c \) back into the equation
Now we substitute \( c \) back into the equation \( \frac{dy}{dx} = c y \):
\[ \frac{dy}{dx} = \left(\frac{1}{y} \frac{dy}{dx}\right) y \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{dy}{dx} \]
This does not provide a new equation, so we need to manipulate our earlier equations.

### Step 5: Form the final differential equation
From \( \frac{dy}{dx} = c y \), we can express it as:
\[ \frac{dy}{dx} - c y = 0 \]
Since we have \( c = \frac{1}{y} \frac{dy}{dx} \), we can substitute this back into the equation to eliminate \( c \):
\[ \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} y = 0 \]
This simplifies to:
\[ \frac{dy}{dx} - \frac{dy}{dx} = 0 \]
Thus, the differential equation representing the family of curves \( y = e^{cx} \) is:
\[ \frac{dy}{dx} = \frac{y}{y} \]
However, we can express this as:
\[ \frac{dy}{dx} = ky \quad (where \, k \, is \, a \, constant) \]
This is a first-order linear differential equation.

### Final Form
The final form of the differential equation is:
\[ \frac{dy}{dx} = ky \]
where \( k \) is a constant related to \( c \).