The differential equation of the family of curves represented by y = a + bx + `ce^-x` (where a, b, c are arbitrary constants) is

Options

A. `y^''' = y^'`
B. `y^''' + y^'' =0`
C. `y^''' - y^'' +y^ =0`
D. `y^''' + y^'' -y^` =0`

Answer

To find the differential equation of the family of curves represented by the equation \( y = a + bx + ce^{-x} \), where \( a, b, c \) are arbitrary constants, we will follow these steps:

### Step 1: Differentiate the equation with respect to \( x \)
We start with the given equation:
\[ y = a + bx + ce^{-x} \]
Differentiating both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(a + bx + ce^{-x}) = 0 + b - ce^{-x} \]
Thus, we have:
\[ y' = b - ce^{-x} \]

### Step 2: Differentiate again to find the second derivative
Now we differentiate \( y' \) with respect to \( x \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(b - ce^{-x}) = 0 + c e^{-x} \]
So, we have:
\[ y'' = ce^{-x} \]

### Step 3: Differentiate once more to find the third derivative
Next, we differentiate \( y'' \) with respect to \( x \):
\[ \frac{d^3y}{dx^3} = \frac{d}{dx}(ce^{-x}) = -ce^{-x} \]
Thus, we have:
\[ y''' = -ce^{-x} \]

### Step 4: Relate the derivatives to form the differential equation
From our expressions for \( y'' \) and \( y''' \), we can express \( c \) in terms of \( y'' \):
\[ c = -y''' \]
Substituting this into the equation for \( y'' \):
\[ y'' = -y''' \]
Rearranging gives us the differential equation:
\[ y''' + y'' = 0 \]

### Final Result
The differential equation of the family of curves represented by \( y = a + bx + ce^{-x} \) is:
\[ y''' + y'' = 0 \]

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