if `y=ae^(x)+be^(-3x)+c`, then the value of `((d^(3)y)/(dx^(3))+2(d^(2)y)/(dx^(2)))/((dy)/(dx))` is

Explanation

To solve the problem, we need to find the value of the expression

\[ \frac{\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2}}{\frac{dy}{dx}} \]

given that

\[ y = ae^x + be^{-3x} + c. \]

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

We start by differentiating \( y \):

\[ \frac{dy}{dx} = \frac{d}{dx}(ae^x + be^{-3x} + c). \]

Using the chain rule and the fact that the derivative of \( e^x \) is \( e^x \) and the derivative of \( e^{-3x} \) is \( -3e^{-3x} \):

\[ \frac{dy}{dx} = a e^x - 3b e^{-3x}. \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(a e^x - 3b e^{-3x}). \]

Differentiating again:

\[ \frac{d^2y}{dx^2} = a e^x + 9b e^{-3x}. \]

### Step 3: Find the third derivative \( \frac{d^3y}{dx^3} \)

Next, we differentiate \( \frac{d^2y}{dx^2} \):

\[ \frac{d^3y}{dx^3} = \frac{d}{dx}(a e^x + 9b e^{-3x}). \]

Differentiating again:

\[ \frac{d^3y}{dx^3} = a e^x - 27b e^{-3x}. \]

### Step 4: Substitute into the expression

Now we substitute \( \frac{d^3y}{dx^3} \) and \( \frac{d^2y}{dx^2} \) into the original expression:

\[ \frac{\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2}}{\frac{dy}{dx}} = \frac{(a e^x - 27b e^{-3x}) + 2(a e^x + 9b e^{-3x})}{(a e^x - 3b e^{-3x})}. \]

### Step 5: Simplify the numerator

Combine the terms in the numerator:

\[ = \frac{a e^x - 27b e^{-3x} + 2a e^x + 18b e^{-3x}}{a e^x - 3b e^{-3x}}. \]

This simplifies to:

\[ = \frac{(3a e^x - 9b e^{-3x})}{(a e^x - 3b e^{-3x})}. \]

### Step 6: Cancel common factors

Now we can see that we can factor out 3 from the numerator:

\[ = 3 \cdot \frac{(a e^x - 3b e^{-3x})}{(a e^x - 3b e^{-3x})}. \]

Since the numerator and denominator are the same, they cancel out:

\[ = 3. \]

### Final Answer

Thus, the value of the expression is:

\[ \boxed{3}. \]