Form the differential equation of the family of curves :
`y= A e^(2x)+ B e^(-3x)`

Answer

To form the differential equation of the family of curves given by the equation \( y = A e^{2x} + B e^{-3x} \), we will follow these steps:

### Step 1: Differentiate the given equation with respect to \( x \)

We start with the equation:
\[ y = A e^{2x} + B e^{-3x} \]

Differentiating both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(A e^{2x}) + \frac{d}{dx}(B e^{-3x}) \]
Using the chain rule:
\[ \frac{dy}{dx} = 2A e^{2x} - 3B e^{-3x} \]

### Step 2: Differentiate again to find the second derivative

Now, we differentiate \( \frac{dy}{dx} \) again with respect to \( x \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2A e^{2x}) + \frac{d}{dx}(-3B e^{-3x}) \]
Applying the chain rule again:
\[ \frac{d^2y}{dx^2} = 4A e^{2x} + 9B e^{-3x} \]

### Step 3: Express \( A \) and \( B \) in terms of \( y \) and its derivatives

From the first derivative, we have:
\[ \frac{dy}{dx} = 2A e^{2x} - 3B e^{-3x} \quad \text{(1)} \]
From the second derivative, we have:
\[ \frac{d^2y}{dx^2} = 4A e^{2x} + 9B e^{-3x} \quad \text{(2)} \]

### Step 4: Solve for \( A e^{2x} \) and \( B e^{-3x} \)

We can express \( A e^{2x} \) and \( B e^{-3x} \) from equations (1) and (2):
1. From equation (1):
\[ 2A e^{2x} = \frac{dy}{dx} + 3B e^{-3x} \]
Thus,
\[ A e^{2x} = \frac{1}{2} \left( \frac{dy}{dx} + 3B e^{-3x} \right) \]

2. From equation (2):
\[ 4A e^{2x} = \frac{d^2y}{dx^2} - 9B e^{-3x} \]
Thus,
\[ A e^{2x} = \frac{1}{4} \left( \frac{d^2y}{dx^2} - 9B e^{-3x} \right) \]

### Step 5: Substitute \( A e^{2x} \) and \( B e^{-3x} \) into one of the equations

Now, we can set the two expressions for \( A e^{2x} \) equal to each other:
\[ \frac{1}{2} \left( \frac{dy}{dx} + 3B e^{-3x} \right) = \frac{1}{4} \left( \frac{d^2y}{dx^2} - 9B e^{-3x} \right) \]

### Step 6: Simplify the equation

Multiply through by 4 to eliminate the fractions:
\[ 2 \left( \frac{dy}{dx} + 3B e^{-3x} \right) = \frac{d^2y}{dx^2} - 9B e^{-3x} \]
Rearranging gives:
\[ \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} - 3B e^{-3x} - 9B e^{-3x} = 0 \]
Combine like terms:
\[ \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 6B e^{-3x} = 0 \]

### Step 7: Eliminate \( B e^{-3x} \)

Since \( B e^{-3x} \) can be expressed in terms of \( y \):
From the original equation:
\[ B e^{-3x} = y - A e^{2x} \]

Substituting this back into the equation gives us the final form of the differential equation:
\[ \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} - 6(y - A e^{2x}) = 0 \]

### Final Differential Equation

After simplification, we arrive at the final differential equation:
\[ \frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0 \]