The differential equation for `y=Ae^(3x)+Be^(2x)` is

Options

A. `(d^(2)y)/(dx^(2))-5(dy)/(dx)+6y=0`
B. `(d^(2)y)/(dx^(2))+5(dy)/(dx)-6y=0`
C. `(d^(2)y)/(dx^(2))-5(dy)/(dx)-6y=0`
D. `(d^(2)y)/(dx^(2))+5(dy)/(dx)+6y=0`

Answer

To find the differential equation for \( y = Ae^{3x} + Be^{2x} \), we will follow these steps:

### Step 1: Find the first derivative \( \frac{dy}{dx} \)
Given:
\[ y = Ae^{3x} + Be^{2x} \]
Taking the derivative with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(Ae^{3x}) + \frac{d}{dx}(Be^{2x}) \]
Using the chain rule:
\[ \frac{dy}{dx} = 3Ae^{3x} + 2Be^{2x} \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)
Now, we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(3Ae^{3x}) + \frac{d}{dx}(2Be^{2x}) \]
Using the chain rule again:
\[ \frac{d^2y}{dx^2} = 9Ae^{3x} + 4Be^{2x} \]

### Step 3: Formulate the differential equation
We have:
1. \( y = Ae^{3x} + Be^{2x} \) (Equation 1)
2. \( \frac{dy}{dx} = 3Ae^{3x} + 2Be^{2x} \) (Equation 2)
3. \( \frac{d^2y}{dx^2} = 9Ae^{3x} + 4Be^{2x} \) (Equation 3)

Now, we can express \( \frac{d^2y}{dx^2} \) in terms of \( y \) and \( \frac{dy}{dx} \).

### Step 4: Express \( \frac{d^2y}{dx^2} \) in terms of \( y \) and \( \frac{dy}{dx} \)
From Equation 1, we can express \( Be^{2x} \) in terms of \( y \):
\[ Be^{2x} = y - Ae^{3x} \]

Substituting this into Equation 3:
\[ \frac{d^2y}{dx^2} = 9Ae^{3x} + 4(y - Ae^{3x}) \]
Simplifying:
\[ \frac{d^2y}{dx^2} = 9Ae^{3x} + 4y - 4Ae^{3x} = (9A - 4A)e^{3x} + 4y \]
\[ \frac{d^2y}{dx^2} = 5Ae^{3x} + 4y \]

### Step 5: Substitute \( \frac{dy}{dx} \) into the equation
From Equation 2, we can express \( 3Ae^{3x} \) in terms of \( \frac{dy}{dx} \):
\[ 3Ae^{3x} = \frac{dy}{dx} - 2Be^{2x} \]
Substituting this into the previous equation gives us:
\[ \frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0 \]

### Final Differential Equation
Thus, the differential equation is:
\[ \frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0 \]

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