The differential equation satisfied by family of curves `y = Ae ^(x)+Be ^(3x) + Ce^(5x)` where A,B,C ar arbitrary constants is:

Options

A. `(d ^(3)y)/(dx ^(3)) -9 ( d ^(2)y)/(dx ^(2))+23 (dy )/(dx)-15 y =0`
B. `(d ^(3)y)/(dx ^(3)) -9 ( d ^(2)y)/(dx ^(2))-23 (dy )/(dx)+15 y =0`
C. `(d ^(3)y)/(dx ^(3)) +9 ( d ^(2)y)/(dx ^(2))-23 (dy )/(dx)+15 y =0`
D. `(d ^(3)y)/(dx ^(3)) +9 ( d ^(2)y)/(dx ^(2))+23 (dy )/(dx)-15 y =0`

Answer

To find the differential equation satisfied by the family of curves given by \( y = A e^x + B e^{3x} + C e^{5x} \), where \( A, B, C \) are arbitrary constants, we will differentiate the function multiple times and eliminate the constants.

### Step 1: Differentiate the function once
Given:
\[ y = A e^x + B e^{3x} + C e^{5x} \]
Differentiating with respect to \( x \):
\[ y' = \frac{dy}{dx} = A e^x + 3B e^{3x} + 5C e^{5x} \]

### Step 2: Differentiate the function twice
Now, we differentiate \( y' \):
\[ y'' = \frac{d^2y}{dx^2} = A e^x + 9B e^{3x} + 25C e^{5x} \]

### Step 3: Differentiate the function three times
Next, we differentiate \( y'' \):
\[ y''' = \frac{d^3y}{dx^3} = A e^x + 27B e^{3x} + 125C e^{5x} \]

### Step 4: Form a linear combination to eliminate constants
Now we have:
1. \( y = A e^x + B e^{3x} + C e^{5x} \)
2. \( y' = A e^x + 3B e^{3x} + 5C e^{5x} \)
3. \( y'' = A e^x + 9B e^{3x} + 25C e^{5x} \)
4. \( y''' = A e^x + 27B e^{3x} + 125C e^{5x} \)

We can express \( y'' \), \( y' \), and \( y \) in terms of \( y''' \):
- From \( y''' - 27y' + 125y = 0 \)

### Step 5: Write the final differential equation
Thus, the differential equation satisfied by the family of curves is:
\[ y''' - 27y' + 125y = 0 \]

### Summary
The differential equation satisfied by the family of curves \( y = A e^x + B e^{3x} + C e^{5x} \) is:
\[ \frac{d^3y}{dx^3} - 27 \frac{dy}{dx} + 125y = 0 \]