The differential equation satisfied by family of curves `y = Ae ^(x)+Be ^(3x) + Ce^(5x)` where A,B,C are 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 \( y = Ae^{x} + Be^{3x} + Ce^{5x} \), where \( A, B, C \) are arbitrary constants, we will follow these steps:

### Step 1: Differentiate the function \( y \)

We start with the function:
\[ y = Ae^{x} + Be^{3x} + Ce^{5x} \]

Now, we differentiate \( y \) with respect to \( x \) to find \( y' \):
\[ y' = \frac{dy}{dx} = Ae^{x} + 3Be^{3x} + 5Ce^{5x} \]

### Step 2: Differentiate again to find \( y'' \)

Next, we differentiate \( y' \) to find the second derivative \( y'' \):
\[ y'' = \frac{d^2y}{dx^2} = Ae^{x} + 9Be^{3x} + 25Ce^{5x} \]

### Step 3: Differentiate a third time to find \( y''' \)

Now, we differentiate \( y'' \) to find the third derivative \( y''' \):
\[ y''' = \frac{d^3y}{dx^3} = Ae^{x} + 27Be^{3x} + 125Ce^{5x} \]

### Step 4: Form the differential equation

Now we have:
- \( y = Ae^{x} + Be^{3x} + Ce^{5x} \)
- \( y' = Ae^{x} + 3Be^{3x} + 5Ce^{5x} \)
- \( y'' = Ae^{x} + 9Be^{3x} + 25Ce^{5x} \)
- \( y''' = Ae^{x} + 27Be^{3x} + 125Ce^{5x} \)

To eliminate the constants \( A, B, C \), we can form a linear combination of these derivatives. We can express the third derivative \( y''' \) in terms of \( y, y', y'' \).

By observing the coefficients, we can set up the following equation:
\[ y''' - 27y' + 9y'' - 15y = 0 \]

This is the differential equation satisfied by the family of curves.

### Final Differential Equation
Thus, the differential equation is:
\[ y''' - 27y' + 9y'' - 15y = 0 \]

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