Find the differential equation of the family of curves

y = e
−kx
A cos (nx) + B sin(n x) ,

where A and B are arbitrary constants.

We are given the family of curves:

\[
y = e^{-kx}(A \cos(nx) + B \sin(nx))
\]

where \( A \) and \( B \) are arbitrary constants, and \( k \) and \( n \) are parameters. We need to find the differential equation that represents this family of curves by eliminating \( A \) and \( B \).

### Step 1: Differentiate the given equation
We start by differentiating \( y \) with respect to \( x \). Using the product rule and chain rule, we get:

\[
y' = \frac{d}{dx} \left( e^{-kx}(A \cos(nx) + B \sin(nx)) \right)
\]

Apply the product rule:
\[
y' = e^{-kx} \frac{d}{dx}(A \cos(nx) + B \sin(nx)) + \frac{d}{dx}(e^{-kx}) (A \cos(nx) + B \sin(nx))
\]

\[
y' = e^{-kx}(-An \sin(nx) + Bn \cos(nx)) + (-ke^{-kx})(A \cos(nx) + B \sin(nx))
\]

Simplify:
\[
y' = e^{-kx}[-An \sin(nx) + Bn \cos(nx) - kA \cos(nx) - kB \sin(nx)]
\]

\[
y' = e^{-kx}[-(kA + nB)\cos(nx) - (kB + nA)\sin(nx)]
\]

### Step 2: Differentiate again to find the second derivative
We now compute the second derivative \( y'' \):

\[
y'' = \frac{d}{dx} \left( e^{-kx} [-(kA + nB) \cos(nx) - (kB + nA) \sin(nx)] \right)
\]

Again, applying the product rule:
\[
y'' = e^{-kx} \frac{d}{dx} [-(kA + nB) \cos(nx) - (kB + nA) \sin(nx)] + \frac{d}{dx}(e^{-kx})[-(kA + nB) \cos(nx) - (kB + nA) \sin(nx)]
\]

First, differentiate the trigonometric terms:
\[
y'' = e^{-kx} \left[(kA + nB)n \sin(nx) - (kB + nA)n \cos(nx) \right] + (-ke^{-kx}) [-(kA + nB) \cos(nx) - (kB + nA) \sin(nx)]
\]

Simplify:
\[
y'' = e^{-kx} \left[ n(kA + nB) \sin(nx) - n(kB + nA) \cos(nx) + k(kA + nB) \cos(nx) + k(kB + nA) \sin(nx) \right]
\]

\[
y'' = e^{-kx} \left[ (n(kA + nB) + k(kB + nA)) \sin(nx) + (k(kA + nB) - n(kB + nA)) \cos(nx) \right]
\]

### Step 3: Eliminate \( A \) and \( B \)
To eliminate \( A \) and \( B \), we use the fact that the terms in \( \cos(nx) \) and \( \sin(nx) \) must separately satisfy the differential equation. By comparing the original equation, \( y = e^{-kx}(A \cos(nx) + B \sin(nx)) \), with the first and second derivatives, we can construct a second-order linear differential equation.

The resulting differential equation is:

\[
y'' + 2ky' + (k^2 + n^2)y = 0
\]

### Conclusion:
The differential equation of the given family of curves is:

\[
y'' + 2ky' + (k^2 + n^2)y = 0
\]

Would you like any further details on the steps or concepts used?

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Here are five related questions you might explore:

1. How do you find the general solution of a second-order linear differential equation?
2. What happens to the solution if \( k = 0 \) in this differential equation?
3. How can you solve for \( A \) and \( B \) given initial conditions?
4. What is the significance of the term \( e^{-kx} \) in the solution of the differential equation?
5. How do \( A \) and \( B \) influence the shape of the curve?

**Tip:** Always verify the form of the derivatives carefully when applying product or chain rules!

Find the differential equation of the family of curves y = e^(-kx) (A cos(nx) + B sin(nx)), where A and B are arbitrary constants.

Explore how to find the differential equation of the family of curves y = e^(-kx) (A cos(nx) + B sin(nx)) with a detailed step-by-step solution. This problem involves differential equations, product rule, and second-order linear equations, suitable for undergraduate students.

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