Math Problem Statement

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.

Solution

To find the differential equation of the given family of curves:

y=ekx[Acos(nx)+Bsin(nx)],y = e^{-kx}[A \cos(nx) + B \sin(nx)], where AA and BB are arbitrary constants, we need to eliminate these constants to form a differential equation.

Steps:

  1. Differentiate yy with respect to xx several times.
  2. Eliminate the constants AA and BB.

Step 1: First derivative of yy

We apply the product rule to differentiate y=ekx[Acos(nx)+Bsin(nx)]y = e^{-kx}[A \cos(nx) + B \sin(nx)]:

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

First, we compute each term separately:

  • ddx(ekx)=kekx\frac{d}{dx}(e^{-kx}) = -k e^{-kx},
  • ddx(Acos(nx)+Bsin(nx))=Ansin(nx)+Bncos(nx)\frac{d}{dx}(A \cos(nx) + B \sin(nx)) = -An \sin(nx) + Bn \cos(nx).

Thus, the first derivative becomes:

[ \frac{dy}{dx} = -k e^{-kx} [A \cos(nx) + B \sin(nx)] + e^{-kx} [-An \sin(nx) + Bn \cos(nx)]. ]

Step 2: Second derivative of yy

Next, we differentiate dydx\frac{dy}{dx} with respect to xx again:

[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( -k e^{-kx} [A \cos(nx) + B \sin(nx)] + e^{-kx} [-An \sin(nx) + Bn \cos(nx)] \right). ]

This will give us the second-order differential equation. Let me calculate this next step.The first and second derivatives of yy with respect to xx are:

First derivative:

[ \frac{dy}{dx} = -k e^{-kx} [A \cos(nx) + B \sin(nx)] + e^{-kx} [-An \sin(nx) + Bn \cos(nx)]. ]

Second derivative:

[ \frac{d^2y}{dx^2} = k^2 e^{-kx} [A \cos(nx) + B \sin(nx)] - 2k e^{-kx} [-An \sin(nx) + Bn \cos(nx)] + e^{-kx} [-An^2 \cos(nx) - Bn^2 \sin(nx)]. ]

Step 3: Form the differential equation

Now, we can form a differential equation by eliminating the constants AA and BB. This involves combining the expressions for yy, dydx\frac{dy}{dx}, and d2ydx2\frac{d^2y}{dx^2}.

Let me proceed with forming the differential equation.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like to continue by manually solving the differential equation, or would you prefer an explanation of the steps involved so far? Let me know how you'd like to proceed!

Here are some related questions that may expand on the topic:

  1. How do we interpret arbitrary constants in the solution of a differential equation?
  2. What is the significance of the exponential term ekxe^{-kx} in differential equations?
  3. How do we derive the order of a differential equation from a family of solutions?
  4. Can this type of method be applied to other functions like polynomials?
  5. How do boundary conditions influence the form of differential equations?

Tip: Always check for initial conditions or boundary conditions to find specific solutions to differential equations.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Calculus
Exponential Functions
Trigonometric Functions

Formulas

y = e^{-kx}[A cos(nx) + B sin(nx)]
First derivative: dy/dx = -k e^{-kx} [A cos(nx) + B sin(nx)] + e^{-kx} [-An sin(nx) + Bn cos(nx)]
Second derivative: d^2y/dx^2 = k^2 e^{-kx} [A cos(nx) + B sin(nx)] - 2k e^{-kx} [-An sin(nx) + Bn cos(nx)] + e^{-kx} [-An^2 cos(nx) - Bn^2 sin(nx)]

Theorems

Product Rule
Chain Rule

Suitable Grade Level

Undergraduate Calculus/Advanced High School