Question

Show that \( y = c_1 e^{ax} \cos(bx) + c_2 e^{ax} \sin(bx) \), where \( c_1, c_2 \) are constants, is a solution of the differential equation \( \frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0 \).

Hint:

When differentiating complex products multiple times, look for opportunities to substitute back the original function (y) or its lower-order derivatives to simplify the expressions. This can significantly reduce the complexity of the algebra.

Answer 1

Step 1: Understanding the Concept:
To show that a given function is a solution to a differential equation, we must find the necessary derivatives of the function, substitute them into the equation, and verify that the equation holds true (i.e., it simplifies to \( 0=0 \)).
Step 2: Key Formula or Approach:
1. Factor the given function for easier differentiation: \( y = e^{ax}(c_1 \cos(bx) + c_2 \sin(bx)) \).
2. Find the first derivative \( \frac{dy}{dx} \) using the product rule.
3. Find the second derivative \( \frac{d^2y}{dx^2} \) by differentiating the first derivative.
4. Substitute \( y, \frac{dy}{dx}, \) and \( \frac{d^2y}{dx^2} \) into the given differential equation and simplify.
Step 3: Detailed Explanation or Calculation:
The given function is \( y = e^{ax}(c_1 \cos(bx) + c_2 \sin(bx)) \).
First Derivative:
Using the product rule: \[ \frac{dy}{dx} = (ae^{ax})(c_1 \cos(bx) + c_2 \sin(bx)) + (e^{ax})(-bc_1 \sin(bx) + bc_2 \cos(bx)) \] We can rewrite this by substituting \( y \): \[ \frac{dy}{dx} = ay + e^{ax}(-bc_1 \sin(bx) + bc_2 \cos(bx)) . (1) \] Second Derivative:
Differentiate equation (1) with respect to x: \[ \frac{d^2y}{dx^2} = a\frac{dy}{dx} + \frac{d}{dx} [e^{ax}(-bc_1 \sin(bx) + bc_2 \cos(bx))] \] Apply the product rule to the second term: \[ \frac{d^2y}{dx^2} = a\frac{dy}{dx} + [ (ae^{ax})(-bc_1 \sin(bx) + bc_2 \cos(bx)) + e^{ax}(-b^2c_1 \cos(bx) - b^2c_2 \sin(bx)) ] \] From equation (1), we know that \( e^{ax}(-bc_1 \sin(bx) + bc_2 \cos(bx)) = \frac{dy}{dx} - ay \). Substitute this into the first part of the bracket: \[ \frac{d^2y}{dx^2} = a\frac{dy}{dx} + \left[ a(\frac{dy}{dx} - ay) + e^{ax}(-b^2(c_1 \cos(bx) + c_2 \sin(bx))) \right] \] Recognize that \( e^{ax}(c_1 \cos(bx) + c_2 \sin(bx)) \) is \( y \): \[ \frac{d^2y}{dx^2} = a\frac{dy}{dx} + a\frac{dy}{dx} - a^2y - b^2y \] \[ \frac{d^2y}{dx^2} = 2a\frac{dy}{dx} - (a^2 + b^2)y \] Verification:
Rearrange the equation for the second derivative: \[ \frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0 \] This is identical to the given differential equation. Therefore, the function \( y \) is a solution.
Step 4: Final Answer:
By finding the first and second derivatives and rearranging the terms, we have shown that the given function \( y \) satisfies the differential equation \( \frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0 \). Hence proved.