Find the General Solution of the given Second-Order Differential Equation y'' − y' − 12y = 0

We will be using the concept of second-order differential equations to solve this.

Answer: The General Solution of the given Second-Order Differential Equation y'' − y' − 12y = 0 is y = A e^{-4x} + B e^{3x}

Let's solve this step by step.

Explanation:

Given that, y'' − y' − 12y = 0

The characteristic equation for the given equation y'' − y' − 12y = 0 is r² - r - 12 = 0

Use the quadratic equation formula, r = [−b ± √(b² − 4ac)] / 2a

Here, a = 1, b = -1, c = -12.

r = [−{-1} ± √{-1}² + 4 × 1 × 12)] / 2

r = [1 ± √1 + 48] / 2

r = [1 ± 7] / 2

r = 4, -3

We know that the general solution of a second-order differential equations with real roots a and b is given by y = A e^ax + B e^bx

Substitute a = 4 and b = -3

⇒ y = A e^-4x + B e^3x

Thus, The General Solution of the given Second-Order Differential Equation y'' − y' − 12y = 0 is y = A e^-4x + B e^3x