Find the General Solution of the given Second-Order Differential Equation y'' − 8y' + 17y = 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'' − 8y' + 17y = 0 is y = e^{4x} ( A cos(x) + B sin(x) ).

Let's solve this step by step.

Explanation:

Given that, y'' − 8y' + 17y = 0

The characteristic equation for the given equation y'' − 8y' + 17y = 0 is r² - 8r + 17 = 0

Use the quadratic equation formula,

r = [−b ± √(b² − 4ac)] / 2a

Here, a = 1, b = -8, c = 17.

r = [−{-8} ± √{-8}² − 4 × 1 × 17)] / 2{1}

r = [8 ± √64 − 68] / 2

r = [8 ± 2i] / 2

r = 4 ± i

We know that general solution of a second order differential equations with complex roots v ± wi is given by y = e^vx ( A cos(wx) + B sin(wx) )

Substitute v = 4 and w = 1.

⇒ y = e^4x ( A cos(x) + B sin(x) )

Thus, The General Solution of the given Second-Order Differential Equation y'' − 8y' + 17y = 0 is y = e^4x ( A cos(x) + B sin(x) ).