# Find the General Solution of the given Second-Order Differential Equation y'' − 6y' + 10y = 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'' − 6y' + 10y = 0 is y = e^{3x} ( C cos(x) + D sin(x) )

Let's solve this step by step.

**Explanation:**

Given that, y'' − 6y' + 10y = 0

The characteristic equation for the given equation y'' − 6y' + 10y = 0 is r^{2} - 6r + 10 = 0

Use the quadratic equation formula,

r = [−b ± √(b^{2 }− 4ac)] / 2a

Where, a = 1, b = -6, c = 10.

r = [−{-6} ± √{-6}^{2 }− 4 × 1 × 10)] / 2{1}

r = [6 ± √36^{ }− 40] / 2

r = [6 ± 2i] / 2

r = 3 ± i

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

Substitute v = 3 and w = 1.

⇒ y = e^{3x} ( C cos(x) + D sin(x) )