# Find values of m so that the function y = e^{mx} is a solution of the given differential equation.

# (i) y + 2y = 0 (ii) y” – 5y’ + 6y = 0

A differential equation is an equation that involves the derivative (derivatives) of the dependent variable with respect to the independent variable (variables) is called a differential equation.

## Answer: The values of m so that the function y = e^{mx} is a solution of the given differential equation is: (i) m = -2, (ii) m = 3, 2

Let's look into the steps below

**Explanation:**

Given: y = e^{mx} -------------- (1)

(i) y + 2y = 0

Let's differentiate (1) with respect to x

⇒ dy/dx = y' = me^{mx} --------------- (2)

⇒ y' = m × y, using (1)

⇒ y' - my = 0

Comparing y' - my = 0 with the given equation y + 2y = 0,

We observe that,

m = -2

(ii) y” – 5y’ + 6y = 0

Let's differentiate (2) with respect to x again

⇒ d^{2}y / dx^{2 }= y'' = m^{2}e^{mx} --------------- (3)

Substituting (1), (2) and (3) in the given equation y” – 5y’ + 6y = 0 we get,

⇒ m^{2}e^{mx} - 5me^{mx} + 6e^{mx} = 0

Dividing the above equation throughout by e^{mx} we get,

⇒ m^{2} - 5m + 6 = 0

By splitting the middle term we get,

⇒ m^{2} - 3m - 2m + 6 = 0

⇒ m(m - 3) - 2(m - 3) = 0

⇒ (m - 3) (m - 2) = 0

Thus, m = 3, 2