For what values of r does the function y = e^rx satisfy the differential equation 2y'' + y' - y = 0.

To find the values of r for the function f (x) = e^rx, we will use the chain rule of differentiation and the concept of the differential equation.

Answer: The values of r that saisfies the differential equation 2y'' + y' - y = 0 is 1/ 2 or - 1.

Let's find the values of r.

Explanation:

Let f ( x ) = e^rx = y --------------- (1)

⇒ 2y'' + y' - y = 0 --------------- (2)

By using the chain rule, let's differentiate (1),

⇒ y = e^rx

⇒ y' = r e^rx

By again differentiating the equation, we get

⇒ y'' = r²e^rx

By substitution the values of y, y' and y'' in (2), we get

⇒ 2y'' + y' - y = 0

⇒ 2 (r²e^rx ) + re^rx - e^rx = 0

⇒ e^rx ( 2r² + r - 1) = 0

⇒ 2r² + r - 1 = 0

Factorise the above equation to find the value of r.

⇒ 2r² + 2r - r - 1 = 0

⇒ 2 r ( r + 1) - 1 ( r + 1) = 0

⇒ ( 2r - 1) ( r + 1) = 0

⇒ ( 2r - 1) = 0 or ( r + 1) = 0

⇒ r = 1/ 2 or - 1

We can use Cuemath's Online Derivative Calculator to find the derivative of a function.