# 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'' = r^{2}e^{rx}

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

⇒ 2y'' + y' - y = 0

⇒ 2 (r^{2}e^{rx} ) + re^{rx} - e^{rx} = 0

⇒ e^{rx} ( 2r^{2} + r - 1) = 0

⇒ 2r^{2} + r - 1 = 0

Factorise the above equation to find the value of r.

⇒ 2r^{2} + 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.

### Thus, the values of r that saisfies the differential equation 2y'' + y' - y = 0 is 1/ 2 or - 1.

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