Question

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

y' + 2y =0

Answer 1

m = - 2 is the value of m that the function y= e^mx is a solution of the differential equation, y' + 2y = 0.

Explanation:

To determine all values of m so that the function y= e^mx is a solution of the given differential equation.

First, we will find y'.

From, y= e^mx

`y = e^(mx)`

But,
`y' = (d)/(dx)y`

Hence,

`y' = (d)/(dx)e^(mx)\\`

∴
`y' = me^(mx)`

Now, we will put the values of y' and y into the given differential equation y' + 2y =0

From the question,

`y = e^(mx)`

and

`y' = me^(mx)`

Then,
`y' + 2y =0\\` becomes

`me^(mx) + 2(e^(mx)) = 0`

Then,
`me^(mx) = - 2(e^(mx)) \\\\`

`m = (-2(e^(mx)) )/(e^(mx) ) \\`

∴
`m = -2`

Hence, m = - 2 is the value of m that the function y= e^mx is a solution of the differential equation, y' + 2y = 0.