Question

What is the constant `m`, if`y = e^(mx)` satisfies `(d^2y)/(dx^2) - 5dy/dx + 6y = 0` ?

Answer 1

`m = 2` or `m= 3`

Explanation:

Given: `y = e^(mx)`

`dy/dx = me^(mx)`

`(d^2y)/(dx^2) = m^2e^(mx)`

Substituting into the equation `(d^2y)/(dx^2) - 5dy/dx + 6y = 0`:

`m^2e^(mx) - 5me^(mx) + 6e^(mx) = 0`

Dividing both sides by `e^(mx)`

`m^2 - 5m + 6 = 0`

Factor:

`(m -2)(m-3)=0`

`m = 2` or `m= 3`

Answer 2

`m=3, m=2`

Explanation:

`y=e^(mx)`

`dy/dx=me^(mx)`

`(d^2y)/(dx^2)=m^2e^(mx)`

`"Differential equation is"`

`(d^2y)/(dx^2)-5(dy)/dx+6y=0`

`"Substituting"`

`m^2e^(mx)-5me^(mx)+6e^(mx)=0`

`(m^2-5m+6)e^(mx)=0`

`m^2-5m+6=0`

`m^2-2m-3m+6=0`

`m(m-3)-2(m-3)=0`

`(m-3)(m-2)=0`

`m-3=0, or m-2=0`

`m=3, m=2`

Answer 3

m=3 or 2

Explanation:

`(dy)/dx=me^(mx)`

`(d^2y)/dx^2=m^2e^(mx)`

Substituting is equation,

`(d^2y)/dx^2 -5(dy)/dx +6y=0`

`m^2e^(mx) -5me^(mx)+6e^(mx)=0`

Dividing both sides by `e^(mx)`,

`m^2-5m+6=0`

`(m-3)(m-2)=0`

`m=3,2`