Show that y1=e^-3x and y2=e^4x are the solutions of d^2y/dx^2-dy/dx-12y=0

Expert Answers

An illustration of the letter 'A' in a speech bubbles

It has to be shown that `y1 = e^(-3x) and y2 = e^(4x)` are the solutions of `(d^2y)/dx^2 - dy/dx - 12*y = 0`

y = e^-3x

=> dy/dx = -3*e^-3x and `(d^2y)/dx^2 = 9*e^(-3x)`

`(d^2y)/dx^2 - dy/dx - 12y`

=> `9*e^(-3x) + 3*e^(-3x) - 12*e^(-3x)`

=> 0

Similarly, if...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

It has to be shown that `y1 = e^(-3x) and y2 = e^(4x)` are the solutions of `(d^2y)/dx^2 - dy/dx - 12*y = 0`

y = e^-3x

=> dy/dx = -3*e^-3x and `(d^2y)/dx^2 = 9*e^(-3x)`

`(d^2y)/dx^2 - dy/dx - 12y`

=> `9*e^(-3x) + 3*e^(-3x) - 12*e^(-3x)`

=> 0

Similarly, if y = e^4x

`dy/dx =4*e^(4x)`

`(d^2y)/(dx^2) = 16*e^(4x)`

`(d^2y)/dx^2 - dy/dx - 12y`

=> `16*e^(4x) - 4*e^(4x) - 12*e^(4x)`

=> 0

This proves that `y1 = e^(-3x) and y2 = e^(4x)` are the solutions of `(d^2y)/dx^2 - dy/dx - 12y = 0`

Approved by eNotes Editorial Team