I know this is an simple problem but for the life of me Im lost. I am confused on the second order deriv I dont know how to handle it.
Verify that y is a solution of the differential equation.
$\displaystyle y = e^-2x; y'' + y' -2y = 0$
I know this is an simple problem but for the life of me Im lost. I am confused on the second order deriv I dont know how to handle it.
Verify that y is a solution of the differential equation.
$\displaystyle y = e^-2x; y'' + y' -2y = 0$
$\displaystyle y = e^{-2x}$
$\displaystyle y' = -2e^{-2x}$
$\displaystyle y'' = 4e^{-2x}$
Now just sub into the differential equation and verify that the left hand side is 0.
You could even do it like this:
$\displaystyle y' = -2y$
$\displaystyle y'' = 4y$
So
$\displaystyle y'' + y' -2y = 0$
$\displaystyle 4y - 2y - 2y = 0$
-Dan