# Differential Equation Help!

• Jul 12th 2008, 08:10 AM
vodka
Differential Equation Help!
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\$
• Jul 12th 2008, 08:23 AM
topsquark
Quote:

Originally Posted by vodka
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
• Jul 12th 2008, 08:26 AM
vodka
Quote:

Originally Posted by topsquark
\$\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

Duh thanks im an idiot its been awhile since I looked over these.

Cheers Dan