Show that the function
y = 2ex + 2e−x/2
satisfies the differential equation
2y'' − y' − y = 0.

Question

Show that the function 
y = 2ex + 2e−x/2
 satisfies the differential equation 
2y'' − y' − y = 0.

accessdenied · Answer

What have you tried thus far?

anonymous · Answer

I honestly don't know how to do it. My professor did not teach us this.

accessdenied · Answer

Ah, okay.
What we should notice in the differential equation given is that we have y,  the function's derivative y', and the second derivative y".
So, we know y.  If we can find y' and y" through regular differentiation, we'd be able to substitute all three into the differential equation and see if the result gets us an identity -- 0 = 0.

accessdenied · Answer

So our first step becomes:   Find y'  and y" from the function y.
Does this seem better?

anonymous · Answer

Oh okay, yes that makes a lot more sense. Let me do that real quick.

anonymous · Answer

Oh and the ex and 2e-x/2 should be e^x and 2e^(-e/x). Sorry about that.

accessdenied · Answer

Yep, no problem.  :)

anonymous · Answer

$$y'=2e ^{x}-e ^{-x/2}$$
$$y''=2e ^{x}+\frac{ 1 }{ 2 }e^{-x/2}$$

accessdenied · Answer

Looks good to me!
Then we just have to directly substitute those for the y, y', and y" in the differential equation.  We are trying to show that it satisfies the equation, so inevitably we should obtain a result that everything cancels and leaves us with 0 = 0.  :)

anonymous · Answer

I got that! 0=0.
So thank you very much for your help!

accessdenied · Answer

You're welcome!  Glad to help.  :D