Verify y1 solves 2nd order differential equation

[accountkiller]

Homework Statement

Verify that y1 = (e^x) * (cos x) solves the linear, homogeneous, 2nd order differential equation y'' - 2y' + 2y = 0

Homework Equations

The Attempt at a Solution

So I used the product rule but just kept going in circles with the cos x. I remember there's a way to do this so you don't go in circles, but I forgot! I'd appreciate any help.

 

[rock.freak667]

What did you get when you differentiated it?

 

[accountkiller]

So the product rule is fg' = fg - int(gf').
I said f = cos x, so f' = -sin x dx, and g' = e^x, so g = e^x.
This gives me e^x(cos x) - int(e^x(-sin x) dx) = e^x(cos x) + int(e^x(sin x) dx).
So then I have to use the product rule again for that integral, with f = e^x, f' = e^x dx, g' = sin x, and g = -cos x. This gives me -e^x(cos x) + int(e^x(cos x) dx) for that integral, and putting that together with what was in front of it:
y1' = e^x(cos x) - e^x(cos x) - int(e^x(cos x) dx)
y1' = -int (e^x(cos x) dx)

So I'm pretty much back where I started.

 

[rock.freak667]

You do not need to integrate anything.

Just find y₁'', y₁' and you have y, put back into the DE and see if you get zero. If you do, it satisfies the DE.

 

[accountkiller]

OH! I was using integration by parts instead of the product rule. Oi, I shouldn't get integration and differentiation mixed up :P

Thanks!