# Verify y1 solves 2nd order differential equation

#### accountkiller

1. The problem statement, all variables and given/known data
Verify that y1 = (e^x) * (cos x) solves the linear, homogeneous, 2nd order differential equation y'' - 2y' + 2y = 0

2. Relevant equations

3. 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

Homework Helper
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

Homework Helper
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!

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