Y'' = -y

1. Jan 13, 2008

snipez90

Hey guys, I was going through my differential equations packet and an example exercise illustrated that the differential equation y'' = -y has solution y = A*sin(x) + B*cos(x) for constants A and B. I understand why this is true but I was wondering if this was the only solution or at the very least if is most popular one. This also leads me to wonder if the diff eq y' = y has y = Ce^x as its only solution for a constant C.

2. Jan 13, 2008

nicksauce

Yes that is the only solution.
Yes, Ce^x is the only solution of y' = y.

3. Jan 13, 2008

Rainbow Child

Every differential equation

$$y^{(n)}(x)=f\left(x,y(x),\dots,y^{(n-1)}(x)\right)$$

has n independent solutions. Thus

$$y''(x)+y(x)=0 \Rightarrow y(x)=A\,\cos x+B\,\sin x \rightarrow \text{General Solution}$$

$$y'(x)-y(x)=0 \Rightarrow y(x)=C\,e^x \rightarrow \text{General Solution}$$

4. Jan 13, 2008

kdv

there is another popular but completely equivalent way to write the solution: $$y= A sin (x + \phi)$$ where phi is the phase constant. (Of course, you may as well use cos instead of sine). That's a form often used in phyiscs because it shows clearly that the solution is a simple sinusoidal with phase constant and amplitude determined by the initial conditions.

5. Jan 13, 2008

snipez90

whoa, thanks for the replies. the solution to y'' = -y seems much less obvious than y' = y. very nice result though.