Differential Equation Solutions - y'' = -y & y' = y

[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.

 

[nicksauce]

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

 

[Rainbow Child]

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.

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}

 

[kdv]

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.

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.

 

[snipez90]

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