Solving a simple Ordinary, first order differential equation

I have been having a problem with this simple equation. It is asking me this: Find all values of $k$ for which the function $y=\sin(kt)$ satisfies the differential equation $y′′+ 7y = 0$.

I have found the second derivative, plugged it back into the differential equation, and found out $\sqrt{7}$ checked, but $-\sqrt{7}$ did not. Please help me solve this question. WebWork says I am incorrect and simply do not know where.

• "$\sqrt{7}$ checked" I don't know what this means. – user223391 Sep 30 '15 at 6:02
• Meaning when I checked for that value of k, the equation was satisfied, both sides are equal to 0. – Juan Sep 30 '15 at 6:06
• So should $-\sqrt{7}$ show full working out please. – Chinny84 Sep 30 '15 at 6:11
• In practice, we usually don't use $-\sqrt 7$, and instead use $-\sin (\sqrt 7 t$), using the fact that sin is odd. (Because in practice, we usually put arbitrary constants in front of the functions, which can be positive or negative) – Alan Sep 30 '15 at 6:26

$$(\sin(kt))''+7\sin(kt)=(7-k^2)\sin(kt).$$
• @Juan Another way of looking at it is that this is an autonomous differential equation, so it behaves the same at any point in time. Therefore any phase shift of the sin will also be a solution to the differential equation, but because you are only allowed to choose $k$, then the only phase shift you can choose is integer multiples of 180°. – Kwin van der Veen Sep 30 '15 at 9:57
• @fibonatic: mh, well observed, but the problem statement does not allow a phase shift and it is rather indirect reasoning that $\sin(\sqrt7(x+\pi/\sqrt7))=\sin(-\sqrt7x)$, isn't it ? – Yves Daoust Sep 30 '15 at 10:04