I'm currently studying ODE's using the textbook Advanced Engineering Mathematics 10e (Kreyszig, 2019) and had a question regarding solving ODE's. In case anyone's wondering, this is exercise problem 6 on page 8.

Solve the ODE by integration or by remembering a differentiation formula.

$$y^{''} = -y$$

The solution that I got is $y = \sin(x) + C$, since to me it seemed obvious that if you differentiate $y = \sin(x)$ twice you get $-\sin{x}$, or $-y$. However, the solution that I've found here states that the solution is actually:

$$y = C_1\sin(x) + C_2\cos(x)$$

Perhaps this is due to the fact that my calculus background is relatively weak, but I'm having trouble understanding how I managed to get a solution that's drastically different from the correct one.

Would anybody be kind enough to help me understand this process? Thank you.

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    $\begingroup$ You need to find all solutions. It is also obvious that $y=\cos x$ is a solution as well, and it is not included in your solution. $\endgroup$ – A.Γ. Oct 12 '19 at 7:19
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    $\begingroup$ Note that not only is your solution incomplete (missing the cosine part), it's also incorrect if $C \neq 0$, since $y'' = -\sin x \neq -(\sin x + C) = -y$. $\endgroup$ – Hans Lundmark Oct 12 '19 at 8:57

The equation is linear, so with $\sin x$ a solution, also $C\sin x$ is a solution.

The equation is autonomous, thus invariant under time shifts. So also $C\sin(x+D)$ is a solution. This is especially true for $D=\frac\pi2$, so that also $C\cos x$ are solutions

By linearity, also $C_1\cos x+C_2\sin x$ are solutions. Note that these can also be expressed as the previous form $C\sin(x+D)$.

By a dimension argument about order and independent parameters in a general solution of a linear DE, these are all possible solutions.


This can be solved by looking at the characteristic equation, which yields: $$y''=-y \implies r^2+1=0 \implies r=\pm i \implies y(t) = c_1\cos(t)+c_2\sin(t) $$

For a better reference, see section 2.2 of Kreyszig's book


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