# Differential equation with constant inside a trig function

I was doing practice exercise for differential equations and couldn't wrap my head around this one:

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

We need to find a differential equation that $y$ satisfies and the $ODE$ should not contain any constants. If the $C_1$ was outside $\sin$, then the answer would be much easier. I tried using different identities, inlcuding: $$\sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y)$$ $$\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y)$$

But I just could not eliminate the constant because it is inside $\sin$. Any tips would be much appreciated.

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Differentiate twice... – copper.hat Feb 16 '14 at 17:13
But it only has one constant, doesn't it mean the differential equation is only first-order, hence only differentiate once? – SomeGuyOutThere Feb 16 '14 at 17:15
How about $y'=\sqrt{1-y^2}$? – Artem Feb 16 '14 at 17:20

$$y=\sin(x+c)\implies y'=\cos(x+c) \implies y''=-\sin(x+c) \implies y''+y=0$$

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For this problem, differential equation does NOT have to be first order linear. There are more than an ODE that works.

Observe that by the trigonometric identity you provided, $$y(x) = \sin(x + \mbox{c}_1) = \cos(\mbox{c}_1)\sin(x) + \sin(\mbox{c}_1)\cos(x)$$

By calculus, we have the following pattern: \begin{aligned} y'(x) &= \cos(\mbox{c}_1)\cos(x) - \sin(\mbox{c}_1)\sin(x)\\ y''(x) &= -\cos(\mbox{c}_1)\sin(x) - \sin(\mbox{c}_1)\cos(x)\\ y'''(x) &= -\cos(\mbox{c}_1)\cos(x) + \sin(\mbox{c}_1)\sin(x)\\ y^{(4)}(x) &= \cos(\mbox{c}_1)\sin(x) + \sin(\mbox{c}_1)\cos(x) \end{aligned}

So one of the ODEs that works (and also easy to find for this type of problem) is $$y^{(4)} = y$$

Remark: You can also differentiate $y(x) = \sin(x + \mbox{c}_1)$ without using the identity. Using chain rule four times, you get $y^{(4)} = y$.

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Thank you. I thought it was limited to linear ODE. – SomeGuyOutThere Feb 16 '14 at 17:31
Those are linear. – Paul Safier Feb 16 '14 at 18:32