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I am reading an article, which i have to do a similar work. But my knowledge in maths is not advanced. There are two time series equation first: \begin{align} x(t) &= \sin(wt) \qquad \tag{ equation 1} \\ y(t) &= \sin(wt + \phi) \tag{ equation 2} \\ y(t) &= \sin(wt)\cos(\phi) + \cos(wt)\sin(\phi) \tag{ equation 3} \end{align}

using equation $1$, $t = (1/w)\arcsin(x)$, replacing this in equation 3, gives equation 4: $$ y = x\cos(\phi) + \cos[\arcsin(x)]\sin(\phi) \qquad \tag{ equation 4}$$

Then, the author just produce the next equation by saying, he derived it from equation 3 by doing an ordinary differential equation in y by eliminating time, thus he got the following:

$$dy(t)/dt - y(t)[w\cot(\phi)] = -w\csc(\phi)x(t) \tag{ equation 5}$$

My question is I am unable to obtain equation $5$ and I don't understand how he did it. Did he used equation $4$ instead of $3$ to get it as we have the $x$ variable in equation.

Can someone please write the full derivation of equation $5$ from the above equations for me.

note: csc is cosecant and cot is cotangent

Thank you.

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2 Answers 2

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First differential equation 3 to get $\frac{dy\left(t\right)}{dt}=w\cos\left(wt\right)\cos\phi-w\sin\left(wt\right)\sin\phi$. Then subtract $y\left(t\right)w\mathrm{ctg}\phi$ from both sides of the equation and simplify. You'll get a cancelation of the $w\cos\left(wt\right)\cos\phi$ term, and finally use the trig identity $\sin^{2}\phi+\cos^{2}\phi=1$ to simplify the remaining 2 terms.

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  • $\begingroup$ Thanks a lot, i was able to do it as you explained. $\endgroup$
    – MathNewbie
    Commented Mar 7, 2019 at 17:40
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So, we have $$ y(t) = \sin(\omega t + \phi) = \sin(\omega t)\cos(\phi) + \cos(\omega t)\sin(\phi). $$ Differentiating this with respect to $t$ gives $$ y'(t) = \omega\cos(\omega t + \phi) = \omega\cos(\omega t)\cos(\phi)-\omega\sin(\omega t)\sin(\phi). $$ At this point, the author wants to eliminate $\cos(\omega t)$ from the equations, as it has no simple expression in terms of $y$ and $x$. Some quick linear algebra gives $$ \sin(\phi)y'(t) - \omega\cos(\phi)y(t) = -\omega\sin(\omega t)\left[\cos(\phi)^2+ \sin(\phi)^2\right] = -\omega \sin(\omega t). $$ Now all that remains is to substitute $\sin(\omega t) = x(t)$ and divide through by $\sin(\phi)$ to get $$ y'(t) - \omega\cot(\phi)y(t) = -\omega\csc(\phi)x(t). $$

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