# Problem in deriving an ordinary differential equation from a trigonometric equation

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.

• – cqfd
Mar 7, 2019 at 17:06

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.
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 $$y'(t) - \omega\cot(\phi)y(t) = -\omega\csc(\phi)x(t).$$