Notes on Differential Equations (2) —— $\frac{dy}{dx}=ay+\cos\omega t$

Here is the part that we get oscillation input, the $\cos\omega t$ part. Or we can think that's a vibration part, or any thing related to circular motion.

$$\frac{dy}{dx}=ay+\cos\omega t$$ when $y=y(0)$ at $t=0$. Look for $y_p(t)=M\cos\omega t+N\sin\omega t$.

The reason that there is a $\sin$ term in the particular is when we take the derivative of $\cos$, it becomes $\sin$. Therefore there will be a $sin$ term. As for the previous one, the derivative of exponential is itself, so only some constant multiples an exponential term in previous form.

1. The first form of the particular solution

First, substitute $y_p(t)$ into the original equation and find $M$ and $N$. After taking derivative of the lefthand side, we have

$$-\omega M\sin\omega t+\omega N\cos\omega t=aM\cos\omega t+aN\sin\omega t+cos\omega t$$

Now the coefficients for $cos$ and $sin$ terms should match at both sides, which is

\begin{align}\left\{\begin{aligned} -aM+\omega N&=1\\ -\omega M-aN&=0 \end{aligned}\right.\end{align}

With simple computation, $M$ and $N$ are

\begin{align}\left\{\begin{aligned} M&=\frac{-a}{\omega^2+a^a}\\ N&=\frac{w}{\omega^2+a^a} \end{aligned}\right.\end{align}

2. The solution in polar coordinate form

Here we get one form of the particular solution, but it can be simplified further. Given that we're dealing with $cos$ and $sin$, it's natural to think of phase and the polar coordinate form, which can combine the $cos$ and $sin$ into a single $cos$ term.

$$y_p(t)=G\cos(\omega t-\alpha)$$

where $G$ denotes the gain, the magnitude, or the amplitude, $\alpha$ is the phase shift, or the time lag.

To find the $G$ and $\alpha$, we expand the $cos$ term

\begin{align*}y_p(t)&=G\cos(\omega t-\alpha)\\ &=G(\cos\omega t\cos \alpha +\sin\omega t\sin \alpha)\\ &=G\cos \alpha\cos\omega t + G\sin \alpha\sin\omega t \end{align*}

It's easy to see that this is basically the same form as in the previous form of particular solution,

\begin{align}\left\{\begin{aligned} M&=G\cos\alpha\\ N&=G\sin\alpha \end{aligned}\right.\end{align}

Now the trick is simple. Remember that we have $\sin^2+\cos^2=1$, so we square both sides,

\begin{align}\left\{\begin{aligned} M^2&=G^2\cos^2\alpha\\ N^2&=G^2\sin^2\alpha \end{aligned}\right.\end{align}

And add up both equations,

\begin{align*}M^2+N^2&=G^2(\cos^2\alpha+\sin^2\alpha)\\ &=G^2\\ G&=\sqrt{M^2+N^2} \end{align*}

To find the value of $\alpha$, we can exploit the equations again, by taking the ratio so that $G$ can get cancelled,

\begin{align}\frac{M}{N}&=\frac{G\cos\alpha}{G\sin\alpha}\\ &=\frac{\cos\alpha}{\sin\alpha}\\ &=\tan\alpha \end{align}

3. Complex number form of solution

\begin{align*}\frac{dy}{dt}&=ay_c+\cos\omega t+i\sin\omega t\\ &=ay_c+e^{i\omega t} \end{align*}

where the subscript $c$ in $y_c$ stands for complex number. Here we look for

$$y_c(t)=Ye^{i\omega t}$$

After substitution and taking the derivative, we have

\begin{align*}i\omega Ye^{i\omega t}&=aYe^{i\omega t}+e^{i\omega t}\\ i\omega Y&=aY+1\\ (i\omega - a)Y&=1\\ Y&=\frac{1}{i\omega - a} \end{align*}

Rewrite the complex number $i\omega - a$ to the form of $re^{i\alpha}$, we get $r=\sqrt{a^2+\omega^2}$, thus


Substitute this result back into $Y$,

\begin{align*}Y&=\frac{1}{i\omega - a}\\ &=\frac{1}{\sqrt{a^2+\omega^2}}e^{-i\alpha} \end{align*}

And now we can plug the $Y$ back into the $y_c(t)=Ye^{i\omega t}$

\begin{align*}y_c(t)&=Ye^{i\omega t}\\ &=\frac{1}{\sqrt{a^2+\omega^2}}e^{-i\alpha}e^{i\omega t}\\ &=\frac{1}{\sqrt{a^2+\omega^2}}e^{i(\omega t-\alpha)}\\ \end{align*}

We take the real part of the solution to match the $\cos\omega t$ term, and of course the imaginary part will match the $\sin\omega t$ term. And the real part will be

\begin{align*}y_c(t)&=\frac{1}{\sqrt{a^2+\omega^2}}e^{i(\omega t-\alpha)}\\ &=\frac{1}{\sqrt{a^2+\omega^2}}\cos(\omega t-\alpha)\\ &=G\cos(\omega t - \alpha) \end{align*}

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