# A non-homogeneous second order differential equation

Below is a problem from the book Differential Equations written by K.A. Stroud and Dexter Booth. It can be found on page 65 of the book. My answer matches the book's answer but I am not sure I did it correctly, especially the last part.
Thanks,
Bob

Problem:
Obtain the general solution of the equation $$\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 5y = 6 \sin t$$ and determine the amplitude and frequency of the steady-state function.
Our approach is to find $$y_c$$ which is the complimentary solution and then find $$y_p$$ which is the particular solution. Then our solution will be $$y = y_c + y_p$$. We have the following characteristic equation: $$m^2 + 4m - 5 = 0$$ We then solve it by applying the quadratic formula. \begin{align*} m &= \frac{-4 \pm \sqrt{16-4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{16-20}}{2} \\ m &= \frac{-4 \pm \sqrt{-4}}{2} \\ m &= -2 \pm i \\ y_c &= e^{-2t}(C_1 \cos t + C_2 \sin t) \\ \end{align*} \begin{align*} y_p &= A \sin t + B \cos t \\ y'_p &= A \cos t - B \sin t \\ y''_p &= -A \sin t - B \cos t \\ \end{align*} \begin{align*} -A \sin t - B \cos t + 4( A \cos t - B \sin t ) + 5( A \sin t + B \cos t ) &= 6 \sin t \\ -A \sin t - B \cos t + 4A \cos t - 4B \sin t + 5A \sin t + 5B \cos t &= 6 \sin t \\ 4A \sin t - B \cos t + 4A \cos t - 4B \sin t + 5B \cos t &= 6 \sin t \\ \end{align*} Now we solve for $$A$$ and $$B$$. \begin{align*} 4A - 4B &= 6 \\ -B + 4A + 5B &= 0 \\ 4A + 4B &= 0 \\ A &= -B \\ 4(-B) - 4B &= B \\ -8B &= 6 \\ B &= -\frac{3}{4} \\ A &= \frac{3}{4} \\ y_c &= \frac{3}{4} \sin t - \frac{3}{4} \cos t \\ \end{align*} Hence the solution we seek is: $$y = e^{-2t}(C_1 \cos x + C_2 \sin x) + \frac{3}{4} \sin t - \frac{3}{4} \cos t$$ Now the problem asks for the amplitude of the steady-state function. Observe that: $$\lim_{t \rightarrow \infty} y = \frac{3}{4} \sin t - \frac{3}{4} \cos t$$ Let $$y_1 = \frac{3}{4} \sin t - \frac{3}{4} \cos t$$. To find the amplitude, I want to find the minimum and maximum of $$y_1$$. \begin{align*} y'_1 &= \frac{3}{4} \cos t + \frac{3}{4} \sin t = 0 \\ \cos t + \sin t &= 0 \\ \cos t + \sqrt{ 1 - \cos^2{t} } &= 0 \\ \cos t &= - \sqrt{ 1 - \cos^2 t } \\ \cos ^2 t &= 1 - \cos^2 t \\ 2 \cos^2 t &= 1 \\ \cos t &= \pm \frac{ \sqrt{2} } {2} \\ % t &= \pm \frac{\pi}{4} \\ % y_1( \frac{\pi}{4} ) &= \frac{3}{4} \sin{ \frac{\pi}{4} } - \frac{3}{4} \cos { \frac{\pi}{4} } \\ % y_1( \frac{\pi}{4} ) &= \frac{3 \sqrt{2}}{4(2)} - \frac{3 \sqrt{2} }{4(2)} \\ % y_1( -\frac{\pi}{4} ) &= \frac{3}{4} \sin{ fixme } - \frac{3}{4} \cos { fixme } \\ \end{align*} Now we need to consider the following four values for $$t$$ to find the maximum and minimum of $$y$$: $$\frac{\pi}{4} \, , \, \frac{3\pi}{4} \, , \, \frac{5\pi}{4} \, , \, -\frac{\pi}{4}$$ \begin{align*} y_1 \left( \frac{\pi}{4} \right) &= \frac{3}{4} \sin{ \frac{\pi}{4} } - \frac{3}{4} \cos { \frac{\pi}{4} } \\ y_1 \left( \frac{\pi}{4} \right) &= \frac{3 \sqrt{2}}{4(2)} - \frac{3 \sqrt{2} }{4(2)} = 0 \\ % y_1 \left( \frac{3\pi}{4} \right) &= \frac{3}{4} \sin{ \frac{3\pi}{4} } - \frac{3}{4} \cos { \frac{3\pi}{4} } \\ y_1 \left( \frac{3\pi}{4} \right) &= \frac{3 \sqrt{2}}{4(2)} - \frac{-3 \sqrt{2} }{4(2)} \\ y_1 \left( \frac{3\pi}{4} \right) &= \frac{3 \sqrt{2}}{4(2)} + \frac{3 \sqrt{2} }{4(2)} = \frac{3\sqrt{2}}{4 } \\ % y_1 \left( \frac{5\pi}{4} \right) &= \frac{3}{4} \sin{ \frac{5\pi}{4} } - \frac{3}{4} \cos { \frac{5\pi}{4} } \\ y_1 \left( \frac{5\pi}{4} \right) &= -\frac{3 \sqrt{2}}{4(2)} - \frac{-3 \sqrt{2} }{4(2)} \\ y_1 \left( \frac{5\pi}{4} \right) &= 0 \\ % y_1 \left( -\frac{\pi}{4} \right) &= \frac{3}{4} \sin{ -\frac{\pi}{4} } - \frac{3}{4} \cos { -\frac{\pi}{4} } \\ y_1 \left( -\frac{\pi}{4} \right) &= \frac{-3 \sqrt{2}}{4(2)} - \frac{3 \sqrt{2} }{4(2)} \\ y_1 \left( -\frac{\pi}{4} \right) &= -\frac{3\sqrt{2}}{4 } \\ \end{align*} Hence, we have a minimum at $$t = -\frac{\pi}{4}$$ and a maximum at $$t = \frac{3\pi}{4}$$. The amplitude is one half of the difference of the function at its maximum and the value of the function at its minimum. In this case the amplitude is: $$\frac{ 3\sqrt{2} }{4}$$

Now we need to find the frequency of the function $$y_1$$. The period of both $$\sin$$ and $$\cos$$ is both $$2\pi$$. Our function is the difference of two functions with a period of $$2 \pi$$. Therefore the period of $$y_1$$ is $$\frac{1}{2 \pi}$$

• I think most of what you have looks correct. I would look at how to calculate the period. $1/2\pi$ doesn't sound right to me. One easy way to check yourself is to plot your function to see what the period is. It should be $2\pi$. – D.B. Jan 13 at 23:18
• @D.B. When I wrote period, I meant frequency. Does a frequency of $\frac{1}{2 \pi}$ seem right to you? – Bob Jan 13 at 23:57
• Indeed. Once complete cycle for every $2\pi$ units. – D.B. Jan 13 at 23:58

You're overthinking it

$$y_\infty (t) = \frac34 (\sin t - \cos t) = \frac{3}{2\sqrt{2}}\left(\frac{1}{\sqrt 2}\sin t - \frac{1}{\sqrt 2}\cos t \right) = \frac{3}{2\sqrt 2}\sin\left(t - \frac{\pi}{4} \right)$$

So the amplitude of the steady-state is $$\frac{3}{2\sqrt 2}$$. In general, you may use

$$A\sin t + B\cos t = \sqrt{A^2+B^2}\left(\cos\phi\sin t +\sin\phi\cos t\right) = \sqrt{A^2+B^2}\sin\left(t + \phi\right)$$

where $$\cos\phi = \frac{A}{\sqrt{A^2+B^2}}$$ and $$\sin \phi = \frac{B}{\sqrt{A^2+B^2}}$$

The amplitude/frequency is straightforward.

• Yes. I'm agree with this too.... – Mostafa Ayaz Jan 14 at 9:56