This is a homework problem that I have for an engineering mathematics class. The problem is as follows:

Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized:

$$y'' - 2y' + y = \sin(t)$$

The way that I solved this doesn't involve Euler's formula, and I was wondering how I might use the formula here.

My approach:

The formula can be written as $y(t) = y_h(t) + y_p(t)$ where $y_h(t)$ is the "homogeneous version" of the ODE and $y_p(t)$ is the particular solution that we'll obtain via the basic rule of the method of undetermined coefficients.


Putting $r(t) = \sin(t) = 0$ in the original equation, the ODE we need to solve is:

$$y'' - 2y' + y = 0$$

where we can set the general solution as $y = e^{\lambda t}$ and obtain the characteristic equation:

$$\lambda^2 - 2\lambda + 1 = 0$$

which has a real double root, hence giving us the solution:

$$y_h(t) = (c_1 + c_2t)e^t$$


Judging by the fact that $r(t)$ is shape $k\sin(\omega t)$ and we know that $\omega = 1$ we can set the general solution to be of form:

$$ \begin{align} y_p(t) & = \phantom{-}K\cos(t) + M\sin(t) \\ y_p'(t) & = -K\sin(t) + M\cos(t) \\ y_p''(t) & = -K\cos(t) - M\sin(t) \end{align}$$

substituting these equations into the original equation and then simplifying gives us:

$$y_p(t) = \frac{1}{2} \cos(t)$$

And in conclusion, we can write that the solution to the given ODE is:

$$ \begin{align} y(t) & = y_h(t) + y_p(t) \\ & = (c_1 + c_2 t)e^t + \frac{1}{2}\cos(t) \end{align} $$

How would we be able to derive this conclusion via Euler's formula? Thanks in advance.

  • $\begingroup$ What is Euler's formula? $\endgroup$ – Dr Zafar Ahmed DSc Nov 29 '19 at 12:58
  • $\begingroup$ Ah, sorry. I'm referring to this: en.wikipedia.org/wiki/Euler's_formula $\endgroup$ – Seankala Nov 29 '19 at 12:58
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    $\begingroup$ Maybe the question wants you to apply $\sin{(x)}=-\frac{i}2(e^{ix}-e^{-ix})$ to find the particular solution in terms of $C_1e^{ix}+C_2e^{-ix}$? $\endgroup$ – Peter Foreman Nov 29 '19 at 13:07

So, here Euler's formula means using $\sin t= =\frac{e^{it}-e^{-it}}{2i}$ The particular integral will be found as $$f(D)y=e^{at}\implies y_p(t)=\frac{ e^{at}}{f(a)}$$ Where we have $$(D-1)^2y= \frac{e^{it}-e^{-it}}{2i}\implies y_p(t)= (2i)^{-1} \left( \frac{e^{it}}{(i-1)^2} -\frac{e^{-it}}{(-i-1)^2}\right)= \frac{1}{4} [e^{it}+e^{-it}]$$ $$=\frac{1}{2} \cos t.$$ $y_h(t)$ remains the same as you found.

  • $\begingroup$ Thanks for the reply. What does $D$ mean here? $\endgroup$ – Seankala Nov 29 '19 at 13:24
  • $\begingroup$ $D$ should be the differential operator $\frac{d}{dt}$. $\endgroup$ – Representation Nov 29 '19 at 13:41
  • $\begingroup$ Tes very much so. $\endgroup$ – Dr Zafar Ahmed DSc Nov 29 '19 at 13:43

I am answering this based on my understanding from Elementary Differential Equations by Boyce and Diprima.

Consider a new differential equation \begin{equation}\label{1}\tag{1} \frac{d^{2}y}{dt^{2}}-2\frac{dy}{dt}+1=\left(\frac{d^{2}}{dt}-2\frac{d}{dt}+1\right)y=e^{it}. \end{equation} Let $y=e^{it}g(t)$. By the product rule \begin{align*} \frac{dy}{dt}=\frac{d}{dt}\left(e^{it}g(t)\right)=e^{it}\frac{d}{dt}g(t)+\left(\frac{d}{dt}e^{it}\right)g(t)=e^{it}\left(\frac{d}{dt}+i\right)g(t). \end{align*} Then the differential equation (1) becomes \begin{align*} e^{it}\left[\left(\frac{d}{dt}+i\right)^{2}-2\left(\frac{d}{dt}+i\right)+1\right]g(t)=e^{it} \end{align*} so that after cancelling $e^{it}$ from both sides and expanding the expression in parenthesis on the LHS \begin{align*} \left[\left(\frac{d}{dt}+i\right)^{2}-2\left(\frac{d}{dt}+i\right)+1\right]g(t)=\left(\frac{d^{2}}{dt^{2}}+(2i-2)\frac{d}{dt}-2i\right)g(t)=1 \end{align*} so that $g(t)=c$ for some constant $c\in\mathbb{C}$. Therefore, $g(t)=i/2$. A particular solution of $y$ would then be \begin{align*} y=\frac{i}{2}e^{it}=\frac{i}{2}(\cos t+i\sin t). \end{align*} Since $\sin t={\rm Im}(e^{it})$, then a particular solution to \begin{equation} \frac{d^{2}y}{dt^{2}}-2\frac{dy}{dt}+1=\left(\frac{d^{2}}{dt}-2\frac{d}{dt}+1\right)y_{p}=\sin t \end{equation} is $y_{p}={\rm Im}(y)={\rm Im}(\frac{i}{2}(\cos t+i\sin t))=\frac{1}{2}\cos t$.

  • $\begingroup$ Thanks for the answer. What does $\text{Im}$ mean? $\endgroup$ – Seankala Nov 29 '19 at 13:35
  • $\begingroup$ ${\rm Im}(z)$ is the imaginary part of the complex number $z$. In the solution above, I defined $y$ to be a complex valued function. $\endgroup$ – Representation Nov 29 '19 at 13:41

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