Violet Woodward

2022-07-22

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

${y}^{\u2033}-2{y}^{\prime}+y=\mathrm{sin}(t)$

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.

${y}_{h}(t)$:

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

${y}^{\u2033}-2{y}^{\prime}+y=0$

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:

${y}_{h}(t)=({c}_{1}+{c}_{2}t){e}^{t}$

${y}_{p}(t)$:

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

$\begin{array}{rl}{y}_{p}(t)& =\phantom{-}K\mathrm{cos}(t)+M\mathrm{sin}(t)\\ {y}_{p}^{\prime}(t)& =-K\mathrm{sin}(t)+M\mathrm{cos}(t)\\ {y}_{p}^{\u2033}(t)& =-K\mathrm{cos}(t)-M\mathrm{sin}(t)\end{array}$

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

${y}_{p}(t)=\frac{1}{2}\mathrm{cos}(t)$

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

$\begin{array}{rl}y(t)& ={y}_{h}(t)+{y}_{p}(t)\\ & =({c}_{1}+{c}_{2}t){e}^{t}+\frac{1}{2}\mathrm{cos}(t)\end{array}$

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

${y}^{\u2033}-2{y}^{\prime}+y=\mathrm{sin}(t)$

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.

${y}_{h}(t)$:

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

${y}^{\u2033}-2{y}^{\prime}+y=0$

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:

${y}_{h}(t)=({c}_{1}+{c}_{2}t){e}^{t}$

${y}_{p}(t)$:

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

$\begin{array}{rl}{y}_{p}(t)& =\phantom{-}K\mathrm{cos}(t)+M\mathrm{sin}(t)\\ {y}_{p}^{\prime}(t)& =-K\mathrm{sin}(t)+M\mathrm{cos}(t)\\ {y}_{p}^{\u2033}(t)& =-K\mathrm{cos}(t)-M\mathrm{sin}(t)\end{array}$

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

${y}_{p}(t)=\frac{1}{2}\mathrm{cos}(t)$

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

$\begin{array}{rl}y(t)& ={y}_{h}(t)+{y}_{p}(t)\\ & =({c}_{1}+{c}_{2}t){e}^{t}+\frac{1}{2}\mathrm{cos}(t)\end{array}$

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

repotasonwf

Beginner2022-07-23Added 12 answers

So, here Euler's formula means using $\mathrm{sin}t==\frac{{e}^{it}-{e}^{-it}}{2i}$ The particular integral will be found as

$f(D)y={e}^{at}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}_{p}(t)=\frac{{e}^{at}}{f(a)}$

Where we have

$(D-1{)}^{2}y=\frac{{e}^{it}-{e}^{-it}}{2i}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}_{p}(t)=(2i{)}^{-1}(\frac{{e}^{it}}{(i-1{)}^{2}}-\frac{{e}^{-it}}{(-i-1{)}^{2}})=\frac{1}{4}[{e}^{it}+{e}^{-it}]$

$=\frac{1}{2}\mathrm{cos}t.$

${y}_{h}(t)$ remains the same as you found.

$f(D)y={e}^{at}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}_{p}(t)=\frac{{e}^{at}}{f(a)}$

Where we have

$(D-1{)}^{2}y=\frac{{e}^{it}-{e}^{-it}}{2i}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}_{p}(t)=(2i{)}^{-1}(\frac{{e}^{it}}{(i-1{)}^{2}}-\frac{{e}^{-it}}{(-i-1{)}^{2}})=\frac{1}{4}[{e}^{it}+{e}^{-it}]$

$=\frac{1}{2}\mathrm{cos}t.$

${y}_{h}(t)$ remains the same as you found.

Luz Stokes

Beginner2022-07-24Added 3 answers

Consider a new differential equation

$\begin{array}{}\text{(1)}& \frac{{d}^{2}y}{d{t}^{2}}-2\frac{dy}{dt}+1=(\frac{{d}^{2}}{dt}-2\frac{d}{dt}+1)y={e}^{it}.\end{array}$

Let $y={e}^{it}g(t)$. By the product rule

$\begin{array}{r}\frac{dy}{dt}=\frac{d}{dt}({e}^{it}g(t))={e}^{it}\frac{d}{dt}g(t)+\left(\frac{d}{dt}{e}^{it}\right)g(t)={e}^{it}(\frac{d}{dt}+i)g(t).\end{array}$

Then the differential equation (1) becomes

$\begin{array}{r}{e}^{it}[{(\frac{d}{dt}+i)}^{2}-2(\frac{d}{dt}+i)+1]g(t)={e}^{it}\end{array}$

so that after cancelling ${e}^{it}$ from both sides and expanding the expression in parenthesis on the LHS

$\begin{array}{r}[{(\frac{d}{dt}+i)}^{2}-2(\frac{d}{dt}+i)+1]g(t)=(\frac{{d}^{2}}{d{t}^{2}}+(2i-2)\frac{d}{dt}-2i)g(t)=1\end{array}$

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

$\begin{array}{r}y=\frac{i}{2}{e}^{it}=\frac{i}{2}(\mathrm{cos}t+i\mathrm{sin}t).\end{array}$

Since $\mathrm{sin}t=\mathrm{I}\mathrm{m}({e}^{it})$, then a particular solution to

$\frac{{d}^{2}y}{d{t}^{2}}-2\frac{dy}{dt}+1=(\frac{{d}^{2}}{dt}-2\frac{d}{dt}+1){y}_{p}=\mathrm{sin}t$

is ${y}_{p}=\mathrm{I}\mathrm{m}(y)=\mathrm{I}\mathrm{m}(\frac{i}{2}(\mathrm{cos}t+i\mathrm{sin}t))=\frac{1}{2}\mathrm{cos}t$

$\begin{array}{}\text{(1)}& \frac{{d}^{2}y}{d{t}^{2}}-2\frac{dy}{dt}+1=(\frac{{d}^{2}}{dt}-2\frac{d}{dt}+1)y={e}^{it}.\end{array}$

Let $y={e}^{it}g(t)$. By the product rule

$\begin{array}{r}\frac{dy}{dt}=\frac{d}{dt}({e}^{it}g(t))={e}^{it}\frac{d}{dt}g(t)+\left(\frac{d}{dt}{e}^{it}\right)g(t)={e}^{it}(\frac{d}{dt}+i)g(t).\end{array}$

Then the differential equation (1) becomes

$\begin{array}{r}{e}^{it}[{(\frac{d}{dt}+i)}^{2}-2(\frac{d}{dt}+i)+1]g(t)={e}^{it}\end{array}$

so that after cancelling ${e}^{it}$ from both sides and expanding the expression in parenthesis on the LHS

$\begin{array}{r}[{(\frac{d}{dt}+i)}^{2}-2(\frac{d}{dt}+i)+1]g(t)=(\frac{{d}^{2}}{d{t}^{2}}+(2i-2)\frac{d}{dt}-2i)g(t)=1\end{array}$

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

$\begin{array}{r}y=\frac{i}{2}{e}^{it}=\frac{i}{2}(\mathrm{cos}t+i\mathrm{sin}t).\end{array}$

Since $\mathrm{sin}t=\mathrm{I}\mathrm{m}({e}^{it})$, then a particular solution to

$\frac{{d}^{2}y}{d{t}^{2}}-2\frac{dy}{dt}+1=(\frac{{d}^{2}}{dt}-2\frac{d}{dt}+1){y}_{p}=\mathrm{sin}t$

is ${y}_{p}=\mathrm{I}\mathrm{m}(y)=\mathrm{I}\mathrm{m}(\frac{i}{2}(\mathrm{cos}t+i\mathrm{sin}t))=\frac{1}{2}\mathrm{cos}t$

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