【General guidance】The answer provided below has been developed in a clear step by step manner.

Step1/2

To find a particular solution to the differential equation \( \mathrm{{y}{''}-{6}{y}'+{9}{y}=\frac{{-{3.5}{e}^{{{3}{t}}}}}{{{t}^{{2}}+{1}}}} \), we can use the method of undetermined coefficients with Laplace transforms.

First, we take the Laplace transform of both sides of the differential equation:\( \mathrm{{\mathcal{{{L}}}}{\left\lbrace{y}{''}\right\rbrace}-{6}{\mathcal{{{L}}}}{\left\lbrace{y}'\right\rbrace}+{9}{\mathcal{{{L}}}}{\left\lbrace{y}\right\rbrace}={\mathcal{{{L}}}}{\left\lbrace{\frac{{-{3.5}{e}^{{{3}{t}}}}}{{{t}^{{2}}+{1}}}}\right\rbrace}} \)

Using the property \( \mathrm{{\mathcal{{{L}}}}{\left\lbrace{y}{''}\right\rbrace}={s}^{{2}}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}{\quad\text{and}\quad}{\mathcal{{{L}}}}{\left\lbrace{y}'\right\rbrace}={s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}} \), and solving for \( \mathrm{{\mathcal{{{L}}}}{\left\lbrace{y}\right\rbrace}={Y}{\left({s}\right)}} \), we get:

\( \begin{align*} \mathrm{{\left({s}^{{2}}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\right)}-{6}{\left({s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}\right)}+{9}{Y}{\left({s}\right)}} &= \mathrm{{\frac{{-{3.5}}}{{{s}-{3}{i}}}}\cdot{e}^{{{3}{i}{\arctan{{\left({t}\right)}}}}}} \end{align*} \)

Simplifying, we get:

\( \begin{align*} \mathrm{{Y}{\left({s}\right)}} &= \mathrm{{\frac{{-{3.5}}}{{{\left({s}-{3}{i}\right)}{\left({s}^{{2}}-{6}{s}+{9}\right)}}}}\cdot{e}^{{{3}{i}{\arctan{{\left({t}\right)}}}}}+{\frac{{{s}{y}{\left({0}\right)}+{y}'{\left({0}\right)}+{6}{y}{\left({0}\right)}}}{{{s}^{{2}}-{6}{s}+{9}}}}} \end{align*} \)

Explanation: Please refer to solution in this step.

Step2/2

Explanation

Now, we need to find the inverse Laplace transform of the first term on the right-hand side.

We notice that the denominator \( \mathrm{{s}^{{2}}-{6}{s}+{9}={\left({s}-{3}\right)}^{{2}}} \) has a repeated root at \( \mathrm{{s}={3}} \), so we can use partial fraction decomposition to simplify the first term:

\( \begin{align*} \mathrm{{\frac{{-{3.5}}}{{{\left({s}-{3}{i}\right)}{\left({s}^{{2}}-{6}{s}+{9}\right)}}}}} &= \mathrm{{\frac{{{A}}}{{{s}-{3}{i}}}}+{\frac{{{B}{s}+{C}}}{{{\left({s}-{3}\right)}^{{2}}}}}} \end{align*} \)

Multiplying both sides by the denominator and solving for A, B, and C, we get \( \mathrm{{A}=-{0.5}{i}} \), \( \mathrm{{B}=-{1.75}} \), and \( \mathrm{{C}={1.25}{i}} \).

Substituting these values back into the equation for Y(s), we get:\( \mathrm{{Y}{\left({s}\right)}={\frac{{-{0.5}{i}}}{{{s}-{3}{i}}}}-{\frac{{{1.75}{s}+{1.25}{i}}}{{{\left({s}-{3}\right)}^{{2}}}}}+{\frac{{{s}{y}{\left({0}\right)}+{y}'{\left({0}\right)}+{6}{y}{\left({0}\right)}}}{{{\left({s}-{3}\right)}^{{2}}}}}} \)

Now, we take the inverse Laplace transform of Y(s) using tables of Laplace transforms:

\( \begin{align*} \mathrm{{y}{\left({t}\right)}} &= \mathrm{{\mathcal{{{L}}}}^{{-{1}}}{\left\lbrace{\frac{{-{0.5}{i}}}{{{s}-{3}{i}}}}\right\rbrace}{\left({t}\right)}-{\mathcal{{{L}}}}^{{-{1}}}{\left\lbrace{\frac{{{1.75}{s}+{1.25}{i}}}{{{\left({s}-{3}\right)}^{{2}}}}}\right\rbrace}{\left({t}\right)}+{\mathcal{{{L}}}}^{{-{1}}}{\left\lbrace{\frac{{{s}{y}{\left({0}\right)}+{y}'{\left({0}\right)}+{6}{y}{\left({0}\right)}}}{{{\left({s}-{3}\right)}^{{2}}}}}\right\rbrace}{\left({t}\right)}} \end{align*} \)

The inverse Laplace transform of the first term can be found directly from tables:\( \mathrm{{\mathcal{{{L}}}}^{{-{1}}}{\left\lbrace{\frac{{-{3.5}{e}^{{{3}{t}}}}}{{{t}^{{2}}+{1}}}}\right\rbrace}=-{3.5}{\mathcal{{{L}}}}^{{-{1}}}{\left\lbrace{\frac{{{e}^{{{3}{t}}}}}{{{t}^{{2}}+{1}}}}\right\rbrace}} \)

Using the table of Laplace transforms, we see that">\( \mathrm{{\mathcal{{{L}}}}{\left\lbrace{f{{\left({t}\right)}}}\right\rbrace}={\frac{{\pi}}{{{2}}}}{e}^{{-{a}{s}}}{o}{p}{e}{r}{a}{t}{o}{r}{n}{a}{m}{e}{\left\lbrace{e}{r}{f}{c}\right\rbrace}{\left({\frac{{{a}-{t}}}{{\sqrt{{{4}{b}{t}}}}}}\right)}} \)

where a>0 and b>0 are constants, and \( \mathrm{{o}{p}{e}{r}{a}{t}{o}{r}{n}{a}{m}{e}{\left\lbrace{e}{r}{f}{c}\right\rbrace}{\left({z}\right)}={\frac{{{2}}}{{\sqrt{{\pi}}}}}{\int_{{z}}^{\infty}}{e}^{{-{t}^{{2}}}}{m}{a}{t}{h}{r}{m}{\left\lbrace{d}\right\rbrace}{t}} \) is the complementary error function. In our case, we have a=3 and b=1/4,

so

\( \begin{align*} \mathrm{{\mathcal{{{L}}}}^{{-{1}}}{\left\lbrace{\frac{{-{3.5}{e}^{{{3}{t}}}}}{{{t}^{{2}}+{1}}}}\right\rbrace}} &= \mathrm{-{3.5}{\mathcal{{{L}}}}^{{-{1}}}{\left\lbrace{\frac{{{e}^{{{3}{t}}}}}{{{t}^{{2}}+{1}}}}\right\rbrace}=-{3.5}\cdot{\frac{{\pi}}{{{2}}}}{e}^{{{3}{s}}}{o}{p}{e}{r}{a}{t}{o}{r}{n}{a}{m}{e}{\left\lbrace{e}{r}{f}{c}\right\rbrace}{\left({\frac{{\sqrt{{{3}}}-{s}}}{{{\frac{{{1}}}{{{2}}}}}}}\right)}} \end{align*} \)

Explanation: Please refer to solution in this step.

Find a particular solution to y'' - 6y' + 9y = (-3.5e^(3t))/(t^2+1). Prefer via Laplace Transform please. I just need the particular solution, not the complimentary or the general solution. I'd like to understand it via Laplace Transform, but if it can't be done then I understand.

Public Answer

Find a particular solution to y'' - 6y' + 9y = (-3.5e^(3t))/(t^2+1).

Prefer via Laplace Transform please. I just need the particular solution, not the complimentary or the general solution. I'd like to understand it via Laplace Transform, but if it can't be done then I understand.