$t{y}^{\u2033}-t{y}^{\prime}+y=1$

where $y(0)=0$ and ${y}^{\prime}(0)=2$

Widersinnby7
2022-11-20
Answered

Solving the IVP

$t{y}^{\u2033}-t{y}^{\prime}+y=1$

where $y(0)=0$ and ${y}^{\prime}(0)=2$

$t{y}^{\u2033}-t{y}^{\prime}+y=1$

where $y(0)=0$ and ${y}^{\prime}(0)=2$

You can still ask an expert for help

grizintimbp

Answered 2022-11-21
Author has **16** answers

$t{y}^{\u2033}-t{y}^{\prime}+y=1$

$-{({s}^{2}\mathcal{L}(y)-sy(0)-{y}^{\prime}(0))}^{\prime}+{(s\mathcal{L}(y)-y(0))}^{\prime}+\mathcal{L}(y)={\displaystyle \frac{1}{s}}$

$(s-{s}^{2}){\mathcal{L}}^{\prime}(y)+(2-2s)\mathcal{L}(y)={\displaystyle \frac{1}{s}}$

${\mathcal{L}}^{\prime}(y)+{\displaystyle \frac{2}{s}}\mathcal{L}(y)={\displaystyle \frac{1}{{s}^{2}(1-s)}}$

with integrating factor ${s}^{2}$ we have

$\begin{array}{rl}({s}^{2}\mathcal{L}(y){)}^{\prime}& ={\displaystyle \frac{1}{1-s}}\\ y& ={\mathcal{L}}^{-1}{\displaystyle \frac{-\mathrm{ln}(1-s)}{{s}^{2}}}+Ct\\ & ={\int}_{0}^{t}{\displaystyle \frac{{e}^{x}}{x}}(t-x)dx+Ct\\ & =t{\int}_{0}^{t}{\displaystyle \frac{{e}^{x}}{x}}dx-{e}^{t}+1+Ct\\ & =t\mathrm{Ei}(t)-{e}^{t}+1+Ct\end{array}$

$-{({s}^{2}\mathcal{L}(y)-sy(0)-{y}^{\prime}(0))}^{\prime}+{(s\mathcal{L}(y)-y(0))}^{\prime}+\mathcal{L}(y)={\displaystyle \frac{1}{s}}$

$(s-{s}^{2}){\mathcal{L}}^{\prime}(y)+(2-2s)\mathcal{L}(y)={\displaystyle \frac{1}{s}}$

${\mathcal{L}}^{\prime}(y)+{\displaystyle \frac{2}{s}}\mathcal{L}(y)={\displaystyle \frac{1}{{s}^{2}(1-s)}}$

with integrating factor ${s}^{2}$ we have

$\begin{array}{rl}({s}^{2}\mathcal{L}(y){)}^{\prime}& ={\displaystyle \frac{1}{1-s}}\\ y& ={\mathcal{L}}^{-1}{\displaystyle \frac{-\mathrm{ln}(1-s)}{{s}^{2}}}+Ct\\ & ={\int}_{0}^{t}{\displaystyle \frac{{e}^{x}}{x}}(t-x)dx+Ct\\ & =t{\int}_{0}^{t}{\displaystyle \frac{{e}^{x}}{x}}dx-{e}^{t}+1+Ct\\ & =t\mathrm{Ei}(t)-{e}^{t}+1+Ct\end{array}$

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$\underset{n\to \mathrm{\infty}}{lim}{\int}_{0}^{n}\frac{1}{(a-s{)}^{2}2{e}^{(a-s)t}}dt={\underset{n\to \mathrm{\infty}}{lim}\frac{2}{(a-s{)}^{3}}{e}^{(a-s)t}|}_{0}^{n}.$

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try to prove that

$\mathcal{L}\{{t}^{2}{e}^{at}\}=\frac{2}{(s-a{)}^{3}}.$

I've gotten to the last integration by parts where

$\underset{n\to \mathrm{\infty}}{lim}{\int}_{0}^{n}\frac{1}{(a-s{)}^{2}2{e}^{(a-s)t}}dt={\underset{n\to \mathrm{\infty}}{lim}\frac{2}{(a-s{)}^{3}}{e}^{(a-s)t}|}_{0}^{n}.$

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