Solve by Laplace transformation method the following D.E. (1) y"-3y'+2y=4^2e^t given that y(0)=-3 , y'(0)=5

Yasmin 2021-09-26 Answered

Solve by Laplace transformation method the following D.E.
(1) \(y''-3y'+2y=4^2e^t\) given that \(y(0)=-3 , y'(0)=5\)​​​​​​​

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Expert Answer

tabuordg
Answered 2021-09-27 Author has 7367 answers

Step 1
Question 1:
To solve the given differential equation using Laplace transformation:
\(y''-3y'+2y=4e^{2t}\) with \(y(0)=-3 , y'(0)=5\)
Step 2
\(y''-3y'+2y=4e^{2t}\)
\(L\left\{y"-3y+2y\right\}=L\left\{4e^{2t}\right\}\)
\(\displaystyle{\left({s}^{{3}}+{3}{s}+{2}\right)}{Y}{\left({s}\right)}={\frac{{{4}}}{{{s}-{2}}}}\)
\((\text{ Since }, L\left\{\delta\left\{4e^{2z}\right\} \right\}=\frac{4}{s-2})\)
\(\displaystyle{s}^{{2}}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}-{3}{\left[{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}\right]}+{2}{Y}{\left({s}\right)}={\frac{{{4}}}{{{s}-{2}}}}\)
\(\displaystyle\text{put }\ {y}{\left({0}\right)}=-{3},{y}'{\left({0}\right)}={5}\)
\(\displaystyle{s}^{{2}}{Y}{\left({s}\right)}+{3}{s}-{5}-{3}{s}{Y}{\left({s}\right)}-{9}+{2}{Y}{\left({s}\right)}={\frac{{{4}}}{{{s}-{2}}}}\)
\(\displaystyle{\left({s}^{{2}}-{3}{s}+{2}\right)}{Y}{\left({s}\right)}={\frac{{{4}}}{{{s}-{2}}}}+{14}-{3}{s}\)
\(\displaystyle{Y}{\left({s}\right)}={\frac{{-{3}{s}^{{2}}+{20}{s}-{24}}}{{{\left({s}-{2}\right)}{\left({s}^{{2}}-{3}{s}+{2}\right)}}}}\)
Step 3
Now simplify Y(s) in order to apply inverse Laplace to find the solution of differential equation, as:
\(\displaystyle{Y}{\left({s}\right)}={\frac{{-{3}{s}^{{2}}+{20}{s}-{24}}}{{{\left({s}-{2}\right)}{\left({s}^{{2}}-{3}{s}+{2}\right)}}}}={\frac{{-{3}{s}^{{2}}+{20}{s}-{24}}}{{{\left({s}-{2}\right)}^{{2}}{\left({s}-{1}\right)}}}}\)
\(\displaystyle{\frac{{-{3}{s}^{{2}}+{20}{s}-{24}}}{{{\left({s}-{2}\right)}^{{2}}{\left({s}-{1}\right)}}}}={\frac{{{A}}}{{{s}-{2}}}}+{\frac{{{B}}}{{{\left({s}-{2}\right)}^{{2}}}}}+{\frac{{{C}}}{{{s}-{1}}}}\)
\(\displaystyle-{3}{s}^{{2}}+{20}{s}-{24}={A}{\left({s}-{2}\right)}{\left({s}-{1}\right)}+{B}{\left({s}-{1}\right)}+{C}{\left({s}-{2}\right)}^{{2}}\)
\(\displaystyle-{3}{s}^{{2}}+{20}{s}-{24}={A}{s}^{{2}}-{3}{A}{s}+{2}{A}+{B}{s}-{B}+{C}{s}^{{2}}-{4}{C}{s}+{4}{C}\)
\(\displaystyle-{3}{s}^{{3}}+{20}{s}-{24}={\left({A}+{C}\right)}{s}^{{2}}+{\left(-{3}{A}+{B}-{4}{C}\right)}{s}+{\left({2}{A}-{B}+{4}{c}\right)}\)
Comparing the coefficients,
\(\displaystyle{A}+{C}=-{3},-{3}{A}+{B}-{4}{C}={20},{\quad\text{and}\quad}{2}{A}-{B}+{4}{c}=-{24}\)
Solving above gives \(A=4 , B=4 , C=-7\)
\(\displaystyle{Y}{\left({s}\right)}={\frac{{-{3}{s}^{{2}}+{20}{s}-{24}}}{{{\left({s}-{2}\right)}{\left({s}^{{2}}-{3}{s}+{2}\right)}}}}={\frac{{{4}}}{{{s}-{2}}}}+{\frac{{{4}}}{{{\left({s}-{2}\right)}^{{2}}}}}-{\frac{{{7}}}{{{s}-{1}}}}\)
\(\displaystyle\Rightarrow{Y}{\left({S}\right)}={\frac{{{4}}}{{{s}-{2}}}}+{\frac{{{4}}}{{{\left({s}-{2}\right)}^{{2}}}}}-{\frac{{{7}}}{{{s}-{1}}}}\)
Step 4
Now using inverse Laplace transform solution of differential equation is find as:
\(y(t)=L^{-1}\left\{Y(s)\right\}\)
\(y(t)=L^{-1}\left\{\frac{4}{s-2}+\frac{4}{(s-2)^2}-\frac{7}{s-1}\right\}\)
\(y(t)=4L^{-1}\left\{\frac{4}{s-2}\right\}(t)+4L^{-1}\left\{\frac{1}{s-2}\right\}(t)-7L^{-1}\left\{\frac{1}{s-1}\right\}(t)\)
\(\displaystyle{y}{\left({t}\right)}={4}{e}^{{{2}{t}}}+{4}{t}{e}^{{{2}{t}}}-{7}{e}^{{t}}\)
Thus, solution is
\(\displaystyle{y}{\left({t}\right)}={4}{e}^{{{2}{t}}}+{4}{t}{e}^{{{2}{t}}}-{7}{e}^{{t}}\)

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