Use Laplace transforms to solve the following initial value problem. x^((3))+x"-6x'=0, x(0)=0 , x'(0)=x"(0)=7. The solution is x(t)=?

alesterp 2021-09-22 Answered
Use Laplace transforms to solve the following initial value problem.
\(\displaystyle{x}^{{{\left({3}\right)}}}+{x}\text{-6x'=0, x(0)=0 , x'(0)=x}{\left({0}\right)}={7}\)
The solution is x(t)=?

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Expert Answer

irwchh
Answered 2021-09-23 Author has 20034 answers
Step 1
Let's solve the initial value problem by Laplace transforms,
\(\displaystyle{X}\text{'+X}-{6}{X}'={0}\)
X(0)=0
X'(0)=7
X"(0)=7
here,
\(\displaystyle{L}\le{f}{t}{\left\lbrace{X}{\left({t}\right)}{r}{i}{g}{h}{t}\right\rbrace}={X}{\left({s}\right)}\)
\(\displaystyle{L}\le{f}{t}{\left\lbrace{X}'{\left({t}\right)}{r}{i}{g}{h}{t}\right\rbrace}={s}{X}{\left({s}\right)}-{X}{\left({0}\right)}={s}{X}{\left({s}\right)}-{0}={s}{X}{\left({s}\right)}\)
\(\displaystyle{L}\le{f}{t}{\left\lbrace{X}\text{}{t}\right)}{r}{i}{g}{h}{t}\rbrace={s}^{{2}}{X}{\left({s}\right)}-{s}{X}{\left({0}\right)}-{X}'{\left({0}\right)}={s}^{{2}}{X}{\left({s}\right)}-{0}-{7}={s}^{{2}}{X}{\left({s}\right)}-{7}\)
\(\displaystyle{L}\le{f}{t}{\left\lbrace{X}{'''}{\left({t}\right)}{r}{i}{g}{h}{t}\right\rbrace}={s}^{{3}}{X}{\left({s}\right)}-{s}^{{2}}{X}{\left({0}\right)}-{s}{X}'{\left({0}\right)}-{X}\text{}{0}{)}={s}^{{3}}{X}{\left({s}\right)}-{7}{s}-{7}\)
Step 2
Now,
\(\displaystyle{X}{'''}+{X}\text{}{6}{X}'={0}\)
\(\displaystyle{s}^{{3}}{X}{\left({s}\right)}-{7}{s}-{7}+{s}^{{2}}{X}{\left({s}\right)}-{7}-{6}{s}{X}{\left({s}\right)}={0}\)
\(\displaystyle{X}{\left({s}\right)}{\left[{s}^{{3}}+{s}^{{2}}-{6}{s}\right]}-{7}{s}-{14}={0}\)
\(\displaystyle{X}{\left({s}\right)}{\left[{s}^{{3}}+{s}^{{2}}-{6}{s}\right]}={7}{s}+{14}\)
\(\displaystyle{X}{\left({s}\right)}={\frac{{{7}{s}+{14}}}{{{s}^{{3}}+{s}^{{2}}-{6}{s}}}}+{\frac{{{7}{\left({s}+{2}\right)}}}{{{s}^{{3}}+{s}^{{2}}-{6}{s}}}}\)
By partial fractions,
\(\displaystyle{\frac{{{s}+{2}}}{{{s}{\left({s}^{{2}}+{s}-{6}\right)}}}}={\frac{{{s}+{2}}}{{{s}{\left({s}^{{2}}+{3}{s}-{2}{s}-{6}\right)}}}}\)
\(\displaystyle={\frac{{{s}+{2}}}{{{s}{\left({s}-{2}\right)}{\left({s}+{3}\right)}}}}\)
Step 3
\(\displaystyle{\frac{{{s}+{2}}}{{{s}{\left({s}-{2}\right)}{\left({s}+{3}\right)}}}}={\frac{{{A}}}{{{s}}}}+{\frac{{{B}}}{{{s}-{2}}}}+{\frac{{{c}}}{{{s}+{3}}}}\)
On solving \(\displaystyle{A}={\frac{{-{1}}}{{{3}}}},{B}={\frac{{-{1}}}{{{15}}}},{C}={\frac{{{2}}}{{{5}}}}\)
\(\displaystyle\therefore{\frac{{{s}+{2}}}{{{s}{\left({s}-{2}\right)}{\left({s}+{3}\right)}}}}={\frac{{{\frac{{-{1}}}{{{3}}}}}}{{{s}}}}+{\frac{{{\frac{{-{1}}}{{{15}}}}}}{{{s}-{2}}}}+{\frac{{{\frac{{{2}}}{{{5}}}}}}{{{s}+{3}}}}\)
\(\displaystyle={\frac{{-{1}}}{{{3}{s}}}}-{\frac{{{1}}}{{{15}{\left({s}-{2}\right)}}}}+{\frac{{{2}}}{{{5}{\left({s}+{3}\right)}}}}\)
\(\displaystyle\therefore{\frac{{{7}{\left({s}+{2}\right)}}}{{{s}{\left({s}-{2}\right)}{\left({s}+{3}\right)}}}}={7}{\left({\frac{{{\frac{{-{1}}}{{{3}}}}}}{{{s}}}}+{\frac{{{\frac{{-{1}}}{{{15}}}}}}{{{s}-{2}}}}+{\frac{{{\frac{{{2}}}{{{5}}}}}}{{{s}+{3}}}}\right)}\)
\(\displaystyle={\frac{{-{7}}}{{{3}{s}}}}-{\frac{{{7}}}{{{15}{\left({s}-{2}\right)}}}}+{\frac{{{14}}}{{{5}{\left({s}+{3}\right)}}}}\)
Step 4
\(\displaystyle{x}{\left({t}\right)}={L}^{{-{1}}}\le{f}{t}{\left\lbrace{\frac{{-{7}}}{{{3}{s}}}}-{\frac{{{7}}}{{{15}{\left({s}-{2}\right)}}}}+{\frac{{{14}}}{{{5}{\left({s}+{3}\right)}}}}{r}{i}{g}{h}{t}\right\rbrace}\)
\(\displaystyle={\frac{{-{7}}}{{{3}}}}-{\frac{{{7}{e}^{{{2}{t}}}}}{{{15}}}}+{\frac{{{14}{e}^{{{3}{t}}}}}{{{5}}}}\)
\(\displaystyle\therefore{x}{\left({t}\right)}={\frac{{-{7}}}{{{3}}}}-{\frac{{{7}{e}^{{{2}{t}}}}}{{{15}}}}+{\frac{{{14}{e}^{{{3}{t}}}}}{{{5}}}}\)
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