Transform the given differential equation or system into an equivalent system of first-order differential equations. t^{3}x^{(3)}-2t^{2}x''+3tx'+5x=\l

avissidep 2021-06-09 Answered

Transform the given differential equation or system into an equivalent system of first-order differential equations.
\(\displaystyle{t}^{{{3}}}{x}^{{{\left({3}\right)}}}-{2}{t}^{{{2}}}{x}{''}+{3}{t}{x}'+{5}{x}={\ln{{t}}}\)

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Expert Answer

Bella
Answered 2021-06-10 Author has 6535 answers
Consider the differential equation
\(\displaystyle{t}^{{{3}}}{x}^{{{\left({3}\right)}}}-{2}{t}^{{{2}}}{x}{''}+{3}{t}{x}'+{5}{x}={\ln{{t}}}\)
The objective is to transform tha given differential equetion into an equivalent system of first-order differeential equations. The third-order equation
\(\displaystyle{t}^{{{3}}}{x}^{{{\left({3}\right)}}}-{2}{t}^{{{2}}}{x}{''}+{3}{t}{x}'+{5}{x}={\ln{{t}}}\)
is equivalent to
\(\displaystyle{f{{\left({t},{x},{x}',{x}{''}\right)}}}={\frac{{{1}}}{{{t}^{{{3}}}}}}{\ln{{t}}}+{\frac{{{2}}}{{{t}}}}{x}{''}-{\frac{{{3}}}{{{t}^{{{2}}}}}}{x}'-{\frac{{{5}}}{{{t}^{{{3}}}}}}{x}\)
Now, let's be
\(\displaystyle{x}={x}_{{1}}\)
\(\displaystyle{x}'={x}_{{2}}={x}'_{{1}}\)
\(\displaystyle{x}{''}={x}_{{3}}={x}'{2}={x}{''}_{{1}}\)
The above substitution yields the system as written below
\(\displaystyle{x}'_{{1}}={x}_{{2}}\)
\(\displaystyle{x}'_{{2}}={x}_{{3}}\)
\(\displaystyle{x}'_{{3}}={\frac{{{1}}}{{{t}^{{{3}}}}}}{\ln{{t}}}+{\frac{{{2}}}{{{t}}}}{x}{''}-{\frac{{{3}}}{{{t}^{{{2}}}}}}{x}'-{\frac{{{5}}}{{{t}^{{{3}}}}}}{x}\)
are the three first order differential equations.
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