# 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

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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Bella
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.