# Transform the given differential equation or system into an equivalent system of first-order differential equations. x^{(3)}-2x''+x'=1+te^{t}

Transform the given differential equation or system into an equivalent system of first-order differential equations.
$$\displaystyle{x}^{{{\left({3}\right)}}}-{2}{x}{''}+{x}'={1}+{t}{e}^{{{t}}}$$

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

Khribechy
Transform the given differential equation or system into an equivalent system of first-order differential equations.
$$\displaystyle{x}^{{{\left({3}\right)}}}-{2}{x}{''}+{x}'={1}+{t}{e}^{{{t}}}$$ The third-order equation $$\displaystyle{x}^{{{\left({3}\right)}}}-{2}{x}{''}+{x}'={1}+{t}{e}^{{{t}}}$$
is equivalent to system
$$\displaystyle{f{{\left({t},{x},{x}',{x}{''},{x}{''}\right)}}}={1}+{t}{e}^{{{t}}}-{x}'+{2}{x}{''}$$
Hense the substitutions $$\displaystyle{x}_{{1}}={x},{x}_{{2}}={x}'={x}'_{{1}},{x}_{{3}}={x}{''}={x}'_{{2}}$$
yield the system
$$\displaystyle{x}'_{{1}}={x}_{{2}}$$
$$\displaystyle{x}'_{{2}}={x}_{{3}}$$
$$\displaystyle{x}'_{{3}}={1}+{t}{e}^{{{t}}}-{x}_{{2}}+{2}{x}_{{3}}$$ is a system of first-order equation.