 # How can the following second-order linear equation be converted into chanyingsauu7 2021-11-21 Answered
How can the following second-order linear equation be converted into a first-order linear equation?
This is our second-order equation:
$$\displaystyle{y}{''}-{2}{y}'+{2}{y}={e}^{{{2}{t}}}{\sin{{t}}}$$

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Note that if $$\displaystyle\alpha$$ and $$\displaystyle\beta$$ are the roots of $$\displaystyle{x}^{{2}}+{a}{x}+{b}={0}$$, we have $$\displaystyle\alpha=-{\left(\alpha+\beta\right)}$$ and $$\displaystyle{b}=\alpha\beta$$.
The equation $$\displaystyle{y}{''}+{a}{y}'+{b}{y}={f{{\left({t}\right)}}}={y}{''}-{\left(\alpha+\beta\right)}{y}'+\alpha\beta{y}$$
We now rewrite the left-hand side as $$\displaystyle{y}{''}-\alpha{y}'-\beta{y}'+\alpha\beta{y}={\left({y}{''}-\alpha{y}'\right)}-\beta{\left({y}'-\alpha{y}\right)}={f{{\left({t}\right)}}}$$ and substitute $$\displaystyle{z}={y}'-\alpha{y}$$ so that $$\displaystyle{z}'={y}{''}-\alpha{y}'$$ and finally the transformed equation becomes
$$\displaystyle{z}'-\beta{z}={f{{\left({t}\right)}}}$$
Once $$\displaystyle{z}={g{{\left({t}\right)}}}$$ is known, y is obtained from $$\displaystyle{y}'-\alpha{y}={g{{\left({t}\right)}}}$$
So one second order equation can be solved by solving two first order equations.
###### Have a similar question? Huses1969
Look at the underlying homogeneous equation $$\displaystyle{y}{''}-{2}{y}'+{2}={0}$$; it's characteristic polynomial equation is the quadratic $$\displaystyle\lambda^{{2}}-{2}\lambda+{2}={0}$$; the discriminant of this quadratic is $$\displaystyle{\left(-{2}\right)}^{{2}}-{4}{\left({2}\right)}={4}-{8}=-{4}{<}{0}$$</span>, so the equation has a pair of complex confugate roots, they are in fact $$\displaystyle{1}\pm{i}$$. Thus the (real) dimension of the solution space is 2, and it can't be reduced. Therefore, since the equation is intrinsically possessed of two degrees of freedom, typically manifested as $$\displaystyle{y}{\left({0}\right)}$$ and $$\displaystyle{y}'{\left({0}\right)}$$, the only way to lower the order is by expressing the original equation as a first order system on the two-dimensional space $$\displaystyle{\mathbb{{{R}}}}^{{2}}$$. Thus we set
$$\displaystyle{z}{\left({t}\right)}={y}'{\left({t}\right)}$$
so that
$$\displaystyle{z}'{\left({t}\right)}={2}{z}{\left({t}\right)}-{2}{y}{\left({t}\right)}+{e}^{{{2}{t}}}{\sin{{t}}}$$
In vector-matrix form this may be written as
$\left(\begin{array}{c}y(t)\\z(t)\end{array}\right)'=\begin{bmatrix}0 & 1 \\-2 & 2 \end{bmatrix}\left(\begin{array}{c}y(t)\\ z(t)\end{array}\right)+\left(\begin{array}{c}0\\ e^{2t}\sin t\end{array}\right)$
A first order system equivalent in all ways to the original second order ODE