TITLE: Converting Second Order Linear Equations to First Order Linear Equations
QUESTION [5 upvotes]: $\color{green}{\text{Question}}$:
How can the following $\color{blue}{\text{second-order linear equation}}$ be converted into a $\color{blue}{\text{first-order linear equation}}$?

This is our second-order linear Equation:
  $${y}''-2y'+2y=e^{2t}\sin t$$

$\color{green}{\text{I think}\ldots}$

This equation contains the dependent variable and the independent variables.
Equation is not complete.
Equation is not homogeneous.

How can I convert it to a first-order linear equation?
Thank you for any hint.

REPLY [4 votes]: Note that if $\alpha$ and $\beta$ are the roots of $x^2+ax+b=0$, we have $a=-(\alpha+\beta)$ and $b=\alpha\beta$.
The equation $y''+ay'+by=f(t)=y''-(\alpha+\beta)y'+\alpha\beta y$.
We now rewrite the left-hand side as $y''-\alpha y'-\beta y' +\alpha\beta y= (y''-\alpha y')-\beta(y' -\alpha y)=f(t)$ and substitute $z=y'-\alpha y$   so that $z'=y''-\alpha y'$ and finally the transformed equation becomes $$z'-\beta z=f(t).$$
Once $z=g(t)$ is known, $y$ is obtained from $y'-\alpha y=g(t)$.
So one second order equation can be solved by solving two first order equations.