# [SOLVED]30.1 - Convert a 2nd order homogeneous ODE into a system of first order ODEs

#### karush

##### Well-known member
given the differential equation
$\quad y''+5y'+6y=0$
(a)convert into a system of first order (homogeneous) differential equation
(b)solve the system.

ok just look at an example the first step would be
$\quad u=y'$

then
$\quad u'+5u+6=0$

so far perhaps???

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#### HallsofIvy

##### Well-known member
MHB Math Helper
Not quite. Your system of first order equations is
y'= u
u'+ 5u+ 6y= 0. You forgot the "y" in "6y"!

You could also write that as the matrix equation $\begin{pmatrix}y' \\ u'\end{pmatrix}= \begin{pmatrix}0 & 1 \\ -6 &-5 \end{pmatrix}\begin{pmatrix}y \\ u \end{pmatrix}$

#### karush

##### Well-known member
Not quite. Your system of first order equations is
y'= u
u'+ 5u+ 6y= 0. You forgot the "y" in "6y"!

You could also write that as the matrix equation $\begin{pmatrix}y' \\ u'\end{pmatrix}= \begin{pmatrix}0 & 1 \\ -6 &-5 \end{pmatrix}\begin{pmatrix}y \\ u \end{pmatrix}$
so are you suggesting the u substitution is not the best way to go
in then now we have 2 variables u and y

#### HallsofIvy

##### Well-known member
MHB Math Helper
Since it is, initially, a "second order linear equation with constant coefficients", about the easiest kind of equation there is, I personally would not change it to two first order equations. The "characteristic equation" is $$r^2+ 5r+ 6= (r+ 2)(r+ 3)= 0$$ which has solution r= -2 and r= -3. The general solution to the equation is $$y(x)= Ae^{-2x}+ Be^{-3x}$$.

But I know that some people just feel more comfortable with systems of first order equations or matrix equations (I suspect those are engineers who use computer solvers). The method I first learned to solve a pair of equations like $$y'= u$$, $$u'= -5u- 6y$$ was to differentiate the first equation again, to get $$y''= u'$$ and then replace u' and u in the second equation to get $$y''= -5y'+ 6y$$ again!

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#### HallsofIvy

##### Well-known member
MHB Math Helper
so are you suggesting the u substitution is not the best way to go
in then now we have 2 variables u and y
Perhaps you are not understanding what the question is asking. There is no "u substitution" in the "Calculus" sense. We are introducing the new variable u so that instead or a single (second order) equation in y we have two equations (first order) in y and u. That was what the problem asked you to do- change the single second order equation to two first order equations. In order to have two equations we have to have two "unknowns".

#### karush

##### Well-known member
Not quite. Your system of first order equations is
y'= u
u'+ 5u+ 6y= 0. You forgot the "y" in "6y"!

You could also write that as the matrix equation $\begin{pmatrix}y' \\ u'\end{pmatrix}= \begin{pmatrix}0 & 1 \\ -6 &-5 \end{pmatrix}\begin{pmatrix}y \\ u \end{pmatrix}$
so I was going to the eiganvalues of the matrix but we have y' and u'

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#### HallsofIvy

##### Well-known member
MHB Math Helper
Why is that a problem? Yes, find the eigenvalues of the matrix. Calling them $\lambda_1$ and $\lambda_2$, the general solution is $y(x)= C_1e^{\lambda_1 x}+ C_2e^{\lambda_2 x}$. You don't need to find u.

#### karush

##### Well-known member
$\left[ \begin{array}{rr} - \lambda & 1 \\ -6 & - \lambda - 5 \end{array} \right] =- \lambda \left(- \lambda - 5\right) + 6=\lambda^2+5\lambda+6=(\lambda+3)(\lambda+2)$
the roots are
$\lambda_1=-3$
$\lambda_2=-2$

so the general solution is

$$\displaystyle y(x)= Ae^{-2x}+ Be^{-3x}$$

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#### HallsofIvy

##### Well-known member
MHB Math Helper
Yes, that is correct. Recall that the original second order differential equation was y''+ 5y'+ 6y= 0. If you "try" a solution of the form $$y(x)= e^{\lambda x}$$ the equation becomes $$\lambda^2 e^{\lambda x}+ 5\lambda e^{\lambda x}+ 6e^{\lambda x}= (\lambda^2+ 5\lambda+ 6)e^{\lambda x}= 0$$ and since $$e^{\lambda x}$$ is never 0 so we must have $$\lambda^2+ 5\lambda+ 6= 0$$, the same as the "characteristic equation" (or "eigenvalue equation") for that matrix. That equation, $$(\lambda+ 3)(\lambda+ 2)= 0$$, has roots $$\lambda= -3$$ and $$\lambda= -2$$ so $$e^{-3x}$$ and $$e^{-2x}$$ are solutions and the "general solution" is, as you say, $$y(x)= Ae^{-3x}+ Be^{-2x}$$.