-a.3.2.96 Convert a 2nd order homogeneous ODE into a system of first order ODEs

• MHB
• karush
In summary, a 2nd order homogeneous ODE is a mathematical equation involving a dependent variable, its derivatives, and independent variables where the highest derivative is second order and all terms are homogeneous. Converting it into a system of first order ODEs separates the second order derivative into two first order derivatives, making it easier to solve using numerical methods or specialized software. This process also provides a clearer understanding of the system and its dynamics. However, it can result in a larger and more complex system of equations and may not be suitable for all equations. Careful consideration is necessary before converting into a system of first order ODEs.
karush
Gold Member
MHB
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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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 $\begin{pmatrix}y' \\ u'\end{pmatrix}= \begin{pmatrix}0 & 1 \\ -6 &-5 \end{pmatrix}\begin{pmatrix}y \\ u \end{pmatrix}$

HallsofIvy said:
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

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!

Last edited by a moderator:
karush said:
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".

HallsofIvy said:
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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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.

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

so the general solution is

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

Last edited:
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, $$y(x)= Ae^{-3x}+ Be^{-2x}$$.

1. What is a 2nd order homogeneous ODE?

A 2nd order homogeneous ODE (ordinary differential equation) is a mathematical equation that describes the relationship between a function and its derivatives up to the second order, where the function and its derivatives are all of the same degree.

2. What does it mean to convert an ODE into a system of first order ODEs?

Converting an ODE into a system of first order ODEs involves rewriting the original equation as a set of first order equations, where each equation represents one of the derivatives of the original function. This allows for easier numerical or analytical solution of the equation.

3. Why would one want to convert a 2nd order homogeneous ODE into a system of first order ODEs?

Converting a 2nd order homogeneous ODE into a system of first order ODEs can make the equation easier to solve, especially for numerical methods. It also allows for more flexibility in manipulating and analyzing the equation.

4. How do you convert a 2nd order homogeneous ODE into a system of first order ODEs?

To convert a 2nd order homogeneous ODE into a system of first order ODEs, you can introduce new variables to represent the derivatives of the original function. Then, rewrite the original equation in terms of these new variables and their derivatives. This will result in a system of first order equations.

5. Are there any limitations or drawbacks to converting a 2nd order homogeneous ODE into a system of first order ODEs?

One limitation of converting a 2nd order homogeneous ODE into a system of first order ODEs is that it can result in a larger system of equations, which may be more complex to solve. Additionally, the process of converting the equation may introduce errors or inaccuracies. It is important to carefully check the resulting system of equations to ensure accuracy.

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