# Thread: transform out 1st order term from 2nd order linear ode

1. ## transform out 1st order term from 2nd order linear ode

Hello,
I am a PHD student in engineering with a strong background in ODEs and PDEs.
I am working through a text called "Combustion Physics" by C. K. Law. On p.327, Law says:

"It is, however, well known that in a second order linear ordinary differential equation, the first order differential can be transformed away."

Such a technique is NOT well known to me and I cannot find anything like it in my undergrad ODEs text (Boyce and DiPrima). Unfortunately, Law doesn't provide a reference, nor does the example he provides explain the method. Can someone refer me to a book where I can learn more about this?
Thanks!

2. ## Re: transform out 1st order term from 2nd order linear ode

Consider

$y'' + p(x) y' + q(x) y = f(x)$

Let $y = a(x) u$ so

$y' = a u ' + a' u,$
$y'' = au'' + 2 a' u' + a''u$

Substitute. This gives

$au'' + 2 a' u' + a''u + p\left(a u ' + a' u \right) + q a u = f$

Re-grouping gives

$au'' + \left(2a' + p a \right)u' + \left(a'' + pa' + qa \right)u = f$

Now pick $a$ such that $2a' + pa = 0$. This gives an ODE absent of the term $u'$.

3. ## Re: transform out 1st order term from 2nd order linear ode

Danny,

Ah, a clever solution! What if, following your nomenclature, q(x) = 0 and in the end, I wish to have only a second order term? In other words, is there a method by which I can eliminate the first order differential from

y" + p(x) y' = f(x)

leaving only a second order differential and no u term?

Thanks,
T

4. ## Re: transform out 1st order term from 2nd order linear ode

Only for certain $p's$ because what you want is

$au'' + \left(2a' + p a \right)u' + \left(a'' + pa' \right)u = f$

to become

$au'' = f$

meaning that both $2a' + pa = 0$ and $a'' + pa' = 0$ must be satisfied. However, why go to all that trouble. If you have

$y'' + p(x) y' = f(x)$

then letting $u = y'$ gives a linear ODE in $u$