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!

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

Consider

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

Let $\displaystyle y = a(x) u$ so

$\displaystyle y' = a u ' + a' u,$

$\displaystyle y'' = au'' + 2 a' u' + a''u$

Substitute. This gives

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

Re-grouping gives

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

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

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

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

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

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

to become

$\displaystyle au'' = f$

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

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

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