# Obtaining particular solution of second order linear DE from first

1. Feb 18, 2012

### K29

In cullen-zill chapter 6 equation 23 it says that

$y_{2}(x)=y_{1}(x)\int\frac{e^{-\int P(x)dx}}{y_{1}^{2}(x)}dx$
is a solution of
$y''+P(x)y'+Q(x)y=0$
whenever $y_{1}(x)$ is a known solution

Where does this come from? I would like to be able to prove this or find a proof somewhere.

My first thought is that since the general solution solution is $y_{h}=C_{1}y_{1}+C_{2}y_{2}$ then
$y_{2}=v(x)y_{1}$ where v(x) is just a function of x.
Thought maybe I could substitute that into the general form of the DE, but it doesn't seem to help much

2. Feb 18, 2012

### HallsofIvy

Staff Emeritus
Yes, that method, "variation of parameters", will work. With $y_2= v(x)y_1$, $y_2'= v'(x)y_1+ v(x)y_1'$ and $y_2''= v''(x)y_1+ 2v'(x)y_1'+ v(x)y_1''$. Putting those into the equation,
$$v''y_1+ 2v'y'+ vy_1''+ P(x)v'y_1+ P(x)vy_1'+ Q(x)vy_1= y_1v''+ 2v'y_1'+ P(x)v'y_1+ v(y_1''+ P(x)y_1'+ Q(x)y_1)= 0$$

Of course, that last term in parentheses is 0 so we have $y_1v''+ (2y_1+ P(x)y_1)v'= 0$. Notice that there is no "v(x)" (undifferentiated) so if we let u= v', we have $y_1u'+ (2y_1+ P(x)y_1)u= 0$, a separable first order equation.

Last edited: Feb 19, 2012
3. Feb 19, 2012

### K29

Solved. Thank you.