Obtaining particular solution of second order linear DE from first

[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

 

[HallsofIvy]

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.

 

[K29]

Solved. Thank you.