# Multiplicative reduction of order

I am working through my textbook on differential equations (Bender & Orszag). I am looking specifically at the following question:

Show that a reduction of order identical to $y(x) = u(x)f(x)$ can be used to eliminate the $y'(x)$ term from a second order homogeneous linear differential equation with variable coefficients, reducing it to a (time independent) Schrodinger equation.

Now, I begin with $$y''(x) + a(x)y'(x) + b(x)y(x) = 0$$ and suppose we have one known solution, namely $y_1(x)$ such that $$y_1(x)'' + a(x)y_1'(x) + b(x)y_1(x) = 0$$ and make the substitution $y_2(x) = u(x)y_1(x)$ such that $$y_2'(x) = u'(x)y_1(x) + u(x)y_1'(x)$$ and then $$y_2''(x) = u''(x)y_1(x) + u(x)y_1''(x) + 2u'(x)y_1(x)$$ I can no substitute these derivatives into the original differential equation (suppressing the arguments from now on), such that we get $$u''y_1 + uy_1'' + 2u'y_1' + a(u'y_1 + uy_1') + buy_1 = 0$$ But this can be rephrased in the following way: $$\Rightarrow u(y_1'' + ay_1' + by_1) + u''y_1 + 2u'y_1' + au'y_1 = 0$$ Since the term inside the first set of brackets is defined to be equal to zero (i.e., it is a solution to the ode!) we get $$\Rightarrow u''y_1 + 2u'y_1' + au'y_1 = 0$$ Am I any closer to getting rid of the first derivatives in the differential equation?

Insert $y'=u'f+uf'$ and $y''=u''f+2u'f'+uf''$ and sort for derivatives of $u$ to get $$0=u''f+2u'f'+uf''+a(u'f+uf')+buf \\ =fu''+(2f'+af)u'+(f''+af'+bf)u$$ From that you see that you need to chose $2f'+af=0$.