Background
1.4.1 Reduction of Order
If a solution $y_1(x)$ of the homogeneous Linear ODE (1.2) is known, then a second linearly independent solution $y_2(x)$ can be found by solving a first-order linear differential equation. A solution of (1.2) is sought in the form
$$y=u(x)y_1(x),\tag{1.4}$$
where $u(x)$ is a new unknown function. Inserting (1.4) into the homogeneous Linear ODE (1.2) and collecting terms with $u,u'$ and $u''$, we find
$$u''y_1+u'(2y'_1+py_1)+u[y''_1+py'_1+qy_1]=0.$$
The last term on the left-hand side is zero because $y_1(x)$ is a solution of (1.2), and we arrive at a first-order differential equation with respect to $U(x)=u'(x):$
$$y''+p(x)y'+q(x)y=0$$ $$y(x)=y_1(x)u(x)$$ $$u''y_1+u'(2y'_1+py_1)+u(y_1''+py_1'+qy_1)=0$$
The Question
Can someone please explain in layman's terms why $y_1(x)$ being a solution means that the last term on the left hand side is zero.
