# Second order linear differential ODEs Reduction of order method

## 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.

You have a differential equation: $$y''+py'+qy=0$$ Then $$y_1$$ is a solution of the DE this means: $$y_1''+py_1'+qy_1=0$$ This is why $$u(y_1''+py_1'+qy_1)=0$$.