Second order, homogeneous, linear differential equation transformation with a substitution

Question

Show that using the substitution: $$ y(x) = u(x)v(x) \quad $$where $$ v(x) = exp(-\frac12 \int p(x)\ dx)$$ transforms the DE: $$y''+p(x)y'+q(x)y=0$$ into: $$u''+f(x)u=0$$

Answer 1

If we start with the equation

$y'' + p(x)y' + q(x)y = 0, \tag 1$

and define

$v(x) = \exp \left (-\dfrac{1}{2} \displaystyle \int p(x) \; dx \right), \tag 2$

and define $u(x)$ via

$y(x) = v(x) u(x), \tag 3$

which is always possible since $v(x) \ne 0; \tag 4$

we note that

$v'(x) = -\dfrac{1}{2}p(x) \exp \left ( -\dfrac{1}{2} \displaystyle \int p(x) \; dx \right ) = -\dfrac{1}{2}p(x)v(x); \tag 5$

thus

$y'(x) = v'(x) u(x) + v(x) u'(x)$ $= -\dfrac{1}{2}p(x) v(x) u(x) + v(x) u'(x) = (-\dfrac{1}{2}p(x)u(x) + u'(x))v(x); \tag 6$

$y''(x) = (-\dfrac{1}{2}p(x)u(x) + u'(x))'v(x) + (-\dfrac{1}{2}p(x)u(x) + u'(x)) v'(x) = (-\dfrac{1}{2}p'(x)u(x) - \dfrac{1}{2}p(x)u'(x) + u''(x))v(x) + (-\dfrac{1}{2}p(x)u(x) + u'(x)) (-\dfrac{1}{2}p(x)v(x))$ $= (-\dfrac{1}{2}p'(x)u(x) - \dfrac{1}{2}p(x)u'(x) + u''(x))v(x) + (\dfrac{1}{4}p^2(x)u(x) - \dfrac{1}{2} p(x)u'(x)) v(x)$ $= (\dfrac{1}{4}p^2(x) u(x) - \dfrac{1}{2}p'(x)u(x) - p(x) u'(x) + u''(x))v(x); \tag{7}$

$y''(x) + p(x)y'(x)$ $= (\dfrac{1}{4}p^2(x) u(x) - \dfrac{1}{2}p'(x)u(x) - p(x) u'(x) + u''(x))v(x) + (-\dfrac{1}{2}p^2(x)u(x) + p(x)u'(x))v(x)$$ = (-\dfrac{1}{4}p^2(x) u(x) - \dfrac{1}{2}p'(x)u(x) + u''(x))v(x); \tag{8}$

$y''(x) + p(x)y'(x) + q(x)y(x)$ $= (-\dfrac{1}{4}p^2(x) u(x) - \dfrac{1}{2}p'(x)u(x) + u''(x))v(x) + q(x) u(x) v(x)$ $= (u''(x) - \dfrac{1}{2}p'(x)u(x) - \dfrac{1}{4}p^2(x)u(x) + q(x)u(x))v(x); \tag{9}$

if we now set

$f(x) = q(x) - \dfrac{1}{2}p'(x) - \dfrac{1}{4}p^2(x), \tag{10}$

then (9) becomes

$y''(x) + p(x)y'(x) + q(x)y(x) = (u''(x)+ f(x)u(x))v(x); \tag{11}$

we now use (1) and the fact that $v(x) \ne 0$ and conclude that

$u''(x) + f(x)u(x) = 0. \tag{12}$

Answer 2

$$ v(x) = \exp(-\frac12 \int p(x)\ dx)$$ $v'=-\frac12 p(x)\exp(-\frac12 \int p(x)\ dx) = -\frac12 pv$

$v''=-\frac12( p'v+pv')= -\frac12 p'v+\frac14 p^2v$

$$y'=(uv)'=v'u+vu'=-\frac12 pvu+vu'$$ $$y''=v''u+2v'u'+vu''=-\frac12 p'vu+\frac14 p^2vu-pvu'+vu''$$ $y''+py'+qy=0$

$\left(-\frac12 p'vu+\frac14 p^2vu-pvu'+vu''\right)+p\left(-\frac12 pvu+vu' \right)+q(uv)=0$

After simplification :

$\left(-\frac12 p'u+\frac14 p^2u-pu'+u''-\frac12 p^2u+pu' +qu\right)v=0$ $$u''+\left(-\frac12 p'-\frac14 p^2 +q\right)u=0$$ Let $\quad -\frac12 \frac{dp}{dx}-\frac14 p(x)^2 +q(x)=f(x)$ $$u''+f(x)u=0$$