Thread: Second Order Linear Differential Equation

1. Second Order Linear Differential Equation

If anyone can guide me through this problem, it would be much helpful. Thank you.

Let u be a nonzero solution of the second-order equation

y'' + P(x)y' + Q(x)y = 0

Show that the substitution y = uv converts the equation

y'' + P(x)y' + Q(x)y = R(x)

into a first-order linear equation for v'

2. Originally Posted by hashshashin715
If anyone can guide me through this problem, it would be much helpful. Thank you.

Let u be a nonzero solution of the second-order equation

y'' + P(x)y' + Q(x)y = 0

Show that the substitution y = uv converts the equation

y'' + P(x)y' + Q(x)y = R(x)

into a first-order linear equation for v'
let

$y=uv$
then
$y'=u \frac{dv}{dx}+\frac{du}{dx}v$
and
$y''=u \frac{d^2v}{dx^2}+2\frac{du}{dx}\frac{dv}{dx}+\fra c{d^2u}{dx^2}v$
Substitute these values in the given equation and determine u

3. Ok, but if I do that, I will still have u' and u'' left over. What do I do with them? Aren't I only suppose to express the equation in terms of v's and their derivatives only?

4. Originally Posted by hashshashin715
Ok, but if I do that, I will still have u' and u'' left over. What do I do with them? Aren't I only suppose to express the equation in terms of v's and their derivatives only?

After Substituting, we get,

$(u''v+2u'v'+uv'')+P(u'v+uv')+Quv=R$

$\therefore uv''+(ru'+Pu)v'+(u''+Pu'+Qu)v=R$

Since $u$is a nonzero solution, then,

$u''+Pu'+Qu=0$

$\therefore uv''+(2u'+Pu)v'=R$

$\therefore v''+\left(\frac{2}{u}u'+P\right)v'=\frac{R}{u}$

Now put $v'=p$, so that, $v''=\frac{dp}{dx}$, then we have,

$\frac{dp}{dx}+\left(\frac{2}{u}u'+P\right)p=\frac{ R}{u}$

Which is a linear equation in p,

$\therefore$ I.F.= $e^{\int \left(\frac{2}{u}u'+P\right)dx}$ $=u^2e^{\int P dx}$

$\therefore pu^2e^{\int P dx}=\int \left(\frac{R}{u}u^2e^{\int P dx}\right)dx+c_1$

5. Oh, I see. Thank you