Question

The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2, y2 = y1(x) e−∫P(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). y'' − 12y' + 36y = 0; y1 = e6x

Answer 1

Given $y_1=e^{6x}$, assume a second solution of the form $y_2=vy_1$, with derivatives

${y_2}'=v'y_1+v{y_1}'$

${y_2}''=v''y_1+2v'{y_1}'+v{y_1}''$

With $y_1=e^{6x}$, you have ${y_1}'=6e^{6x}$ and ${y_1}''=36e^{6x}$.

Substitute these into the ODE and you get

$(e^{6x}v''+12e^{6x}v'+36e^{6x}v)-12(e^{6x}v'+6e^{6x}v)+36e^{6x}v=0$

$v''+12v'=0$

Now substitute $w=v'$, so that $w'=v''$ and you have a linear first-order ODE:

$w'+12w=0\implies e^{12x}w'+12e^{12x}w=(e^{12x}w)'=0\implies e^{12x}w=C$

$\implies w=v'=Ce^{-12x}$

$\implies v=C_1e^{-12x}+C_2$

$\implies y_2=(C_1e^{-12x}+C_2)e^{6x}=C_1e^{-6x}+C_2e^{6x}$

But $y_1=e^{6x}$ is already accounted for, so the second fundamental solution to the ODE is $y_2=e^{-6x}$.