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Quiz 6 solution
Use reduction of order to find a second linearly independent solution to the differential equation $$ty''+(1-2t)y'+(t-1)y=0,$$ on $(0,\infty)$ where $y_1(t)=e^t$ is a known solution. Write a general solution.
Solution: The reduction of order technique assumes the second independent solution $y_2$ takes the form $y_2(t)=u(t)y_1(t)$. In this case, the product rule shows that $y_2'=u'y_1+uy_1'$ and $$y_2''=u''y_1+u'y_1'+u'y_1'+uy_1''= uy_1''+2u'y_1'+u''y_1.$$ For us, $y_1(t)=e^t$, so $y_2'(t)=\Big( u(t)+u'(t)\Big)e^t$ and $$y_2(t)=u''(t)e^t+2u'(t)e^t+u(t)e^t=\Big( u''(t)+2u'(t)+u(t) \Big)e^t.$$ Substituting $y=y_2$ into the differential equation yields $$t\Big( u''(t)+2u'(t)+u(t) \Big)e^t+(1-2t)\Big( u'(t)+u(t)\Big)e^t +(t-1)u(t)e^t=0.$$ Divide by $e^t$ and expand the products to get $$tu''+2tu'+tu+u'-2tu'+u-2tu+tu-u=0.$$ Simplify to get $$tu''+u'=0.$$ To solve this equation, let $w=u'$ so that $w'=u''$. This turns the differential equation for $u$ into $tw'+w=0$. This is a first-order linear differential equation. To solve it, one can use the integrating factor method to ultimately obtain $(tw)'=0$. Integration yields $tw(t)=C$ for some constant $C$. Therefore $w(t)=\dfrac{C}{t}$. But since $u'(t)=w(t)=\dfrac{C}{t}$, we integrate once again to obtain $u(t)=\displaystyle\int \dfrac{C}{t} \mathrm{d}t = C\ln(t)$. Therefore, $y_2(t)=u(t)y_1(t)=C\ln(t)e^t$. Thus the general solution is $$y(t)=c_1e^t+c_2\ln(t)e^t.$$