## Elementary Differential Equations and Boundary Value Problems 9th Edition

$$t^{2} y^{\prime \prime}- t y^{\prime}+5 y=0 , \quad t > 0$$ the general solution of the given differential equation is $$y(t) =t [c_{1} \cos (2 \ln t)+ c_{2} \sin (2 \ln t )]$$ where $c_{1}$ and $c_{2}$ are arbitrary constants.
$$t^{2} y^{\prime \prime}- t y^{\prime}+5 y=0 , \quad t > 0 \quad (1).$$ Let $x = \ln t$ . Then we have $$\frac{d y}{d t}=\frac{d y}{d x} \frac{d x}{d t}=\frac{1}{t} \frac{d y}{d x}$$ $$\frac{d^{2} y}{d t^{2}}=\frac{d}{d t}\left(\frac{1}{t} \frac{d y}{d x}\right)=\frac{-1}{t^{2}} \frac{d y}{d x}+\frac{1}{t} \frac{d}{d t}\left(\frac{d y}{d x}\right)=\frac{1}{t^{2}}\left(-\frac{d y}{d x}+\frac{d^{2} y}{d x^{2}}\right)$$ So if we use $x$ instead of $t$ as the variable, the equation (with unknown $y$ and variable $x$) reads $$\frac{d^{2} y}{d x^{2}}- 2\frac{d y}{d x}+ 5 y=0, \quad (2)$$ We assume that $y = e^{rx}$, and it then follows that $r$ must be a root of the characteristic equation $$r^{2}-2r+5=0,$$ so its roots are $$\:r_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}\quad$$ Thus the possible values of $r$ are $$r_{1}= 1+2i ,\quad r_{2}=1-2i,.$$ Therefore two solutions of Eq. (1) are $$y_{1}(x) = e^{(1+2i )x}= e^{x} ( \cos 2x + i \sin 2 x )$$ and $$y_{2}(x) = e^{(1-2i )x}= e^{x} ( \cos 2x - i \sin 2 x )$$ Thus the general solution of the differential equation is $$y(x) = e^{(x)} ( c_{1} \cos 2x+ c_{2} \sin 2 x ) \quad\quad\quad (2)$$ where $c_{1}$and $c_{2}$ are arbitrary constants. Back to t: $$\begin{split} y(t) & = e^{( \ln t)} ( c_{1} \cos 2 \ln t+ c_{2} \sin 2 \ln t ) \\ & = t [c_{1} \cos (2 \ln t)+ c_{2} \sin (2 \ln t )] \end{split}$$ where $c_{1}$and $c_{2}$ are arbitrary constants.