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

$$t^{2} y^{\prime \prime}+4t y^{\prime}+2y=0 , \quad t > 0$$ the general solution of the given differential equation is $$y(t) =c_{1} t^{-1} +c_{2} t^{-2}$$
$$t^{2} y^{\prime \prime}+4t y^{\prime}+2y=0 , \quad t > 0 \quad (1).$$ 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 $$\frac{d^{2} y}{d x^{2}}+ 3\frac{d y}{d x}+ 2 y=0, \quad (2)$$ We assume that $y = e^{rx}$, and it then follows that $r$ must be a root of the characteristic equation $$r^{2}+3r+2=0,$$ so its roots are $$r_{1}=-1 ,\quad r_{2}= -2 .$$ Therefore the general solution of the differential equation (2) is $$y(x) = c_{1} e^{-x}+c_{2} e^{-2x}$$ where $c_{1}$and $c_{2}$ are arbitrary constants. Back to t: $$\begin{split} y(t) & = c_{1} e^{-( \ln t )}+c_{2} e^{-2 ( \ln t)} \\ & = c_{1} e^{( \ln t^{-1} )}+c_{2} e^{( \ln t^{-2})} \\ & =c_{1} t^{-1} +c_{2} t^{-2} \end{split}$$ where $c_{1}$and $c_{2}$ are arbitrary constants.