Elementary Differential Equations and Boundary Value Problems 9th Edition

$$y^{\prime \prime}-6 y^{\prime}+9 y=0$$ The general solution of that equation is given by $$y=c_{1}t e^{3t} +c_{2} e^{3t}$$ where $c_{1}$ and $c_{2}$ are arbitrary constants.
$$y^{\prime \prime}-6 y^{\prime}+9 y=0 \quad \quad (1)$$ We assume that $y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation $$r^{2}- 6r+9=(r-3)^{2}=0,$$ so $r_{1}=r_{2}=3$. Therefore one solution of Eq. (1) is $y_{1}=e^{3t}$. To find the general solution of Eq. (1), we need a second solution that is not a multiple of $y_{1}$. This second solution can be found, by using D’Alembert's method, by replacing $c$ by a function $v(t)$ and then trying to determine $v(t)$ so that the product $v(t) y_{1}(t)$ is also a solution of Eq. (1). Now we substitute $y = v(t)y_{1}(t)$ in Eq. (1) and use the resulting equation to find $v(t)$. $$y=v(t) y_{1}(t)=v(t) e^{ 3t} \quad \quad (2)$$ we have $$y^{\prime}=v^{\prime}(t) e^{3t}+3 v(t) e^{3 t} \quad \quad (3)$$ and $$y^{\prime \prime}=v^{\prime \prime}(t) e^{ 3t}+6 v^{\prime}(t) e^{3t}+9 v(t) e^{ 3t} \quad \quad (4)$$ By substituting the expressions in Eqs. (2), (3), and (4) in Eq. (1) and collecting terms, we obtain $$v^{\prime \prime}(t)=0$$ Therefore $$v(t)=c_{1}t+c_{2}$$ where $c_{1}$ and $c_{2}$ are arbitrary constants. Finally, substituting for $v(t)$ in Eq. (2), we obtain $$y=c_{1}t e^{3 t} +c_{2} e^{3 t} \quad \quad (5)$$ The second term on the right side of Eq. (5) corresponds to the original solution $y_{1}(t) = e^{3t}$, but the first term arises from a second solution, namely, $y_{2}(t) = t e^{3t}$. We can verify that these two solutions form a fundamental set by calculating their Wronskian: $W(y_{1}, y_{2})(t)\ne0$ Therefore $y_{1}(t) = e^{3t}$, $y_{2}(t) = t e^{3t}$ form a fundamental set of solutions of Eq. (1), and the general solution of that equation is given by $$y=c_{1}t e^{3t} +c_{2} e^{3t}$$ where $c_{1}$ and $c_{2}$ are arbitrary constants.