## Calculus: Early Transcendentals 8th Edition

For $r=2\pm\sqrt 3$, then the function $y=e^{rx}$ satisfies the differential equation.
$$y=e^{rx}$$ 1) Find the first derivative of $y$ $$y'=\frac{d(e^{rx})}{dx}$$ Apply the Chain Rule: $$y'=\frac{d(e^{rx})}{d(rx)}\frac{rdx}{dx}$$ $$y'=e^{rx}\times r\times1$$ $$y'=re^{rx}$$ 2) Find the second derivative $y$ $$y''=(re^{rx})'$$ $$y''=r(e^{rx})'$$ We already know from part 1) that $(e^{rx})'=re^{rx}$ Therefore, $$y''=r^2e^{rx}$$ 3) Now we apply what we have found above to the differential equation $$y''-4y'+y=0$$ $$r^2e^{rx}-4re^{rx}+e^{rx}=0$$ $$e^{rx}(r^2-4r+1)=0$$ $$e^{rx}=0$$ or $$r^2-4r+1=0$$ *Consider $e^{rx}=0$ However, we know that $e^{rx}\gt0$ for all $x\in R$ Therefore, there cannot be any $x$ that satisfies $e^{rx}=0$ *Consider $r^2-4r+1=0$ $$r^2-4r+1=0$$ Now you can use calculator to deal with the equation. The results would be $$r=2\pm\sqrt 3$$ Therefore, for $r=2\pm\sqrt 3$, then the function $y=e^{rx}$ satisfies the differential equation.