# Chapter 17 - Second-Order Differential Equations - 17.2 Exercises: 3

$y=c_{1}cos(3x)+c_{2}sin(3x)+\frac{1}{13}e^{-2x}$

#### Work Step by Step

$y''+9y=e^{-2x}$ Use auxiliary equation $r^{2}+9=0$ $r^{2}=-9$ $r=±3i$ $α=0$ $β=3$ $y_{c}=e^{αx}(c_{1}cos(βx)+c_{2}sin(βx))$ $y_{c}=e^{0x}(c_{1}cos(3x)+c_{2}sin(3x))$ $y_{c}=c_{1}cos(3x)+c_{2}sin(3x)$ $y_{p}=Ae^{-2x}$ $y_{p}'=-2Ae^{-2x}$ $y_{p}''=4Ae^{-2x}$ Plug into original equation $4Ae^{-2x}+9Ae^{-2x}=e^{-2x}$ $4A+9A=1$ $A=\frac{1}{13}$ $y_{p}=\frac{1}{13}e^{-2x}$ $y=y_{c}+y_{p}$ $y=c_{1}cos(3x)+c_{2}sin(3x)+\frac{1}{13}e^{-2x}$

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