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

$y(t)=1+\frac{1}{4}(t^2+3t)-te^t$
Let $\;\;\;\;\;y=e^{rt}\\\\$ ${y}'''-3{y}''+2{y}'=0 \;\;\;\;\Rightarrow \;\;\;\; r^3e^{rt}-3r^2e^{rt}+2re^{rt}=0\\\\$ $r^3-3r^2+2r=r(r-1)(r-2)=0$ $\rightarrow\;\;\;\;\; r_{1}=0\;\;\;\;\;\;\;or\;\;\;\;,r_{2}=1\;\;\;\;or\;\;\;\;r_{3}=2\\\\$ $\boxed{y_{c}(t)= C_{1}+C_{2}e^{t}+C_{3}e^{2t}}$ Let; $\;\;\;\;Y(t)=At^2+Bt+Cte^t$ ${Y}'=2At+B+Ce^t+Cte^t$ ${Y}''=2A+2Ce^t+Cte^t$ ${Y}'''=3Ce^t+Cte^t$ ${Y}^{(4)}=16Acos(2t)+16Bsin(2t)$ ${y}'''-3{y}''+2{y}'=t+e^t$ $3Ce^t+Cte^t-6A-6Ce^t-3Cte^t+4At+2B+2Ce^t+2Cte^t=t+e^t$ $-Ce^t+4At+(-6A+2B) =t+e^t \;\;\;\;\;\;\Rightarrow \;\;\;A=\frac{1}{4}\;\;\;,\;\;\;B=\frac{3}{4}\;\;\;\;\;,\;\;\;C=-1$ $\boxed{Y(t)=\frac{1}{4}t^2+\frac{3}{4}t-te^t}$ The general solution : $y(t)=y_{c}(t)+Y(t)$ $y(t)=C_{1}+C_{2}e^{t}+C_{3}e^{2t}+\frac{1}{4}t^2+\frac{3}{4}t-te^t$ $y(0)=C_{1}+C_{2}+C_{3}=1 \;\;\;\;\;\; \;\;\;\;$ ${y}'(0)=C_{2}+2C{3}-\frac{1}{4}=\frac{1}{4} \;\;\;\;\;\; \;\;\;\;$ ${y}''(0)=C_{2}+4C_{3}-\frac{-3}{2}=\frac{-3}{2}\;\;\;\;\;\; \;\;\;\;$ $C_{1}=1\;\;\;\;\;,\;\;\;C_{2}=0\;\;\;\;,\;\;\;\;C_{3}=0\;\;\;\;\;\;\;\;$ $y(t)=1+\frac{1}{4}t^2+\frac{3}{4}t-te^t$ $y(t)=1+\frac{1}{4}(t^2+3t)-te^t$