Answer
$y=-3+2t$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
$y^{(4)}-4{y}'''+4{y}''=0 \;\;\;\;\Rightarrow \;\;\;\; r^4e^{rt}-4r^3e^{rt}+4r^2e^{rt}=0\\\\$
$r^2(r^2-4r+4)=r^2(r-2)(r-2)=0 $ $ \rightarrow\;\;\;\;\; r_{1},r_{2}= 0\;\;\;\;\;or\;\;\;r_{3}\;,r_{4}=2 \;\;\;\;\;\;\\\\$
So the 3 roots are: $\;\;\;r_{1},r_{2}=0 \;\;\;,\;\;r_{3},r_{4}=2 $
$y= C_{1}+tC_{2}+C_{3}e^{2t}+C_{4}te^{2t}$
Derivatives of the general solution;
${y}'=C_{2}+2C_{3}e^{2t}+2C_{4}te^{2t}+C_{4}e^{2t}$
${y}''=4C_{3}e^{2t}+4C_{4}te^{2t}+4C_{4}e^{2t}$
${y}'''=8C_{4}e^{2t}+8C_{4}te^{2t}+12C_{4}e^{2t}\\\\$
At;
$y(1)=C_{1}+C_{2}+e^2C_{3}+e^2C_{4}=-1 $
${y}'(1)=C_{2}+2C_{3}e^{2}+3C_{4}e^{2}=2$
${y}''(1)=4C_{3}e^{2}+8C_{4}e^{2}=0 $
${y}'''(1)=8C_{4}e^{2}+20C_{4}e^{2}=0 $
$\;\;\;\;\;\Rightarrow \;\;\;\;C_{3}=C_{4}=0\;\;\;\;\;\;\;C_{2}=2\;\;\;\;\;\;C_{1}=-3$
$\therefore y=-3+2t$
