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

$\;\;\;\;\;\;C_{1}=0\;\;\;\;\;\;\;C_{2}=2\;\;\;\;\;\;\;\;\;C_{3}=1$ $y(t)=2cos(t)+sin(t)+ln[sec(t)+tan(t)]-tcos(t)+sin(t)ln[cos(t)]$
The solution of the differential equation was found out in problem (4) : $y(t)=C_{1}+C_{2}cos(t)+C_{3}sin(t)+ln(sec(t)+tan(t))-tcos(t)+sin(t)ln(cos(t))$ ${y}'=-C_{2}sin(t)+C_{3}cos(t)+tsin(t)-cos(t)-sin(t)tan(t)+cos(t)ln(cos(t))+sec(t)$ ${y}''=-C_{2}cos(t)-C_{3}sin(t)+sin(t)+tcos(t)-cos(t)tan(t)-sin(t)sec^2(t)-sin(t)ln(cos(t))-sec(t)tan(t)$ $y(0)=C_{1}+C_{2}=2$ ${y}'(0)=C_{3}-1+1=1$ ${y}''(0)=-C_{2}=-2$ So; $\;\;\;\;\;\;C_{1}=0\;\;\;\;\;\;\;C_{2}=2\;\;\;\;\;\;\;\;\;C_{3}=1$ $y(t)=2cos(t)+sin(t)+ln(sec(t)+tan(t))-tcos(t)+sin(t)ln(cos(t))$