Solve the Initial value problem

4y′′−12y′+9y=0,y(0)=3,y′(0)=52\displaystyle {4}{y}{''}-{12}{y}'+{9}{y}={0},\quad{y}{\left({0}\right)}={3},{y}'{\left({0}\right)}=\frac{{5}}{{2}}

y(x)=\displaystyle {y}{\left({x}\right)}=  

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