Solve the initial value problem
(x−1)y′+3y=1(x−1)3+sin(x)(x−1)2\displaystyle {\left({x}-{1}\right)}{y}'+{3}{y}=\frac{{1}}{{\left({x}-{1}\right)}^{{3}}}+\frac{{\sin{{\left({x}\right)}}}}{{\left({x}-{1}\right)}^{{2}}}(x−1)y′+3y=(x−1)31+(x−1)2sin(x) with initial condition y(0)=1\displaystyle {y}{\left({0}\right)}={1}y(0)=1
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