Solve the initial-value problem for x\displaystyle {x}x as a function of t\displaystyle {t}t .(2t3−2t2+t−1)dxdt=9,x(2)=0\displaystyle {\left({2}{t}^{{3}}-{2}{t}^{{2}}+{t}-{1}\right)}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={9},{x}{\left({2}\right)}={0}(2t3−2t2+t−1)dtdx=9,x(2)=0
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