Solve the initial-value problem for x\displaystyle {x}x as a function of t\displaystyle {t}t .(t2−6t+8)dxdt=1,(t>4,x(9)=0)\displaystyle {\left({t}^{{2}}-{6}{t}+{8}\right)}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={1},{\left({t}>{4},{x}{\left({9}\right)}={0}\right)}(t2−6t+8)dtdx=1,(t>4,x(9)=0)
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