Question Number 227174 by Spillover last updated on 04/Jan/26
$${Solve}\:{the}\:{equation} \\ $$$$\left(\boldsymbol{{x}}−\mathrm{2}\right)\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}−\boldsymbol{{y}}=\left(\boldsymbol{{x}}−\mathrm{2}\right)^{\mathrm{3}} \\ $$$$\boldsymbol{{given}}\:\boldsymbol{{y}}=\mathrm{10}\:\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{4}. \\ $$$$\mathrm{4}/\mathrm{1}/\mathrm{2026} \\ $$
Answered by AgniMath last updated on 04/Jan/26
$$\left({x}−\mathrm{2}\right)\frac{{dy}}{{dx}}−{y}=\left({x}−\mathrm{2}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}−\frac{\mathrm{1}}{{x}−\mathrm{2}}\:.\:{y}=\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${IF}\:=\:{e}^{−\int\frac{\mathrm{1}}{{x}−\mathrm{2}}{dx}} ={e}^{−{ln}\left({x}−\mathrm{2}\right)} =\frac{\mathrm{1}}{{x}−\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{x}−\mathrm{2}}\:{y}=\int\frac{\mathrm{1}}{{x}−\mathrm{2}}\left({x}−\mathrm{2}\right)^{\mathrm{3}} {dx} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{x}−\mathrm{2}}{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}+{C} \\ $$$$\Rightarrow{y}=\left({x}−\mathrm{2}\right)\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}\right)+\left({x}−\mathrm{2}\right){C} \\ $$$${y}=\mathrm{10}\:{when}\:{x}=\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{2}{C}=\mathrm{10}\:\Rightarrow{C}=\mathrm{5} \\ $$$$\therefore\:{y}=\left({x}−\mathrm{2}\right)\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}\right)+\mathrm{5}\left({x}−\mathrm{2}\right) \\ $$$$=\frac{{x}^{\mathrm{3}} }{\mathrm{2}}−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{10}\:\:\left({Ans}\right) \\ $$
Commented by Spillover last updated on 04/Jan/26
$${correct} \\ $$