Question Number 53890 by shaddie last updated on 27/Jan/19
$$\mathrm{pls}.\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{this} \\ $$$$\mathrm{if}\:\mathrm{y}=\frac{\mathrm{x}−\mathrm{2}}{\mathrm{2x}−\mathrm{1}}\:\mathrm{show}\:\mathrm{that}\:\left(\mathrm{2x}−\mathrm{1}\right)\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\mathrm{4}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{0} \\ $$
Answered by math1967 last updated on 27/Jan/19
$$\Rightarrow{y}\left(\mathrm{2}{x}−\mathrm{1}\right)=\left({x}−\mathrm{2}\right) \\ $$$$\Rightarrow\:\mathrm{2}{y}\:+\left(\mathrm{2}{x}−\mathrm{1}\right)\frac{{dy}}{{dx}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}\frac{{dy}}{{dx}}\:+\mathrm{2}\frac{{dy}}{{dx}}\:+\left(\mathrm{2}{x}−\mathrm{1}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\therefore\left(\mathrm{2}{x}−\mathrm{1}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\mathrm{4}\frac{{dy}}{{dx}}\:=\mathrm{0} \\ $$