Question Number 143008 by mohammad17 last updated on 08/Jun/21
$${Solve}\::\:\:{x}={p}^{\mathrm{3}} −{p}+\mathrm{2}\:\:,\:{y}^{'} ={p} \\ $$
Commented by mohammad17 last updated on 09/Jun/21
$${how}\:{sir}\:{can}\:{you}\:{give}\:{me}\:{steb}\:{by}\:{steb}\:{please}? \\ $$
Answered by ajfour last updated on 09/Jun/21
$$\frac{{dy}}{{dx}}={p} \\ $$$$\frac{{dx}}{{dp}}=\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\:{dy}={pdx} \\ $$$$\Rightarrow\:\:{dy}={p}\left(\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1}\right){dp} \\ $$$$\Rightarrow\:{y}=\frac{\mathrm{3}{p}^{\mathrm{4}} }{\mathrm{4}}−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+{k} \\ $$$${p}^{\mathrm{4}} −\frac{\mathrm{2}}{\mathrm{3}}{p}^{\mathrm{2}} +\frac{\mathrm{4}{k}}{\mathrm{3}}−{y}=\mathrm{0} \\ $$$$\&\:\:{p}^{\mathrm{3}} ={p}+{x}−\mathrm{2} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} +{px}−\mathrm{2}{p}−\frac{\mathrm{2}}{\mathrm{3}}{p}^{\mathrm{2}} +\frac{\mathrm{4}{k}}{\mathrm{3}}−{y}=\mathrm{0} \\ $$$$\Rightarrow\:\frac{{p}^{\mathrm{2}} }{\mathrm{3}}−\mathrm{2}{p}+\frac{\mathrm{4}{k}}{\mathrm{3}}−{y}+{px}=\mathrm{0} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} =\mathrm{6}{p}−\mathrm{4}{k}+\mathrm{3}{y}−\mathrm{3}{px} \\ $$$$\Rightarrow\:\mathrm{6}{p}^{\mathrm{2}} −\mathrm{4}{kp}+\mathrm{3}{py}−\mathrm{3}{p}^{\mathrm{2}} {x} \\ $$$$\:\:\:\:\:\:=\:{p}+{x}−\mathrm{2} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{2}} \left(\mathrm{6}−\mathrm{3}{x}\right)+\left(\mathrm{3}{y}−\mathrm{4}{k}−\mathrm{1}\right){p} \\ $$$$\:\:\:\:\:\:\:\:={x}−\mathrm{2}\:\:\:\:....\left({II}\right) \\ $$$${and}\:\:{p}^{\mathrm{2}} −\left(\mathrm{6}−\mathrm{3}{x}\right){p}=\mathrm{3}{y}−\mathrm{4}{k} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{3}} +\left(\mathrm{3}{y}−\mathrm{4}{k}−\mathrm{1}\right){p} \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\mathrm{3}{y}−\mathrm{4}{k}\right){p} \\ $$$$\Rightarrow\:\:\:{p}^{\mathrm{3}} −{p}=\mathrm{0} \\ $$$${x}=\left({p}^{\mathrm{3}} −{p}\right)+\mathrm{2}\:=\:\mathrm{2} \\ $$$$ \\ $$