Question Number 143165 by mohammad17 last updated on 11/Jun/21
$${solve}\:{the}\:{differention}\:{equation} \\ $$$${x}={p}^{\mathrm{3}} −{p}+\mathrm{2}\:\:\:{since}:{p}={y}' \\ $$
Commented by mohammad17 last updated on 11/Jun/21
$$???????\: \\ $$
Commented by mr W last updated on 11/Jun/21
$${you}\:{can}'{t}\:{expect}\:{an}\:{answer}\:{to}\:{a} \\ $$$${question}\:{which}\:{is}\:{no}\:{question}. \\ $$$${x}={p}^{\mathrm{3}} −{p}+\mathrm{2}\:{is}\:{neither}\:{a}\:{differential} \\ $$$${equation}\:{nor}\:{a}\:{differention}\:{equation}. \\ $$$${it}\:{is}\:{even}\:{not}\:\:{an}\:{equation}\:{at}\:{all}. \\ $$
Commented by mohammad17 last updated on 11/Jun/21
$${sir}\:{its}\:{differential}\:{equation}\:{such}\:{that} \\ $$$$ \\ $$$${p}={y}^{'} \\ $$
Commented by mr W last updated on 11/Jun/21
$${but}\:{you}\:{didn}'{t}\:{give}\:{this}\:{before}. \\ $$$${besides},\:{what}\:{is}\:{meant}\:{with}\:{y}'\:? \\ $$$${is}\:{it}\:{y}'=\frac{{dy}}{{dx}}\:{or}\:{y}'=\frac{{dy}}{{dp}}\:? \\ $$$${in}\:{mathemstics}\:{the}\:{things}\:{must}\:{be} \\ $$$${exact}\:{and}\:{clear}! \\ $$
Answered by Cwesi last updated on 10/Jun/21
$$\frac{{dx}}{{dp}}=\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1} \\ $$
Answered by mr W last updated on 11/Jun/21
$${y}'=\frac{{dy}}{{dx}}={p} \\ $$$$\frac{{dy}}{{dp}}×\frac{{dp}}{{dx}}={p} \\ $$$$\frac{{dy}}{{dp}}={p}\frac{{dx}}{{dp}}={p}\left(\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{3}{p}^{\mathrm{3}} −{p} \\ $$$${y}=\frac{\mathrm{3}{p}^{\mathrm{4}} }{\mathrm{4}}−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\frac{{C}_{\mathrm{1}} }{\mathrm{4}} \\ $$$$\mathrm{3}{p}^{\mathrm{4}} −\mathrm{2}{p}^{\mathrm{2}} +{C}_{\mathrm{1}} −\mathrm{4}{y}=\mathrm{0} \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{3}\left(\mathrm{4}{y}−{C}_{\mathrm{1}} \right)}}{\mathrm{3}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{12}{y}+{C}}}{\mathrm{3}} \\ $$$$\Rightarrow{p}=\pm\sqrt{\frac{\mathrm{1}\pm\sqrt{\mathrm{12}{y}+{C}}}{\mathrm{3}}} \\ $$$$\Rightarrow{x}={p}\left({p}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{2} \\ $$$$\Rightarrow{x}=\pm\sqrt{\frac{\mathrm{1}\pm\sqrt{\mathrm{12}{y}+{C}}}{\mathrm{3}}}\left(\frac{−\mathrm{2}\pm\sqrt{\mathrm{12}{y}+{C}}}{\mathrm{3}}\right)+\mathrm{2} \\ $$