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Solve-x-p-3-p-2-y-p-




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

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