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Question Number 162959 by mkam last updated on 02/Jan/22
solve the differential equation y = x + p^3
$${solve}\:{the}\:{differential}\:{equation}\:{y}\:=\:{x}\:+\:{p}^{\mathrm{3}} \\ $$
Commented by mr W last updated on 02/Jan/22
this is not a differential equation.
$${this}\:{is}\:{not}\:{a}\:{differential}\:{equation}. \\ $$
Commented by Kamel last updated on 02/Jan/22
Mr W p=(dy/dx).
$${Mr}\:{W}\:{p}=\frac{{dy}}{{dx}}. \\ $$
Commented by mr W last updated on 02/Jan/22
what is then p^3  ?  p^3 =((dy/dx))^3  ? or p^3 =y′′′=(d^3 y/dx^3 ) ?
$${what}\:{is}\:{then}\:{p}^{\mathrm{3}} \:? \\ $$$${p}^{\mathrm{3}} =\left(\frac{{dy}}{{dx}}\right)^{\mathrm{3}} \:?\:{or}\:{p}^{\mathrm{3}} ={y}'''=\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:? \\ $$
Commented by Kamel last updated on 02/Jan/22
(E)⇔((dy/dx))^3 +x=y
$$\left({E}\right)\Leftrightarrow\left(\frac{{dy}}{{dx}}\right)^{\mathrm{3}} +{x}={y} \\ $$
Commented by mkam last updated on 02/Jan/22
yes sir
$${yes}\:{sir} \\ $$
Answered by mr W last updated on 02/Jan/22
((dy/dx))^3 +x=y  let u=y−x  (du/dx)=(dy/dx)−1  ((du/dx)+1)^3 =u  (du/dx)=(u)^(1/3) −1  (du/( (u)^(1/3) −1))=dx  ∫(du/( (u)^(1/3) −1))=∫dx  let t=(u)^(1/3) −1  u=(t+1)^3   du=3(t+1)^2 dt  ∫((3(t+1)^2 )/t)dt=∫dx  3∫(t+2+(1/t))dt=∫dx  3((t^2 /2)+2t+ln t)+C=x  (((t+2)^2 )/2)+ln t+C=(x/3)  ((((u)^(1/3) +1)^2 )/2)+ln ((u)^(1/3) −1)+C=(x/3)  ⇒(((((y−x))^(1/3) +1)^2 )/2)+ln (((y−x))^(1/3) −1)+C=(x/3)
$$\left(\frac{{dy}}{{dx}}\right)^{\mathrm{3}} +{x}={y} \\ $$$${let}\:{u}={y}−{x} \\ $$$$\frac{{du}}{{dx}}=\frac{{dy}}{{dx}}−\mathrm{1} \\ $$$$\left(\frac{{du}}{{dx}}+\mathrm{1}\right)^{\mathrm{3}} ={u} \\ $$$$\frac{{du}}{{dx}}=\sqrt[{\mathrm{3}}]{{u}}−\mathrm{1} \\ $$$$\frac{{du}}{\:\sqrt[{\mathrm{3}}]{{u}}−\mathrm{1}}={dx} \\ $$$$\int\frac{{du}}{\:\sqrt[{\mathrm{3}}]{{u}}−\mathrm{1}}=\int{dx} \\ $$$${let}\:{t}=\sqrt[{\mathrm{3}}]{{u}}−\mathrm{1} \\ $$$${u}=\left({t}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$${du}=\mathrm{3}\left({t}+\mathrm{1}\right)^{\mathrm{2}} {dt} \\ $$$$\int\frac{\mathrm{3}\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{{t}}{dt}=\int{dx} \\ $$$$\mathrm{3}\int\left({t}+\mathrm{2}+\frac{\mathrm{1}}{{t}}\right){dt}=\int{dx} \\ $$$$\mathrm{3}\left(\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{t}+\mathrm{ln}\:{t}\right)+{C}={x} \\ $$$$\frac{\left({t}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}}+\mathrm{ln}\:{t}+{C}=\frac{{x}}{\mathrm{3}} \\ $$$$\frac{\left(\sqrt[{\mathrm{3}}]{{u}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}+\mathrm{ln}\:\left(\sqrt[{\mathrm{3}}]{{u}}−\mathrm{1}\right)+{C}=\frac{{x}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\left(\sqrt[{\mathrm{3}}]{{y}−{x}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}+\mathrm{ln}\:\left(\sqrt[{\mathrm{3}}]{{y}−{x}}−\mathrm{1}\right)+{C}=\frac{{x}}{\mathrm{3}} \\ $$
Commented by Kamel last updated on 02/Jan/22
This is a Gauss differential equation  for example y=1+x is a singular solution.  If we consider that p=(dy/dx) we get a linear   differential equation with x, then if we solve it we get  a general solution of equation with   parameter p=Const.
$${This}\:{is}\:{a}\:{Gauss}\:{differential}\:{equation} \\ $$$${for}\:{example}\:{y}=\mathrm{1}+{x}\:{is}\:{a}\:{singular}\:{solution}. \\ $$$${If}\:{we}\:{consider}\:{that}\:{p}=\frac{{dy}}{{dx}}\:{we}\:{get}\:{a}\:{linear} \\ $$$$\:{differential}\:{equation}\:{with}\:{x},\:{then}\:{if}\:{we}\:{solve}\:{it}\:{we}\:{get} \\ $$$${a}\:{general}\:{solution}\:{of}\:{equation}\:{with} \\ $$$$\:{parameter}\:{p}={Const}. \\ $$
Commented by mr W last updated on 02/Jan/22
i don′t understand that much. what  is then the solution? is anything  wrong in my solution?
$${i}\:{don}'{t}\:{understand}\:{that}\:{much}.\:{what} \\ $$$${is}\:{then}\:{the}\:{solution}?\:{is}\:{anything} \\ $$$${wrong}\:{in}\:{my}\:{solution}? \\ $$
Commented by Kamel last updated on 02/Jan/22
No, it′s correct.
$${No},\:{it}'{s}\:{correct}.\: \\ $$
Commented by mr W last updated on 02/Jan/22
thanks for confirming sir!
$${thanks}\:{for}\:{confirming}\:{sir}! \\ $$

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