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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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