If we have an ODE that contains y ,x , and y' . we can take y' in one side and solve the DE . This is fine ..

But :

What about taking y in one side , differentiating both sides wrt x , then solving by separation ?

What about taking x in one side , differentiating both sides wrt y , then solving by separation ?

Do these last 2 methods represent other ways to solve an ODE ? so the solutions of all of the 3 methods are equivalent ? what is the name of these last 2 methods? Can you provide a reference explaining these methods ,when to use them, and the meaning of the solution gotten from each one?

For example, the ODE $$yy'=x$$ First method : separating y' in one side then solve: $$dy/dx=x/y$$ $$\int y dy = \int x dx$$ $$y^2=x^2+2c$$ Second method: separating y in one side , calling y' as p , then differentiating wrt x and separating variables : $$y=x/p$$ $$p=\frac{1}{p}+\frac{-x}{p^2}\frac{dp}{dx}$$ $$\int dx=\int\frac{1-p^2}{p}dp$$ $$x+c=ln(p)-\frac{p^2}{2}$$ $$x+c=ln(x/y)-\frac{x^2/y^2}{2}$$ Third method : separating x in one side , calling y' as p , then differentiating wrt y and separating variables: $$x=yp$$ $$\frac{dx}{dy}=\frac{1}{p}=p+y\frac{dp}{dy}$$ $$\int \frac{dy}{y}=\int\frac{p}{1-p^2}dp$$ after integration and elimination of p , we will get $$y=\frac{1}{c\sqrt{1-(x/y)^2}}$$

  • $\begingroup$ Please write down what you mean, by an example. $\endgroup$ – MrYouMath Oct 5 '17 at 12:24
  • $\begingroup$ I edited the post and added an example $\endgroup$ – MCS Oct 5 '17 at 12:48
  • $\begingroup$ Both your second and third methods are wrong due to incorrect algebra. $\endgroup$ – Teh Rod Oct 5 '17 at 14:39

In your example you have the hypersurface $x=yp$. On this hypersurface you search the curves that correspond to the original DE. The connection is made by the plane field $dy=p\,dx$ which follows from the chain rule. The intersection of the plane field with the tangent planes of the surface will in general result in a line field. The tangent planes are described by $dx = y\,dp+p\,dy$. With all these 3 equations you can now try to eliminate one variable in the hope to get a more simple relation than the original one.

  • Eliminate $x$: $dy=p\,dx=yp\,dp+p^2\,dy$ resulting in $$\frac{dy}y=\frac{p\,dp}{1-p^2}\implies y=\frac{C}{\sqrt{1-p^2}},\quad x=\frac{Cp}{\sqrt{1-p^2}}$$ which gives solutions along the curves $y^2-x^2=C^2$.

  • Eliminate $y$: $p\,dx=yp\,dp+p^2\,dy=x\,dp+p^3\,dx$ resulting in separated equation and integral $$\frac{dx}{x}=\frac{dp}{p(1-p^2)}=\frac{dp}{p}+\frac{p\,dp}{1-p^2}\implies x = \frac{Cp}{\sqrt{1-p^2}}$$ etc.

  • Eliminate $p$: This leads back to the original ODE written as exact DE $x\,dx=y\,dy$ with solutions $y^2=x^2+C$


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