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When getting the envelope of a family of curves that represent the general solution of an ODE, why do we differentiate wrt c and equate to zero? I would like a simple proof or a good reference for this.

I mean for example, the ODE $$y=2xy'+y^2y'^3$$ has a general solution $$y=2c\frac{x}{y}+\frac{c^3}{y}$$ If we differentiate the general solution w.r.t. $c$, we will get the singular solution which is the envelope of the general solution $$y^4=-\frac{32}{27}x^3.$$

TRUSKI
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MCS
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  • In your question, you introduces a symbol $c$ which comes like a fly in the ointment. What is $c$ ? – JJacquelin Oct 17 '17 at 07:10
  • @JJacquelin Sorry for that , I edited the post and added an example to clarify.. c is the constant in the general solution of an ODE – MCS Oct 17 '17 at 09:33
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    This is well explained in : https://en.wikipedia.org/wiki/Envelope_(mathematics) – JJacquelin Oct 17 '17 at 10:16

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Consider the diagram

enter image description here

where the point of intersection is a point near the evelope. The smaller $\mathrm{d}c$, the closer to the envelope we get.

This means that the point on the envelope is at the intersection of $$ f(x,c)=0 $$ and $$ \frac{\partial}{\partial c}f(x,c)=0 $$

Example

The family of lines parametrized by $a$: $$ \frac{x}{1-a}+\frac{y}{a}=1 $$ Take the derivative with respect to $a$: $$ \frac{x}{(1-a)^2}-\frac{y}{a^2}=0 $$ Solve simultaneously $$ x=(1-a)^2\qquad y=a^2 $$ enter image description here

The family of lines is in black and the envelope is in green. The envelope follows the intersection of adjacent curves.

robjohn
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