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The question comprises of three subparts which need to be converted to Clairaut's form through suitable substitutions and then solved :

(a) x p2 - 2yp + x + 2y = 0

(b) x2 p2 + yp (2x + y) + y2 = 0

(c) (x2+y2)(1+p)2-2(x+y)(1+p)(x+yp)+(x+yp)2=0

Note : p = dy/dx

I understand that Reducing to Clairaut's form involves suitable substitution so as to bring it in the form of V = P U + f(P) but i am unable to form any intuition about what such substitutions might be , as the above equations seem complicated with more than one combination of variables and 'p'.

I have added three sub-parts to get a better understanding of the intuition involved in such substitutions. Help would be greatly appreciated. Thanks.

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$$ \color{blue}{\mbox{(a)}\quad xp^2 - 2yp + x + 2y=0} $$ In order to use Clairaut's technique, we write $\color{blue}{\mbox{(a)}}$ as $$ (p^2 + 1)x + 2(1-p)y = 0 \qquad \Longrightarrow \qquad y = \frac{p^2+1}{2(p-1)}x $$ Now, deriving with respect to $x$, $$ \frac{d y}{d x} = p = \frac{d}{dx}\left\{\frac{p^2+1}{2(p-1)}\right\}x + \frac{p^2+1}{2(p-1)}, $$ or $$ p - \frac{p^2+1}{2(p-1)} = \frac{p^2 - 2 p - 1}{2(p-1)^2}p'(x) x. $$ Simplifying, $$ \frac{p'(x)}{p-1} = \frac{1}{x}, $$ which can be integrated, yielding to $p(x) = 1 + c_1 x$, and then $y(x) = \frac{c_1}{2}x^2 + x + c_2$. Now, in order to determine the extra constant, we substitute into $\color{blue}{\mbox{(a)}}$, and then $$ y(x) = \frac{c_1}{2}x^2 + x + \frac{1}{c_1}. $$

$$ \color{blue}{\mbox{(b)}\quad x^2p^2 + yp(2x+ y) + y^2 =0} $$

This might see more complicated but, we can write it as $$ (x p + y)^2 + p y^2 = 0. $$ Now, in order for the equation to be satisfied, $p$ must be negative. Then $$ |x p + y| = (-p)^{1/2} |y|, $$ which is very unfortunate, as we end up with a lot of cases.

Case 1. $\quad y > 0, \quad y + p x > 0$.

$$ x p = \left( (-p)^{1/2} - 1 \right)y. $$

Putting this expression in the form $y = F(p) x$ and derivating, we come up with

$$ \frac{p'}{p\left((-p)^{1/2} - 1\right)} = \frac{2}{x}. $$

Integrating, $$ \frac{1 - (-p)^{1/2}}{(-p)^{1/2}} = c_1 x $$ we have $$ p = - \frac{1}{(1+c_1 x)^2}, $$ and then $$ y(x) = \frac{1}{c_1(1+c_1 x)} + c_2. $$ Evaluating for $c_2$ we end up with $$ y(x) = \frac{1}{c_1(1+c_1 x)}, $$ which is valid for $c_1 > 0$ in the interval $$ -\frac{1}{c_1} < x. $$

EDIT

Well, after reading a few pages of the book, one can see that $\color{blue}{\mbox{(b)}}$ is solved in page 153 (exercise 9).

Taking the change of variables $y = u$ and $v = xy$, you have that $$ \frac{d v}{d x} = x p + y $$ Now, defining $P$ as $$ P = \frac{d v}{d u} = \frac{d v}{d x} \frac{d x}{d u} = \frac{x p + y}{p} $$ it's easy to see that $$ p = \frac{y}{x - P}. $$ Substituting in the ODE, one has $$ x^2 + (x - P)(2x + y) + (x - P)^2 = 0 $$ or $$ v = u P + P^2 $$ which is in the desired Clairaut's form.

$$\color{blue}{\mbox{(c)}\quad (x^2+y^2)(1+p)^2-2(x+y)(1+p)(x+yp)+(x+yp)^2=0}$$

Again, this is exercise 5 on page 154. The suggestion here is to take $x + y = u$ and $x^2 + y^2 = v$. Following $\color{blue}{\mbox{(b)}}$, $$ P = \frac{d v}{d u} = \frac{d v}{d x}\frac{d x}{d u} = \frac{2 x + 2py}{1 + p}, $$ and then $$ p = \frac{P - 2 x}{2 y - P}. $$ Substituting in the ODE and simplifying, $$ (x - y)^2 (P^2 + 4 v - 4 P u) = 0. $$

Can you work the details?

EDIT 2

@Vish.Math: If you read carefully, the author clearly states "There is no general method of deciding about the proper substitution in a certain case. These can be learned only by practice".

That being said, you can see in $\color{blue}{\mbox{(b)}}$ that it can be written as $$ (x + p y)^2 + p y^2 = 0. $$ The substitution $v = xy$ will lead to $v' = x + py$, or $$ \frac{v'^2}{p^2} + \frac{y^2}{p} = 0. $$ Finally, one would wish that the left term to be a derivative of something, ie $$ \frac{d v}{d u} = \frac{v'}{p} = \frac{d v}{d x}\frac{d x}{d y} \quad \Longrightarrow \quad u = y. $$ Then, letting $P = \frac{d v}{d u} = \frac{x + y p}{p}$, we end up with the desired form: $$ v = u P + P^2 $$

In $\color{blue}{\mbox{(c)}}$, we see that the substitution $v = x^2 + y^2$ leads to $v' = 2 x + 2 p y$, which is a term in the ODE, so $$ v - (x + y)\frac{v'}{1+p} + \frac{v'^2}{4 (1+p)^2} = 0 $$ Taking $$ P = \frac{d v}{d u} = \frac{d v}{d x} \frac{d x}{d u} = \frac{2(x + y p)}{1 + p} $$ and then $$ \frac{d u}{d x} = 1 + p \quad \Longrightarrow \quad u = x + y $$ and $$ 4 v = 4 P u - P^2. $$

That's why the author says that only experience can help determining the correct change of variables. If you have terms $p x + y$, choose $v = xy$ and see if a proper $u$ can be constructed. If you have terms $x + p y$, choose $v = x^2 + y^2$ and try to construct the $u$. Further than that, there is no other intuitive explanation that I can give, or accept.

As my advisor sais: "Euler had intuition, the rest of us have experience".

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  • $\begingroup$ thank you for the help. I guess i can take it from there from where you left. But, is there a way to substitute values such as for instance, x+y=u and xy = v so that the entire equation comes in the form of v = U P + F (P) . The book which i am reading now specifically talks about doing such substitutions and solving. Thank you $\endgroup$ – MathMan Jul 5 '13 at 8:11
  • $\begingroup$ @Vish.Math What book is that? $\endgroup$ – Pragabhava Jul 5 '13 at 15:52
  • $\begingroup$ Its differential equations by m.d. raisinghania $\endgroup$ – MathMan Jul 5 '13 at 17:51
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    $\begingroup$ @Vish.Math See the edit. If you find it appropriate, please consider accepting the answer. $\endgroup$ – Pragabhava Jul 8 '13 at 16:13
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    $\begingroup$ @Vish.Math See my last edit but be warned that there is never a satisfactory explanation for intuitive steps. Math is experience. $\endgroup$ – Pragabhava Jul 12 '13 at 19:23

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