# Solve the following differential equations by converting to Clairaut's form through suitable substitutions.

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.

$$\color{blue}{\mbox{(a)}\quad xp^2 - 2yp + x + 2y=0}$$ 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".

• 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 – MathMan Jul 5 '13 at 8:11
• @Vish.Math What book is that? – Pragabhava Jul 5 '13 at 15:52
• Its differential equations by m.d. raisinghania – MathMan Jul 5 '13 at 17:51
• @Vish.Math See the edit. If you find it appropriate, please consider accepting the answer. – Pragabhava Jul 8 '13 at 16:13
• @Vish.Math See my last edit but be warned that there is never a satisfactory explanation for intuitive steps. Math is experience. – Pragabhava Jul 12 '13 at 19:23