Question

\[y'' + x(y')^2 = 0\]

Answer 1

Split it as :

y' = u
u' + xu^2 = 0

second equation is separable

Answer 2

\[u'+xu^2 = 0\\
u'/u^2 = -x\\
-1/u = -(x^2+c)/2\\
1/u = (x^2+c)/2\\
y' = 2/(x^2+c)\\
y = \int\frac2{x^2+c}\mathrm dx\]

Answer 3

but (2/a) arctan (x/a) + b

doesn't solve the original eqation

Answer 4

guess :p
y=c

Answer 5

y = c is only one solution

Answer 6

if its any help :P
its definitely not anything in the form x^n

Answer 7

i think it's inverse hyperbolic

Answer 8

How about using a substitution like z =1/y or something?

Answer 9

Hmm why doesn't ganeshie's method work
http://www.wolframalpha.com/input/?i=y%27%27+%2B+x(y%27)%5E2+%3D+0

Answer 10

the sign of c

Answer 11

You're right uncle rhaukus it's to do with the sign of c. Inverse hyper tan answer

Answer 12

It does solve the original equation.

\[y=\frac{ 2 }{ c _{1} }\arctan (\frac{ x }{ c _{1} })+ c _{2}\]
\[y \prime=\frac{ 2 }{ 1+x^2 }\]
\[y \prime \prime=\frac{ -4x }{ (1+x^2)^2 }\]
\[y \prime \prime+x(y \prime)^2=-\frac{ 4x }{ (1+x^2)^2 }+x(\frac{ 2 }{ (1+x^2) })^2=0\]

Answer 13

One more thing. If c in \[y=\int\limits_{}^{}\frac{ 2 }{ x^2+c }dx \] is a negative than we'll have another answer.

Answer 14

aha yes, it does solve the equation (when i don't forget the -ve sign on the second derivative) ,thank you @yagosik.

The answer provided in the back of the book was
y = c or y = - ln x + c,

but
y = - ln x + c
y' = -1/x
y'' = 1/x^2

y''+x(y')2
= 1/x^2 + x(-1/x)^2
= 1/x^2 + 1/x
≠ 0

Answer 15

Well done @yagosik

Answer 16

y=(2/c1)arctanh (x/c1) + c2

Is also an answer

Answer 17

How can I get the hyperbolic solution from the equation?

Answer 18

arctah(x) = iarctan(-ix)

Answer 19

\[\begin{align}
y &= c_2 - \dfrac{\sqrt{2}\color{red}{arctanh (x/\sqrt{2c_1})}}{\sqrt{c_1}}\\~\\
&= c_2 - \dfrac{\sqrt{2}\color{red}{iarctan (-ix/\sqrt{2c_1})}}{\sqrt{c_1}}\\~\\
\end{align}\]

Answer 20

Let \(-i/\sqrt{2c_1} = 1/c_1'\) and we should be good..

Answer 21

Brilliant