Question

Q: solve the DE

\(y^{\prime\prime}+x(y^\prime)^2=0\)

A: \(y=c\) or \(y=-\ln x+c\)

Answer 1

\[y^{\prime\prime}+x(y^\prime)^2=0\]\[\text{let } y^\prime=p\]\[y^{\prime\prime}=p^\prime\]\[p^\prime+xp^2=0\]\[p^\prime=-xp^2\]\[\frac{\text dp}{\text dx}=-x{p^2}\]\[\int\frac{{-\text dp}}{p^2}=\int x\text dx\]\[\frac{1}{p}=\frac{x^2}{2}+c\]\[\frac{\text dy}{\text dx}=\frac{2}{x^2+2c}\]\[{\text dy}=\frac{2}{x^2+2c}{\text dx}\]\[y=\int\frac{2}{x^2+2c}\text dx\]

Answer 2

im just a bit confuse with the constants, i think i need a negative to get logs when integrating

Answer 3

$$ y = 2\tan^{-1}{\frac{x}{\sqrt{2C_1}}} + C_2$$

Answer 4

i am trying to get to the solution posted in the question box

Answer 5

a second order DE must have two constants in it's general solution.

Answer 6

so the answers for the back of my book are incorrect @experimentX ?

Answer 7

y = c is plainly incorrect.

Answer 8

0=0

Answer 9

as for y=−lnx+c
let's differentiate it
dy/dx = -1/x <--- this is your DE which is of first order.

Answer 10

y = c => dy/dx = 0

Answer 11

wolf says
http://www.wolframalpha.com/input/?i=y%27%27+%2B+x+%28y%27%29^2+%3D+0

Answer 12

yeah, and?

Answer 13

I think that the given answer is quite not right.

Answer 14

well that is the answer i am trying to get to,
what would you have the logarithmic form of the answer be?

Answer 15

what do you mean by logarithmic form??

Answer 16

with logs rather than inverse trig identities

Answer 17

Oh wait ..

Answer 18

the difference has something to do with the domain for x

Answer 19

@Zarkon

Answer 20

for some strange reason wolf is saying
\[ y = \frac {\sqrt 2} { \sqrt {c_1}}\tan^{-1}{\frac{x}{\sqrt{2C_1}}} + C_2 = y(x) = c_2-(\sqrt 2 \tanh^{-1}{x/(\sqrt 2 \sqrt{c_1}))}/\sqrt{c_1}\]

Answer 21

Maybe I am missing something, but when I plug
y=−lnx+c

into the original differential equation, I don't get zero.

Answer 22

that is true @phi\[\frac 1{x^2}+x\times\left(-\frac1x\right)^2\neq0\]

i guess it must be a mistake in the book , i have checked it multiple times,

Answer 23

but
\[y=c\]\[0+x\times(0)^2=0\]

Answer 24

well y = c is one solution but not complete solution.