Question

Please help solving the differential equation:

\[y'' = (y')^{3}\]

Thanks

Answer 1

\[y'' = (y')^{3}\]

Answer 2

I think this is bernoulli..i think let me check..

Answer 3

oh wow this looks so simple and I am not sure how to do this.. Well one solution is definitely:
y=0, but that's trivial

Answer 4

Haha, had the same first impression, I already thought about a substitution z=y', but then the particular Integral becomes a mess

Answer 5

can u name the topic or something where I can look around.. I took ODEs last semestor..

Answer 6

Unfortunately I cannot, have that one from a collection of miscellaneous exam problems.

Answer 7

Ohh this is one of those substitutions.. when x is missing hold on.. and I meant *semester previously

Answer 8

let me just find the section in the book..

Answer 9

Reduction of order that's what it's called..

Answer 10

okay there are two substitutions here..

Answer 11

first one is:
y' = z and y" = z'
2nd one is bernoulli..

Answer 12

trying to solve this on paper..

Answer 13

I tried it that way, when you then try to substitute z back to y you get an integral of the form

\[\int\limits_{}^{}\sqrt{-1/(2x+C)}\] Is there no more elegant way?

Answer 14

thinking.. trying this on paper..

Answer 15

arghh im getting mixed up with all these variables..
y" = d^2y/dx^2
z' = dz/dx
v' = dv/dx am I right?

Answer 16

I have v' = -2 so far so what do I integrate it with respect to??

Answer 17

Ok, you integrate this with respect to x, I finished of and got the result:

y=\[C _{2} - \sqrt{2}-\sqrt{-x-C _{1}}\]

Answer 18

actually i figured something out..

Answer 19

let me just work it out and see what i have.

Answer 20

I checked with wolfram alpha and that is correct, it's just, if I got this answer in an exam, I would be 100% sure to have made a mistake somewhere!

Answer 21

Anyway bahrom, thanks for the effort!

Answer 22

actually im almost done solving..

Answer 23

okay done..

Answer 24

let me take a picture and upload..

Answer 25

here