Question

Solving first order non linear ordinary differential equation

Answer 1

\[(\cfrac{dy}{dx})^2 - x(\cfrac{dy}{dx}) + y = 0\]

I don't know from where to start.

Answer 2

@hartnn

Answer 3

Looks like you might be able to use the quadratic formula... Wait is that a second derivative or first derivative squared?

Answer 4

First derivative squared

Answer 5

Sorry, you said first order in the first post.

Answer 6

That is also correct. It is a first order second degree differential equation.

Answer 7

Anyone have some ideas to throw out at it? @vishweshshrimali5 where did you get this from?

Answer 8

Advanced Engineering Mathematics Book

Answer 9

I would start by factoring it

Answer 10

But I don't know how !!

Answer 11

I noticed that y=0 is a solution. So then I tried y=C to see that wouldn't work. Then y=Ax+B to see and found:

y'=A

plugging in:

\[A^2-xA+(Ax+B)=0\]\[A^2+B=0\]

So you get these two solutions:

y=0 and y=Ax-A^2

Answer 12

Yup I got it. Thanks a lot.

Answer 13

If that hadn't worked I'd have tried the next order up, only motivation I can give for my guesses is the xy' term sort of hinted me off to a polynomial. I had originally thought to use power series, but of course the squared power series seemed not fun. Then Euler's type where you use y=x^n but that sort of came up dry, and this was my third and final guess. Yeah, not a very good method I'm afraid. =/

Answer 14

Well, but it worked.
Thanks a lot

Answer 15

Also, it says that there is a singular solution. How do I find it ?

Answer 16

It says that the singular solution is \(y = \cfrac{x^2}{4}\)

Answer 17

Any idea how to obtain it ?

Answer 18

I'd just say keep guessing up to higher orders past y=Ax+B and guess y=Dx^2 and solve for D. If I wasn't warned that there was a singular solution I probably wouldn't have known/been lucky enough to search for it though.

Honestly, I have no idea how to solve nonlinear differential equations without just going down a list of easiest possible guesses.

Answer 19

Ohh, but how will I guess to put y = Dx^2 and not say, y = Dx^2 + Ex + F

Answer 20

Actually that's exactly what I would do. \[y=Dx^2+Ex+F\]\[y'=2Dx+E\]\[(y')^2=y'=4D^2x^2+4DEx+E^2\]\[(4D^2x^2+4DEx+E^2)-(2Dx+E)+(Dx^2+Ex+F)=0\]
collect like terms.

Answer 21

I accidentally messed up what I wrote on that third line for (y')^2=y'... The y' part shouldn't be there.

Continuing on... one sec.

Answer 22

ok

Answer 23

I think the 4th line is also wrong. It should be:

\[4D^2 x^2 + E^2 + 4DEx - x(2Dx + E) + Dx^2 + Ex + F = 0\]
\[\implies (4D^2 -D)x^2 + 4DEx + (E^2 + F) = 0\]

Answer 24

Ok I forgot to multiply that middle term by -x instead of just -1! lol no wonder I'm confused.

Yes, exactly, now you know each of those terms are =0.

Answer 25

How ?

Answer 26

We don't know whether, D, E or F are +ve, -ve or 0.

Answer 27

No, I mean those parts in front of x^2, x, and 1.

Answer 28

But, why ?

Answer 29

\[(4D^2-D)=0\]\[(4DE)=0\]\[(E^2+F)=0\]

If not, then it won't satisfy the differential equation you see? We have to have the left and right side of it be zero, and we know that since y is a function of x, then that means the x part is NOT 0.

Answer 30

Yes. I got it .

Answer 31

Thanks again. :)

Answer 32

Now we know:

\[D(4D-1)=0\] so \[D=0 \ \ or \ \ D=1/4\]

from this, \[DE=0\] So if D=0, E can be anything. If D=1/4, then E=0.

Taking this further\[E^2=-F\] So if D=1/4, E=0, F=0
BUT we got the other one as well!
D=0 implies E is anything and that F=-E^2

Perfect, we got both results from this.

y=0x^2+Ex-E^2
y=1/4x^2+0x+0

Answer 33

lol I learned from doing this problem, thanks good question @vishweshshrimali5 !

Answer 34

It was my pleasure.