Question

Solve the O.D.E:

y^2 y'' + 4 = 0

Answer 1

Maybe it can be considered in the form
\[ y''+0y'+\frac{4}{y^2}=0\]

Answer 2

I thought of doing the substitution which yields a first order O.D.E:

y' = P

y'' = p dp/dy

y^2 p dp/dy + 4 = 0

pdp = -4/y^2

p^2/2 = 4/y + C

p^2 = 8/y + C

p = sqrt( 8/y + c)

dy/dx = sqrt(8/y + c)

can't seem to find a way to integrate

dy/sqrt(8/y + c)

Answer 3

y'' + 0 y' + 4/y^2 = 0

Ok, then ?

Answer 4

And then, I am not sure thought, we can treat it whole as a second order linear differential equation and can use the general solutions to find its solution

Answer 5

Differential operator ? variation of parameters ? undetermined coefficient ?

Answer 6

y'' = -4/y^2

D^2 y = -4 /y^2

D^2 y = -4 /y^2

To me stalemate ..

Answer 7

Oh no no. Havent you studied the general solutions for some simple differential equations
like y"+y+a =Ae^cx +Be^dx etc?

Answer 8

brah , x ? not y

Answer 9

y'' + 0 y' + 4/y^2 = 0

y'' + y + a = ...

because you let

y = e^bx

y' = be^bx

y'' = b^2 e^bx

Answer 10

And BTW, what you have given is called a complementary function

Answer 11

The general solution is a combination of both a complementary function and a particular solution to the equation.

Answer 12

Yes Its been a while since I have done these so...

Answer 13

So you know those right?

Answer 14

Because this is what I thought to be solved
Maybe it can be considered in the form
\[y''+0y'+1=1-\frac{4}{y^2}\]

Answer 15

Now you have both complimentary function and particular Integral.
@IrishBoy123 Maybe you can help

Answer 16

this is non-linear, right?

Answer 17

True.

Answer 18

so old trick, sorry i have not read the therad, is

\(y^2 \dfrac{d^2 y}{dx^2} = - 4\)

\( y^2 y' \dfrac{d y'}{dy} = -4\)

Answer 19

I did it :P , got:


dy/sqrt(8/y + c) = x + b

Answer 20

i'd ask wolfram, personally

you're changing variable here....

Answer 21

I did so in the second comment. Actually the book final answer is cute.

4 + ay^2 = b^2(x+b)^2

Answer 22

.

Answer 23

that looks do-able

are you struggling to get it, or just sharing the answer?

Answer 24

Struggling to get it.

Answer 25

As a first step,
\[y^2\frac{\mathrm d^2y}{\mathrm dx^2}+4=0\implies \frac{\mathrm d^2y}{\mathrm dx^2}=-\frac{4}{y^2}\implies \frac{\mathrm dy}{\mathrm dx}\frac{\mathrm d^2y}{\mathrm dx^2}=-\frac{4}{y^2}\frac{\mathrm dy}{\mathrm dx}\]Integrating both sides with respect to \(x\) yields
\[\int\frac{\mathrm dy}{\mathrm dx}\frac{\mathrm d^2y}{\mathrm dx^2}\,\mathrm dx=\frac{1}{2}\int\frac{\mathrm dy}{\mathrm dx}\,\mathrm d\left(\frac{\mathrm dy}{\mathrm dx}\right)=\frac{1}{2}\left(\frac{\mathrm dy}{\mathrm dx}\right)^2+C_1\]on the left, while on the right you get
\[-4\int\frac{1}{y^2}\frac{\mathrm dy}{\mathrm dx}\,\mathrm dx=-4\int\frac{\mathrm dy}{y^2}=\frac{4}{y}+C_2\]so you're left with
\[\frac{1}{2}(y')^2=\frac{4}{y}+C\]which appears a bit easier to work with.

Answer 26

Same as the substitution , of P or y'

yields the integration of


dy/sqrt( 8/y + C)

I find it cool to use dy/dx confusing xD

Answer 27

\(y'=\sqrt { \frac{8}{y}+C}\)

\(\int \dfrac{1}{\sqrt { \frac{8}{y}+C}}dy=\int dx\)

\(\int \dfrac{y}{\sqrt { 8+C y^2}}dy= \int dx\)

and pattern match from

\(\dfrac{d}{dx} \sqrt { 8+C y^2} = \dfrac{1/2 * 2 Cy}{\sqrt { 8+C y^2}}\) to solve the LHS

Answer 28

i missed out a \(\pm\) in there but if you square it as they do, that problem goes away

Answer 29

I think you missed a y ?

sqrt(8/y + c) = y/y * sqrt(8/y + c) = sqrt(8y + cy^2) ?

Answer 30

sudden non-scheduled windows upgrade to version 1607

Lagging as hell.
PC shutdown automatically

Answer 31

@IrishBoy123

Answer 32

@TrojanPoem

you're right! :-(


so yes, that was terrible algebra on my part

Answer 33

Stuck with two more ...

I manged to get to a form integrate-able by a trigonometric substitution. It can never be simplified to the book's ANS.