Question

ODE question!

Answer 1

Can anyone tell me what the question is asking me to do exactly?, I am confused by the wording of the question.

Answer 2

hmm i am not familiar of this method o.O

Answer 3

the problem wants u to change variables form x,y to x,z by substituting z=x/y^2

Answer 4

this substitution will make the DE easier to understand and solve

Answer 5

Ok I'll try it now.

Answer 6

So I'll just sub in x=y^2*z and z=x/Y^2 for for everything
.

Answer 7

I assume I need to find y' as well?

Answer 8

yes thats right

Answer 9

I need to use the quotient rule on y=sqrt(z/x)??

Answer 10

Sorry y=sqrt(x/z)

Answer 11

use the problem hint
start by y^2 z=x take derivative of both sides

Answer 12

Oh dx/dz and use the product rule on the RHS?

Answer 13

yes...product rule but why dx/dz?

Answer 14

No wait no 1

Answer 15

It's early here!

Answer 16

I got 1=y^2 dx/dz+z*2yz

Answer 17

u need to evaluate dy/dx from that and then put it in the equation

Answer 18

What do you mean? Can I not just in y' back into in the original eq?

Answer 19

sub in y' *

Answer 20

^2 dx/dz+z*2yz-1

Answer 21

y^2 dx/dz+z*2yz-1 **

Answer 22

\[y^2 z=x\\ 2yy'z+y^2 z'=1\\ 2yy'=\frac{1-y^2 z'}{z}=\frac{1-\frac{x}{z} z'}{z}=\frac{1}{z}-\frac{x}{z^2} z'\]

Answer 23

your equation is
\[2yy' x=y^2-x\]
now put the 2yy' and y^2 in terms of x and z in ur equation

Answer 24

I must have done the product rule wrong! Mine seems slightly different.

Answer 25

sure u can do it .try again

Answer 26

I did it again, Letting u=y^2 and v=z

u*v'+v*u'

Answer 27

y^2 dx/dz + z*2y

Answer 28

I don't know how you got y' in your product rule derivative

Answer 29

well just notice for a function of x like u
\[\frac{d}{dx} u^n =n u' u^{n-1}\]
now let u=y and n=2
what will u get

Answer 30

do u remember that formula?

Answer 31

Well the basic way to derive an x term like nx^n-1

Answer 32

yes

Answer 33

Yes so you are just asking be to derive y^2?

Answer 34

thats right

Answer 35

2y

Answer 36

That's what I have my derivative from my product rule.

Answer 37

u = y^2
u'=2y
v=z
v"=dx/dz

Answer 38

y is a function of x and u have to take the derivative with respect to x

Answer 39

So dy/dz?

Answer 40

No wait dy/dx

Answer 41

Can someone breakdown the product rule for me in this case I am confused about what I am exactly doing wrong.

Answer 42

Let z = x/ y²
-> y² = x/ z
=> 2ydy= (1/z) - (xdz / z²)

Thus:
x [ (1/z) - (xdz / z²) ] = (x/ z) - x