Question

I need to find the general implicit solution to this ODE using substitution

Answer 1

\[(yx+y^2)dx-x^2dy=0\]

Answer 2

I know that this can be rewritten in the form
\[dy/dx = (xy+y^2)/x^2\]

Answer 3

my problem now is my question is asking me to sub u=(y/x)
\[u \prime = (f(u)-u)/x\]

this is where im having problems

Answer 4

\[u=\frac yx\implies u'=\frac{y'x-y}{x^2}\]solve for y' and sub that into the equation

Answer 5

how did you get to that?

Answer 6

The quotient rule\[\frac d{dx}\frac uv=\frac{u'v-v'u}{v^2}\]with \(u=y\) and \(v=x\)

Answer 7

well what exactly is it my problem that my problem wants me to do? I put what it is asking in the third message up there

Answer 8

I honestly don't know, but I am trying to show you how to use the substitution \(u=\frac yx\) to solve the problem.

Answer 9

I understand that, but I cant continue with my problem without first having an answer for this portion in the third message.

Answer 10

This homework system is really weird

Answer 11

what exactly is the question asking?

Answer 12

Next use the substitution u=y/x to write the equation as an ODE with independent variable x and dependent variable u , i.e., in the form

\[u \prime = \frac{ f(u)-u }{ x }\]

Answer 13

perhaps it means this\[\large u'=\frac{y'x-y}{x^2}={y'-\frac yx\over x}\]in which case again, we still need to substitute for y' in terms of u.

Answer 14

where is y prime coming from?

Answer 15

and why does y =y/x here?

Answer 16

u=y/x
and y' comes from taking the derivative of u with respect to x using the quotient rule\[u=\frac yx\implies u'=\frac{\frac d{dx}(y)x-\frac d{dx}(x)y}{x^2}=\frac{y'x-y}{x^2} \]

Answer 17

oh... I see. thank you so much!

Answer 18

welcome, let me know if you need more help :)

Answer 19

I'm still really not sure how this apply's to my answer though...

Answer 20

I cant put y prime into my answer

Answer 21

I understand, what you must do is solve for y' in terms of u.

Answer 22

but then I will have a u prime in my equation. 1 equation 2 unknowns

Answer 23

are you not allowed to enter u' in this part of the question? because you should be able to get this all in terms of u, u', and x, which is a linear (in this case seperable) solveable DE.

Answer 24

No it will only let me use the variables u and x as the problem stated

Answer 25

hmm... well you derived earlier that\[y'=\frac{xy+y^2}{x^2}\]we can rewrite that in terms of u

Answer 26

are you still with me?

Answer 27

yes I am sorry. yes I said that but that brought me to the solution

\[(xy+y^2)/x^4+y/x^2\]

Answer 28

This is not the right answer...

Answer 29

\[y'=\frac{xy+y^2}{x^2}=\frac yx+\frac{y^2}{x^2}\]now substitute \(u=\frac yx\)

Answer 30

once you have \(y'\) in terms of \(u\), sub that into what we got earlier for the derivative\[u'=\frac{y'-\frac yx}x\]again, using \(u=\frac yx\) where possible we can get the expresion for \(u'\) in terms of \(u\) and \(x\).