Question

Solve (xy+y^2+x^2) dx -( x^2 )dy = 0 by substitution, any help?

Answer 1

sorry, not my style :{

Answer 2

it is homogeneous, so try \(\large \mathrm{y = vx}\)

Answer 3

first, how do i verify if it is homogeneous? My book is very vague about that.

Answer 4

is it if equal to 0?

Answer 5

replace `x` by `kx`
and `y` by `ky`

Answer 6

(xy+y^2+x^2) dx -( x^2 )dy = 0

((kx)(ky)+(ky)^2+(kx)^2) dx -( (kx)^2 )dy = 0

k^2 [(xy+y^2+x^2) dx -( x^2 )dy] = 0

Answer 7

Notce that entire k power got factored out, so we can use this for testing quickly whether a DE is homogenerous or not..

Answer 8

ah I see. So then when I replace the whole thing by y=ux, I can then do (x^2u+u^2x^2+x^2)dx - x^2 dy and I can factor out x^2, and be left with u's only. right? also replacing dy = xdu + u dx right?

Answer 9

Exactly !

Answer 10

maybe it is easier to see this by putting it in `dy/dx = f(x,y)` form first

Answer 11

hmmmm I don't know how to do that, or not sure rather*

Answer 12

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

is equivalent to

\[\dfrac{dy}{dx} = \dfrac{xy+y^2+x^2}{x^2}\]

Answer 13

Oh I see what you mean, my bad haha, stressing over an exam isn't helping to see clearly.

Answer 14

but afterwards, I see that my solutions cancel out the x^2 from x^2(u+u^2+1)dx - x^2( x du + u dx) giving (u+u^2+1) dx - (x du + u dx). is it just because it is a comon coefficient between both functions?

Answer 15

and then you can cancel everything and be left with u^2+1 dx - xdu = 0 and use separable equation..

Answer 16

And lastly if I may ask, for substitution, how do you decide to use it? after verifying if it is not exact? and how did you know to use y = vx, just a rule type thing?

Answer 17

It helps to know that there are only two standard ways for solving ANY differential equation :

1) separable
2) linear

We try to change all other types of equations into one of these known forms and solve..

Answer 18

look at your DE in this form :

\[\dfrac{dy}{dx} = \dfrac{xy+y^2+x^2}{x^2}\]

Answer 19

to me it doesn't seem linear with y^2

Answer 20

as you can see the right hand side is pretty complicated with a mix of x and y terms

Answer 21

but if you look closer, you will see that if you replace \(y\) by \(ux\), both \(x\) and \(y\) disappear giving you a rational expression in \(u\) alone :
\[\dfrac{dy}{dx} = \dfrac{xy+y^2+x^2}{x^2}\]

plugin \(y=ux \implies \frac{dy}{dx} = \frac{du}{dx}x+u \)
the eqn becomes :
\[\dfrac{du}{dx}x+u = \dfrac{x(ux)+(ux)^2+x^2}{x^2}\]

\[\dfrac{du}{dx}x+u = \dfrac{x^2(u+u^2+1}{x^2}\]

\[\dfrac{du}{dx}x+u = u+u^2+1\]

which is separable

Answer 22

Here is a full solution

Answer 23

soorry to bother you one last time but where did you get y=ux --> du/dx x+u and for the rest I think i could give you a hug for your patience and help, thank you for allowing me to understand this concept! :)

Answer 24

\(\large \color{magenta}{\heartsuit} \)\[y = ux\]
differentiate both sides with.resepct.to \(x\)

Answer 25

use product rule. .