- #1

huaxue09

- 1

- 0

I tried all ways I can to solve (y^2+x^2+x)y'-y=0, but still cann't find a way to solve it.

Could anybody help on it?

Thank you very much!

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- Thread starter huaxue09
- Start date

- #1

huaxue09

- 1

- 0

I tried all ways I can to solve (y^2+x^2+x)y'-y=0, but still cann't find a way to solve it.

Could anybody help on it?

Thank you very much!

- #2

jackmell

- 1,804

- 53

[tex](y^2+x^2)dy+xdy-ydx=0[/tex]

Ok, divide by x^2 to get:

[tex]\left(1+\frac{y^2}{x^2}\right)dy+\frac{xdy-ydx}{x^2}=0[/tex]

bingo-bango right?

- #3

chief10

- 78

- 0

you wouldn't go one further?

- #4

JJacquelin

- 801

- 34

Another method consists in converting to polar coordinates, which leads to a linear ODE very easy to solve.

The result can be expressed on a parametric form : x=(c-t)/tan(t) , y= c-t

or as x fonction on y :

x = y / tan(c-y)

- #5

jackmell

- 1,804

- 53

you wouldn't go one further?

Well, one is a differential of just y and the other is a differential of the quantity y/x. So then if we have:

[tex]\left(1+\frac{y^2}{x^2}\right)dy+\frac{xdy-ydx}{x^2}=0[/tex]

we could re-write that as:

[tex]dy=\frac{d(y/x)}{1+\left(\frac{y}{x}\right)^2}[/tex]

Now, that right side is equivalent to the expression:

[tex]\frac{du}{f(u)}[/tex]

right? So we just integrate it:

[tex]\int dy=\int\frac{d(y/x)}{1+\left(\frac{y}{x}\right)^2}[/tex]

and get:

[tex]y=\arctan(y/x)+c[/tex]

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