Question

Need help solving this homogeneous equation.
(xy+y^2)dx-x^2dy=0.
This is what I did so far,
dy/dx= (-xy-y^2)/(-x^2)
Let z=(y/x) so multiply the right by (y/x^3)
I got,
((-y/x)^2-(y/x)^3)/(-y/x).
Is this right so far? As I went on things got confusing.

Answer 1

if z = y/x, doesnt y = zx?

Answer 2

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

(x(zx)+(zx)^2)dx -x^2dy=0

(x(zx)+(zx)^2) -x^2 dy/dx=0

well, since zx = y, then z'x + z = y', or =dy/dx

(x(zx)+(zx)^2) -x^2(z'x + z)=0

but then its still not a linear equation is it, since we have a z^2 instead of a y^2

Answer 3

the sub of z = y/x is useful if it makes the setup seperable, does this make it seperable?

Answer 4

it does. After I get dy/dx into the form of (y/x) I can then substitute z and rewrite my equation as x(dz/dx)+z. So I get,
x(dz/dx) = (z^2-z^3)/z - z

Answer 5

does that make sense? I just learned this today so I may be wrong

Answer 6

lets try this

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

dividing by dx

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

and rewriting in the y' notation

(xy+y^2) -x^2 y' = 0

divide by x^2

y/x + (y/x)^2 -y' = 0

so z + z^2 - y' = 0 with the sub

but y' = z'x + z

z + z^2 -z'x - z = 0

z^2 -z'x = 0

ok, its looking better to me now

Answer 7

^^ yeah thats what i was getting.... now its separable:)

Answer 8

that is making much more sense!!
I am having trouble getting the equation into the form of (y/x). Like figuring out what to multiply by to get in that form

Answer 9

its still not linear, but seperable doesnt care much about linearity :)

Answer 10

dividing off the x^2 seemed reasonable to me so i tried it to see what it would do, it was a good call apparently :)

Answer 11

okay! Thank you! I will try this out and see if it works :)

Answer 12

youre welcome

Answer 13

It worked!