Identifying First Order Diff Equations

So the equation is $\displaystyle (xy+y)y' = x-xy$

It says to identify is the equation is Bernoulli, Separable, Linear and Homogeneous

If these definitions are wrong on how to identify which is which, please let me know, because i researched this because my professor doesn't teach well.

Bernoulli - $\displaystyle y' + P(x)y = Q(x)y^n$

Separable - $\displaystyle y' = p(x)h(y)$where p and h are functions (I don't really understand this one too much.)

Homogeneous - Look at the degrees of each grouping of coefficients, if they are all the same means homogeneous (Ex. $\displaystyle xy^2 y' = x^3e^(y/x) - x^2y$, so all the coefficient grouping ignoring y' are 3)

Linear - $\displaystyle (Q)y' + (P)y = Q(x)$ where P and Q are continuous functions

So $\displaystyle (xy+y)y' = x-xy$

Wouldn't I manipulate this to $\displaystyle (xy+y)y' + xy = x$ and that would be linear

Also

$\displaystyle x(1+y^2) + y(1+x^2)y' = 0$

$\displaystyle x(1-y) + y(1+x^2)y' = 0$

How did you come up with the conclusion you did for each one?

Re: Identifying First Order Diff Equations

Quote:

Originally Posted by

**bustacap09** So the equation is $\displaystyle (xy+y)y' = x-xy$

It says to identify is the equation is Bernoulli, Separable, Linear and Homogeneous

If these definitions are wrong on how to identify which is which, please let me know, because i researched this because my professor doesn't teach well.

Bernoulli - $\displaystyle y' + P(x)y = Q(x)y^n$

Separable - $\displaystyle y' = p(x)h(y)$where p and h are functions (I don't really understand this one too much.)

Homogeneous - Look at the degrees of each grouping of coefficients, if they are all the same means homogeneous (Ex. $\displaystyle xy^2 y' = x^3e^(y/x) - x^2y$, so all the coefficient grouping ignoring y' are 3)

Linear - $\displaystyle (Q)y' + (P)y = Q(x)$ where P and Q are continuous functions

So $\displaystyle (xy+y)y' = x-xy$

Wouldn't I manipulate this to $\displaystyle (xy+y)y' + xy = x$ and that would be linear

Also

$\displaystyle x(1+y^2) + y(1+x^2)y' = 0$

$\displaystyle x(1-y) + y(1+x^2)y' = 0$

How did you come up with the conclusion you did for each one?

If you factor the original equation you get

$\displaystyle (x+1)y\frac{dy}{dx}=x(1-y) \iff \frac{ydy}{1-y}=\frac{xdx}{x+1}$

What does this tell you?