# Identifying First Order Diff Equations

• Sep 26th 2012, 09:35 AM
bustacap09
Identifying First Order Diff Equations
So the equation is $(xy+y)y' = x-xy$
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 - $y' + P(x)y = Q(x)y^n$
Separable - $y' = p(x)h(y)$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. $xy^2 y' = x^3e^(y/x) - x^2y$, so all the coefficient grouping ignoring y' are 3)
Linear - $(Q)y' + (P)y = Q(x)$ where P and Q are continuous functions

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

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

How did you come up with the conclusion you did for each one?
• Sep 26th 2012, 09:41 AM
TheEmptySet
Re: Identifying First Order Diff Equations
Quote:

Originally Posted by bustacap09
So the equation is $(xy+y)y' = x-xy$
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 - $y' + P(x)y = Q(x)y^n$
Separable - $y' = p(x)h(y)$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. $xy^2 y' = x^3e^(y/x) - x^2y$, so all the coefficient grouping ignoring y' are 3)
Linear - $(Q)y' + (P)y = Q(x)$ where P and Q are continuous functions

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

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

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

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

If you factor the original equation you get

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

What does this tell you?