Nonlinear Differential Equation that's separable Consider the differential equation $y'=|y|$. 


*

*a) Is the equation linear or nonlinear? Is it separable? Justify. 

*b) what do you observe when you solve the two initial value problems $y'=|y|$, $y(0)=1$ and $y'=y$, $y(0)=1$? 

*c) what do you observe when you solve the two initial value problems $y'=|y|$, $y(0)= -1$ and $y'=y, y(0)= -1$? 


For a), I said it was a first-order nonlinear D.E. and that it is separable and I justified it by saying it can be rewritten in the form of $y'= \sqrt{y^2}$.
How would I solve for b and c, can I get a thorough explanation please
 A: I am going to do a slightly different analysis to see if we can study more details of this problem. I will leave it to you to finish off the problem as everything you need is in this writeup.


*

*The equation is nonlinear (you should state why)

*It is separable (it is asking you to justify that)


What if we wrote:
$$y' = |y| \rightarrow \frac{dy}{|y|} = dt$$
If we integrate both sides, we get:
$$  (\text{sgn}~y) \ln y+ c_1 = t + c_2$$
WLOG, I am going to combine constant terms into a single term, yielding:
$$\tag 1 (\text{sgn}~y) \ln y = t + c$$
What impact does the sgn term have on the solution of $(1)$? 


*

*Case 1: If $y$ is positive, we get:


$$\ln y = t + c \rightarrow y(t) = c e^{t}$$


*

*Case 2: If $y$ is zero, we get:


$$y(t) = 0$$


*

*Case 3: If $y$ is negative, we get


$$-\ln y = t + c \rightarrow y(t) = c e^{-t}$$
What happens to each of those solutions as $t \rightarrow \infty$? It is clear that for negative $y$, we approach zero, for $y = 0$, we have a trivial (critical) fixed point and for positive $y$, we approach $\infty$ as $t \rightarrow \infty$.
Lets look at a phase portrait (my software defines everything in terms of $x$, so please excuse that) that superimposes the direction fields (green) and actual solutions (blue) curves for many initial values and see if that is the case. Notice how curves from $-y$ all approach zero for any initial condition, curves $y = 0$ are a fixed point and curves from $+y$ all approach $\infty$ as we suspected from the analysis above.

What if we used the definition of absolute value to rewrite the differential equation as:
$y' = 
\cases{
~~~y  & \text{if } ~~y\ge 0\cr
-y & \text{if } ~~y\lt 0
}$
Now, you should be in a better position to handle the remaining portions of the question.
