Question

prove that the general equation y'=p(x)+q(x)y+r(x)y^2 has the general solution y(x)=Yp+z(x) where Yp is a particular solution and z(x) is a solution of: z'-(q+2rYp)z=rz^2.

Prove this and then find the general solution of:
y'=y/x+x^3y^2-x^5 (note that Yp=x)

Answer 1

Have you made any progress here so far?

I believe that, if you plug the general solution into the differential equation, you can use the definitions of Yp and z(x) to conclude that the two resulting sides are equal!
--> Yp satisfies the equation: y' = p(x)+q(x)y+r(x)y^2 y=Yp
--> You are given that z(x) is a solution to z' - (q+2 r Yp)z = rz^2.

Answer 2

I don't think I follow. so I should replace all of the y's with Yp, but then where do I use the z? I am sure this is super simple I am just making it complicated and I can't seem to see it just yet

Answer 3

Well, think about what the goal is. We want to show that y = Yp + z(x) is a general solution to the differential equation. The best way to show that is by plugging it in and checking if your end result is a universally true statement -- like "A = A."

We know some information about Yp and z(x). I'll write it here for clarity:
Yp : Yp' = p(x) + q(x) Yp + r(x) Yp^2
By definition of a particular solution of a DE.
z(x) : z' - (q+2rYp)z = rz^2
: z' = rz^2 + (q+2rYp)z
This is a given in the problem.

So, we start with our original differential equation:
y' = p(x) + q(x) y' + r(x) y^2
If we let y = Yp + z(x) and plug that in, our differential equation becomes:
(Yp + z(x))' = p(x) + q(x) (Yp + z(x)) + r(Yp + z(x))^2
Now, we can simplify some things. Differentiation is linear: (f + g)' = f' + g'. We can expand out (a+b)^2 = a^2 + 2ab + b^2. That will give us:
Yp' + z' = p(x) + q(x) Yp + q(x) z + r(x) (Yp^2 + 2 Yp z + z^2)
Yp' + z' = p(x) + q(x) Yp + r Yp^2 + (q(x) + 2r(x) Yp) z + r(x) z^2

You might notice at this point that the known information is actually embedded in the equation right now! In fact if we make a substitution using our known information (we could swap out Yp' and z' for their values, or the other way around; the end result is the same), we should end up with the same thing on both sides!

Answer 4

Or, alternatively, we could add together Yp' and z' from our information and point out that both equations are the same. Just different ways to arrive at the same conclusion.

Answer 5

ok great! so I would just solve the problem below the proof in my original question this same way?

Answer 6

I think you're primarily looking at this part, for the question:
z'-(q+2rYp)z=rz^2.
I believe you would fill in the values for the q,r,Yp functions and solve this DE for z.

Answer 7

The only trick is, I'm not immediately seeing the method of solving that DE! I have to get ready to leave for now, so I'll try to come back to this in a couple hours. If you still need help you ought to bump this question! Good luck!

Answer 8

I will wait for your help because you seem to know what you're talking about! Thanks so much for all of your help!

Answer 9

Thanks :P although there are many OS users that also know a fair bit about DE. I had to refer back to my engineering math text for this solution. :P

Are you familiar with Bernoulli differential equations? That is, of the form:
y' + P(x) y = R(x) y^a
There exists a nifty trick of changing variables to make this into a linear differential equation. Basically, we will change variables from z into v, defined so that:
v = z^(1-a)
v' = (1-a) z^(-a) z' In our case, a = 2.
We convert these two equations into equations for z and z'. Then swap out the z and z' of our equation with the new variable.
z' - (q+2rYp) z = r z^2.

If that's not familiar to you, I could try to find something else.
A side note, It just happens that the first equation we discussed (y' = p(x) + q(x) y + r(x) y^2) has a name as well; the Riccati equation.

Answer 10

ok but how does changing z into V help with the original ricatti equation? Would I put V in place of y?

Answer 11

Sorry I am typing some long replies here, hopefully they are thorough in explanation.
Okay, basically, we'll only use this equation in passing:
y' = y/x + x^3y^2 - x^5 = -x^5 + 1/x y + x^3 y^2
y' = p(x) + q(x)y+ r(x) y^2
to determine what our coefficient functions are.
p(x) = -x^5,
q(x) = 1/x,
r(x) = x^3
We know already that Yp = x (given). So the only thing we need to find the general solution is z(x). (y(x) = Yp + z(x))

Essentially, we're going to plug the information into this equation and solve for the function z.
z' - (q+2rYp) z = r z^2.
z' - (1/x + 2*x^3*x) z = x^3 z^2
z' - (1/x + 2x^4) z = x^3 z^2
Recall our definition for Bernoulli equations: z' + P(x) z = R(x) z^a
In this case, P(x) = -(1/x+2x^4), R(x) = x^3, and a = 2.
The substitution we make is the following:
v = z^(1-a) = z^(1-2) = z^-1
v' = -z^-2 z'
We'll do a little rearranging to get these into substitutions of z.
v = z^-1 ==> v^-1 = z
v' = -z^-2 z' ==> -z^2 v' = z' (z=v^-1) ==> z' = -v^-2 v'.
Substitute:
z' - (1/x + 2x^4) z = x^3 z^2
-v^-2 v' - (1/x + 2x^4) v^-1 = x^3 v^-2
Notice how multiplying by v^2 will change this into a linear DE!
-v' - (1/x + 2x^4) v = x^3
v' + (1/x + 2x^4) v = -x^3

Answer 12

If you are unfamiliar with that strategy, I'll attempt to find something better but nothing really comes to mind at the time. Otherwise you can give that strategy above some thought.

The procedure from the point I left off is:
. solve the linear DE for v
. reverse the substitution to get it back in terms of z(x).
. now we have z(x), we can put 'em in the general solution! y(x) = Yp + z(x) = x + z(x)
Again, best of luck with the Maths! I'll get back to this again tomorrow around 2pm.

Answer 13

so when I did all of that, and I put it back into the equation for z(x), that z'-(q=2rYp)z=rz^2 and I got x=0......so I'm not sure what to do now.