# ODE substitution question

I would appreciate any advice on the following ODE substitution question:
xy' = y + (x^2 + y^2)^.5
Dividing thru by x and using the usual y/x substitution, I get:
y' = v + (1 + v^2)^.5 but I don't know if that is right or how to integrate the left side. The book has the answer of y + (x^2 + y^2)^.5 = Cx^2
Am I not handling the radical correctly?

## Answers and Replies

It looks like you didn't differentiate implicitly correctly. You should have a step that looks like:

v+xdv/dx = v+sqrt(1+v^2)

then

x*dv/dx = sqrt(1+v^2)

You're right, I did leave that out but I think my problem is not knowing how to integrate that radical. I tried entering it in mathematica and it came up with a hyperbolic sine if I remember correctly, is that what you are seeing?

Whenever I see a constant + variable^2 underneath the radical, I think tangent substitution. See if that you takes you somewhere.

dextercioby
Science Advisor
Homework Helper
Yes, the solution is in terms of hyperbolic sine.

Great. We "skipped" those in my Calc 2 class. If you'd be willing to opine on one other one that I am not getting right that I know doesn't involve a wild integral, I'd appreciate it. I'll post it here in a sec in this thread.

The solution doesn't have to be in terms of sinh. It can be a log.

Did you cover trig subs in your Calc2 class? That'll get you there--you don't need hyperbolics for it, though they "clean up" the end result.

x(x+y)y' + y(3x+y) = 0
I did the following steps:
x(x+y)y' = -y(3x+y)
Divide thru...
y' = -v * [ (3x+y)/(x+y) ]
Then I divided what was in the brackets by x, obtaining:
v + xv' = -v * (3 + v)/(1+v)
Speeding up a little, I get:
(1+v)/(3-v^2) dv = dx / x
I'm under the impression I need to split the left side into:
1/ (3-v^2) + v/(3-v^2) but again, I'm not sure I know how to do the integral.
The solutions manual does the whole thing in just a few steps using a method I haven't seen in my class but makes it look more simple than it really is.

dextercioby
Science Advisor
Homework Helper
The first integral can be expressed in terms of argtanh x, while the second one is very easy to integrate by a simple substitution.