Differential Equation substitute-to-make-seperable question

[FocusedWolf]

Homework Statement

Problem #7 in this old exam: http://math.njit.edu/pdfs/222ex1.pdf

... note: i didnt care about getting the IVP solvied... this is just practice and getting the general solution was all i cared about doing

2. The attempt at a solution
http://wolfsfiles.googlepages.com/scan.jpg

I think i confirmed my answer in a cas. But my confussion is if i plug the original differential equation, without making that substitution, that i get a different general solution... and I'm not entirely sure if that's to be expected.

In short, is my answer correct :P

3. Quick question about a different problem
Problem #3-A on that old exam:
"(d/dx)[sec(x)y]=x ...get general solution"

should the left side be expanded first, and then work the problem, or am i not seeing some easy way to separate this thing right away.

 

[rock.freak667]

Well when i did the first one I eventually got

$$\frac{1}{2x}v' = \frac{v^2+1}{v}$$

and I believe it was because of this step

$$y=vx$$ so $$\frac{dy}{dx}=x\frac{dv}{dx} + v$$ you missed out the extra v and put 0 instead

and for $$\frac{d}{dx}{[sec(x)y]=x$$

I think they are telling you that the differential of sec(x)y w.r.t x is equal to x...so you can just integrate both sides w.r.t x

 

[FocusedWolf]

Thanks for the help. Teacher confirmed how you did the substitution, but think we got different answer, or its in a different form.

From just sub-ing in y = vx the general solution from that was
v^2 = CX-1, which was de-substituted to become
(Y/X)^2 = CX - 1 by replacing v with y/x

and y(1) = 1 was the initial condition... you i know i ddidnt want to do the ivp,, but this is answer anyhoot

gives... C = 2

so exact solution is: (y/x)^2 = 2x -1

so you my problem was i differentiated wrong... was trying to do partial derivitives but i think the goal was implicit differentation... and it got the respect-to-x from the original differential equation which i didn't realize to do. Thx for the help.