# Homework Help: Differential Equation substitute-to-make-seperable question

1. Sep 24, 2007

### FocusedWolf

1. The problem statement, all variables and given/known data

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

... 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

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 differant general solution... and i'm not entirely sure if thats to be expected.

In short, is my answer correct :P

3. Quick question about a differant 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 seperate this thing right away.

Last edited by a moderator: May 3, 2017
2. Sep 24, 2007

### 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

3. Sep 25, 2007

### FocusedWolf

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