Solve differential equation:
x²dy + √(y²+4) dx + dy = 0

advise if below is correct solution:
x²dy + dy = - √(y²+4) dx
(x² + 1)/ (x² + 1)dy = - √(y²+4)/ (x² + 1) dx
1dy = - √(y²+4)/ (x² + 1) dx
1/√(y²+4)dy = - 1/ (x² + 1) dx
∫1/√(y²+4)dy = - ∫1/ (x² + 1) dx
sinh¯¹[y(x)/2] = - [tan¯¹x + C]
y = - 2 sinh(tan¯¹x + C)

Question

Solve differential equation:
x²dy + √(y²+4) dx + dy = 0

advise if below is correct solution:
x²dy + dy = - √(y²+4) dx
 (x² + 1)/ (x² + 1)dy = - √(y²+4)/ (x² + 1) dx
1dy = - √(y²+4)/ (x² + 1) dx
1/√(y²+4)dy = - 1/ (x² + 1) dx
∫1/√(y²+4)dy = - ∫1/ (x² + 1)  dx
sinh¯¹[y(x)/2] = - [tan¯¹x + C]
y = - 2 sinh(tan¯¹x + C)

psymon · Answer

I don't see any problem with it personally.

anonymous · Answer

How did you get (x^2+1)/(x^2+1)dy
wouldn't it be (x^2+1)dy...? @Psymon  does my question make sense.

psymon · Answer

@mebs He was dividing by both sides in that step from what it looked like.

anonymous · Answer

Yea I guess he was isolating dy

psymon · Answer

Pretty much. Just looked awkward having it written like thatin the step, but I saw what he was doing.

anonymous · Answer

@Psymon  Do you see something wrong with his substitution?

psymon · Answer

Looking at something in particular?

anonymous · Answer

1/√(y²+4) dy

psymon · Answer

If I remember, I thought that was sinh^-1(u/a)

unklerhaukus · Answer

good work @johnbeukes 
 your solution is correct $\checkmark$