Expert: Paul Klarreich - 10/9/2006

Question
I am studying Calculus in a 1st degree computer  course.  I am trying to understand a worked example in Engineering Mathematics by KA Stroud, 5th edition,  pub palgrave macmillan,
isbn-13 978-0-333-91939-2
isbn-10 0-33-91939-4
page 1049 frame 40.

Can't figure out how line 1 below gives line 2 (tan^-1):

Quest/ solve (x-y)dy/dx = x + y,

.........................
.........................
(1)(integral sign)S{1/1+v^2 - v/1+v^2}dv = lnx+C
(2) therefore tan^-1 v -1/2ln(1+v^2) = lnx+lnA

answer = tan^-1 {y/x} =lnA+lnx+1/2ln{1+ y^2/x^2}

Answer
winston morris Asks in Category Calculus ...
 
Subject:  1st order differential equations
Private:  no
 
Question:  I am studying Calculus in a 1st degree computer  course.  I am trying to understand a worked example in Engineering Mathematics by KA Stroud, 5th edition,  pub palgrave macmillan,
isbn-13 978-0-333-91939-2
isbn-10 0-33-91939-4
page 1049 frame 40.

Can't figure out how line 1 below gives line 2 (tan^-1):

Quest/ solve (x-y)dy/dx = x + y,

.........................
(1)(integral sign)S{1/1+v^2 - v/1+v^2}dv = lnx+C
(2) therefore tan^-1 v -1/2ln(1+v^2) = lnx+lnA

answer = tan^-1 {y/x} =lnA+lnx+1/2ln{1+ y^2/x^2}
...............................................
Hi, Winston,

Alas, I don't have my differential equations books here with me, and while this looks familiar, I can't recall the details.

HOWEVER, if all you want is the explanation of why
(      1         v
| [ ------- - ------- ] dv  =  arctan(v) - 1/2 ln(1 + v^2)
)   1 + v^2   1 + v^2

that's easier. [I have trouble writing '-1', so I prefer 'arc' for online stuff.]

There are two terms in your integral.  The first is a standard form.  If you don't recall it, you can integrate:

(     dv   
|  -------    
)   1 + v^2

using a trigonometric substitution:

Let  v = tan t, so  t = arctan(v)
Then dv = sec^2 t dt
and  1 + v^2 = 1 + tan^2(t) = sec^2(t)

So the integral becomes:

(  sec^2 t dt
|  ----------   =
)  sec^2 t

(
| dt = t = arctan(v)
)
..........................................
As to the second:


(   v dv
| -------
) 1 + v^2

Let  t = 1 + v^2, then
dt = 2v dv
v dv = dt/2

and the integral is:

{  dt
| ---- = 1/2 ln(t) = 1/2 ln(1 + v^2)
} 2 t

Does that do it for you?