I'm supposed to find the solution of ( x y &#x2212;<!-- − --> 1 )

Jayla Christensen

Jayla Christensen

Answered question

2022-06-23

I'm supposed to find the solution of
( x y 1 ) d y d x + y 2 = 0
The solution in the book manipulates it to
d x d y + x y = 1 y 2
and proceeds to use an integrating factor to solve it to
x y = ln y + c
Which makes sense, but I did it by isolating the differential of y, substituting x y = v, and simplifying it to
1 v v d v = d x x
which gives, on solving,
ln y + c = ln x 2 + y x
Which isn't the same.
I can't see any flaw in what I did except for dividing variables throughout the equation (which I see everywhere in methods of solving differential equations, without a clue as to why.)
Can anyone help me with where I got it wrong?
EDIT: I'll add some detail on my steps;
x y = c d y d x = y 2 x y 1 x d v d x = v 2 1 v + v = v 1 v d x x = ( 1 v ) d v v

Answer & Explanation

Xzavier Shelton

Xzavier Shelton

Beginner2022-06-24Added 26 answers

From
1 v v d v = d x x
you should get
ln v v + C = ln x ln ( x y ) x y + C = ln x ln x + ln y x y + C = ln x x y = ln y + C
You probably mistook v for y x .

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?