A 1 Parameter Family of Solutions for a Differential Equation?

[Nano-Passion]

Homework Statement

Find a 1 parameter family of solutions for the differential equation yx^2 dy - y^3 dx = 2x^2 dy

The Attempt at a Solution

yx^2 dy - y^3 dx = 2x^2 dy
x^2 dy - y^2 dx = 2x^2/y dy
dy - y^2/x^2 dx =
- y^2/x^2 dx = 2/y dy - dy
Divide all by y^2
dx/x^2 = (2dy/y - dy ) * -1/y^2
dx/x^2 = -2dy/y^3 + dy/y^2
Integrate
-1/x = y^2 - y + C

Where y ≠ 0 & x ≠ 0

While the book gives an answer of
(cx+1)y^2 = (y -1)x
with the same domain

 

[BruceW]

dx/x^2 = -2dy/y^3 + dy/y^2
This line was correct, but then integrating this does not lead to your next line:
-1/x = y^2 - y + C
I think it was maybe a mistake? I've done similar mistakes loads of times.

 

[SammyS]

Homework Statement

Find a 1 parameter family of solutions for the differential equation yx^2 dy - y^3 dx = 2x^2 dy

The Attempt at a Solution

yx^2 dy - y^3 dx = 2x^2 dy
x^2 dy - y^2 dx = 2x^2/y dy
dy - y^2/x^2 dx =
- y^2/x^2 dx = 2/y dy - dy
Divide all by y^2
dx/x^2 = (2dy/y - dy ) * -1/y^2
dx/x^2 = -2dy/y^3 + dy/y^2
Integrate
-1/x = y^2 - y + C

Where y ≠ 0 & x ≠ 0

While the book gives an answer of
(cx+1)y^2 = (y -1)x
with the same domain

Check the integration for y.

What is $\displaystyle \int (-2y^{-3}+y^{-2})\,dy\ ?$

 

[Nano-Passion]

Thank you Bruce & Sammy. It seems mental integration isn't always reliable.

So my new answer is -1/x = y^2 - y^-1 + C

But I still don't have the answer in the book (cx+1)y^2 = (y -1)x

dx/x^2 = -2dy/y^3 + dy/y^2
This line was correct, but then integrating this does not lead to your next line:
-1/x = y^2 - y + C
I think it was maybe a mistake? I've done similar mistakes loads of times.

Check the integration for y.

What is $\displaystyle \int (-2y^{-3}+y^{-2})\,dy\ ?$

 

[SammyS]

Thank you Bruce & Sammy. It seems mental integration isn't always reliable.

So my new answer is -1/x = y^2 - y^-1 + C

But I still don't have the answer in the book (cx+1)y^2 = (y -1)x

Hopefully, you mean -1/x = y^-2 - y^-1 + C.

 

[Nano-Passion]

Hopefully, you mean -1/x = y^-2 - y^-1 + C.

Hmm, forgot to fix that too!

 

[SammyS]

Hmm, forgot to fix that too!

Does that mean you now have the book's answer?

 

[Nano-Passion]

Does that mean you now have the book's answer?

Yes! By distributing the variables within parentheses and then dividing by x*y^2.

Thank you. I am not sure why the book chose to put it in that form though.

 

[BruceW]

Nice work! The book might have chosen that form to test rearranging skills. Also, there are no reciprocals which is always nice.