Hey all,
I used a v = y/x substitution on this and I got a really weird looking answer so I just want to see how right or wrong I am.
Solve: (x^2 + 2y^2)dx + xydx = 0
v = y/x and dy/dx = v +x(dv/dx)
So: x(dv/dx) = (-1/v) - v
after integrating I get:
(-1/2)ln|1 + v^2| = ln(x) + C
subbing back gives
(1 + y^2/x^2)^2 = x + e^c
Can someone please check this work and then come up with an equation for y? I have an answer but it looks so weird I don't think its right. Thanks.
Thanks Danny,
I have seen the errors and redid it. If you don't mind I would like to stay with the y/x sub instead of the Bernoulli sub since we have not yet covered them ok?
Solve: (x^2 + 2y^2)dx + xydy = 0
xydy = -(x^2 + 2y^2)dx
dy/dx = -(x/y) - (2y/x)
let v = y/x, y = vx, dy/dx = v + xdv/dx
v + x(dv/dx) = -(1/v) - 2v
please check algebra here:
x(dv/dx) = (-1-3v^2)/v
-vdv/(1+3v^2) = dx/x
with a u substitution I get
LHS: -(1/6)ln|1 + 3y^2/x^2|
So:
ln|1 + 3y^2/x^2| = -6ln|x| + C
Is this correct so far? Also please could you show me some skills on how to wrap this up correctly. Thanks so much.
I have this for an answer, this is the process. How right or wrong is this?
ln|1 + 3y^2/x^2| = -6ln|x| + C
1 + 3y^2/x^2 = x^-6 + e^c
3y^2/x^2 = 1/x^6 + e^c -1
3y^2 = 1/x^4 + x^2e^c - x^2
y = sqrt[ (1/3x^4) + (x^2e^c/3) - (x^2/3) ]
Yes, no, maybe?
Hello, pberardi!
I came up with a different answer.
My journey was quite intricate, but the answer came out rather "neat."
We have: .
Divide top and bottom of the fraction by
Let: .
Substitute: .
. . Then: .
Separate variables: .
The left side requires Partial Fractions: .
Integrate: .
Multiply by 2: .
. . . . .
Exponentiate: .
Back-substitute: .
Therefore: .