Differential Equations - Find equation of line

[qw111]

Homework Statement

A curve passing through (3,-2) has a slope given by (x^2 + y^2)/(y^3 - 2xy). Find the equation of the curve.

Homework Equations

The Attempt at a Solution

My first thought was to plug in the points (3,-2) into the slope equation and plug them into the line equation (y - y1) = m(x - x1). Is that wrong? Seemed too easy for it to be a problem for differential equations..

So, I found that i could make the slope equation into an exact equation and solved for it.
I found that the answer i get is:
with c = -17
f(x,y) = x^3/3 + xy^2 - y^4/4 - 17
However that is not the equation of the curve. Is that the equation of the tangent line at (3,-2)?
I am stuck at this point, I am pretty sure I am done with 90% of the work, but I can't seem to figure out the equation of the curve from here on.

 

[Mark44]

Yes, you are pretty close, but you faltered a bit at the end.

The equation you were solving looked something like this:
F_x dx + F_y dy, where F_x = x² + y² and F_y = -y³ + 2xy

The solution to the equation above with the partials is F(x, y) = C, so you should have gotten (1/3)x³ - (1/4)y⁴ + xy² = C.

Substitute y(3) = -2 into the equation above to find the constant C.

 

[qw111]

Yes I did that, realized that C = 17 (not negative). But I am still puzzled, what is the equation of the curve?

 

[Mark44]

(1/3)x³ - (1/4)y⁴ + xy² = 17.