Find an explicit or implicit solutions to the differential equation, what now?

Hello everyone, yet another obscure problem on web work. No examples like this in the book nor did the professor go over it so i was wondering if someone can let me know what exactly they are wanting me to do!

Find an explicit or implicit solutions to the differential equation
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/7b/3a599a5966a2ac65a84b6575a645b61.png [Broken]
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/8d/8616ca1dd6ce230911485b98a57fa31.png [Broken] = ?
What i did was divide through by dx, so i got
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/2c/f2d7bb6b17ac7be8cca5e4969f1a091.png [Broken]y' = 0;
Then
I let (x^2-4xy) = M, and N = x;
My = -4x;
Nx = 1;
Not exact.

Did i f this up or what?

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I believe this is homogeneous, try the substitution y=ux

disregard that

This is linear, try looking at it this way $$(x^2-4xy)=-x\frac{dy}{dx}$$

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hey I figured it out, i wasn't usre if this is what u were talking about. But if anyone was cur8ious this is my work:
http://img430.imageshack.us/img430/2407/lastscan1hu.jpg [Broken]

http://img229.imageshack.us/img229/1413/lastscan23ch.jpg [Broken]

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i see what you did but I just solved it by rewriting it as
$$\frac {dy}{dx}-4y=-x$$

Then use the integrating factor
$$e^{-4x}$$

Then multiply through to get

$$e^{-4x} \frac {dy}{dx}-e^{-4x}4y=-e^{-4x}x$$

Now integrate both sides to get

$$e^{-4x}y=\frac{x}{4}e^{-4x}+\frac{1}{16}e^{-4x}+C$$

multiply through by

$$e^{4x}$$

and I came up with

$$y=\frac{x}{4}+\frac{1}{16}+Ce^{4x}$$

Just seems like a few less steps than yours...same answer though

Do you have to use the exact solution method?

Wow that was alot easier, this problem was under the Exact Equations problems, so yes he makes us do it this way, but really, it only evaluates the answer so I guess i wasn't forced to do it that way. Thanks!