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

[mr_coffee]

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
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/8d/8616ca1dd6ce230911485b98a57fa31.png = ?
What i did was divide through by dx, so i got
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/2c/f2d7bb6b17ac7be8cca5e4969f1a091.png 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?

 

[Valhalla]

I believe this is homogeneous, try the substitution y=uxdisregard that

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

 

[mr_coffee]

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

http://img229.imageshack.us/img229/1413/lastscan23ch.jpg

 

[Valhalla]

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?

 

[mr_coffee]

Wow that was a lot 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!