First Order Homogeneous Equations

[Linday12]

Homework Statement

\frac{dy}{dx} = \frac{3xy}{3x^2+7y^2}, y(1)=1

Express it in the form F(x,y)=0

The Attempt at a Solution

I'm not sure where I'm going wrong. I let v=y/x,

v+x\frac{dv}{dx} = \frac{3x^2v}{3x^2+7x^2v^2}=\frac{x^2(3v)}{x^2(3+7v^2)}= \frac{3v}{3+7v^2}

\Rightarrow x\frac{dv}{dx} = \frac{-7v^3}{3+7v^2}
I think that's right so far, but I'm not entirely sure. Here is where I feel something is going wrong.
\frac{3+7v^2}{-7v^3} \frac{dv}{1} = \frac{dx}{x}

\frac{3}{14v^2}-ln(abs(v))=ln(x)+c

\frac{3x^2}{14y^2}-ln(abs(\frac{y}{x}))-ln(x)+c=0
c=3/14 since y(1)=1

And therefore,
F(x,y)= \frac{3x^2}{14y^2}-ln(abs(\frac{y}{x}))-ln(abs(x))-\frac{3}{14}=0

The answer isn't right. If someone could help me find where I've gone wrong, it would be very appreciated. Sorry about the scripting. I'm not skilled at it yet, but, hopefully it's easy enough to read. Thank you!

 

[Forty-Two]

I got the same answer you did...

What is the answer you are given?

Perhaps the answer is just formatted differently?

Maybe try combining the natural logs?

ln(y/x) + ln(x) = ln(yx/x) = ln(y)?

 

[Linday12]

Thank you, that was exactly what I needed!

 

[Forty-Two]

Glad I could help!