First Order Homogenous Differential Equation

[Josephine]

By using the substitution y=vx, find the general solution of the differential equation 2xy dy/dx = y^2 - 4x^2
Answer: y^2 = (A-4x)x

This is what I've done so far:

2xy dy/dx = y^2 - 4x^2
dy/dx = (y^2 - 4x^2)/2xy

y=vx,
v + x dv/dx = [(vx)^2 - 4x^2]/2x(vx)
= (v^2 - 4)/2v
x dv/dx = (-3v^2 -4)/2v

Then, integrate the equation.
I'll get:

1/3 ln (3v^2-4) = -ln(x) + C
1/3 ln (3(y/x)^2 - 4) = -ln(x) + C

How do i continue to get the following answer:
y^2 = (A-4x)x

 

[skeeter]

\(\displaystyle \L \frac{dy}{dx} = \frac{y^2-4x^2}{2xy}\)

\(\displaystyle \L \frac{d}{dx}(vx) = \frac{v^2x^2 - 4x^2}{2x^2v}\)

\(\displaystyle \L v + x\frac{dv}{dx} = \frac{v^2 - 4}{2v}\)

\(\displaystyle \L x\frac{dv}{dx} = \frac{v^2 - 4}{2v} - \frac{2v^2}{2v}\)

\(\displaystyle \L x\frac{dv}{dx} = -\frac{v^2+4}{2v}\)

\(\displaystyle \L \frac{2v dv}{v^2+4} = -\frac{dx}{x}\)

\(\displaystyle \L \ln{(v^2+4)} = -\ln{(x)} + C\)

\(\displaystyle \L \ln{(v^2+4)} + \ln{(x)} = C\)

\(\displaystyle \L \ln{[x(v^2+4)]} = C\)

\(\displaystyle \L x(v^2+4) = e^C = A\)

\(\displaystyle \L xv^2 + 4x = A\)

\(\displaystyle \L x\frac{y^2}{x^2} + 4x = A\)

\(\displaystyle \L y^2 + 4x^2 = Ax\)

\(\displaystyle \L y^2 = Ax - 4x^2 = x(A - 4x)\)

 

[Josephine]

Thank you

Hi skeeter! I saw where my mistake was. Thank you very much!