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
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