dy/dx= - x*x+3y*y/3x*x+y*y

Question

solve dy / dx= - x*x+3 y*y/3 x*x+y*y

Answer 1

Solve dy/dx= - x*x+3y*y/3x*x+y*y

Assuming that you mean, dy/dx= - (x*x+3y*y) / (3x*x+y*y),

Divide above and below of rhs by x^2, to give

dy/dx = -(1 + 3v^2) / (3 + v^2), where v = y/x, or y  = vx, which gives dy/dx = v + x.dv/dx

So, v + x.dv/dx = dy/dx = -(1 + 3v^2) / (3 + v^2)

x.dv/dx = -(1 + 3v^2) / (3 + v^2) - v.(3 + v^2) / (3 + v^2) = {-1 - 3v^2 - 3v - v^3} / (3 + v^2) = -(1 + v)^3 / (3 + v^2)

(3 + v^2) / (1 + v)^3 dv/dx = -1/x

int (3 + v^2) / (1 + v)^3 dv = - int 1/x dx

Using partial fractions gives us,

int {1/(1 + v) - 2/(1 + v)^2 + 4/(1 + v)^3} dv = - int 1/x dx

ln(1 + v) + 2/(1 + v) - 2/(1 + v)^2 = -ln(Ax)

(2 + 2v)/(1 + v)^2 - 2/(1 + v)^2 = -ln(Ax) - ln(1 + v) = -ln{Ax(1 + v)}

2v/(1 + v)^2 = -ln{Ax(1 + v)}

back-substituting for v = y/x,

2(y/x)/(1 + (y/x))^2 + ln{Ax(1 + (y/x))}

2xy/(x+y)^2 + ln{A(x + y)} = 0