What is the solution for (4x^2+3y^2)dx-(2xy)dy=0 ?

The solution for (4x^2+3y^2)dx-(2xy)dy=0 is y=xsqrt(-4+c_{1)x},y=-xsqrt(-4+c_{1)x}

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(4x^2+3y^2)dx-(2xy)dy=0

(4 x^{2} + 3 y^{2}) d x - (2 x y) d y = 0

y = x - 4 + c_{1} x, y = - x - 4 + c_{1} x

Solution steps

(4 x^{2} + 3 y^{2}) d x - (2 x y) d y = 0

Solve Bernoulli ODE:

y = x - 4 + c_{1} x, y = - x - 4 + c_{1} x

y = x - 4 + c_{1} x, y = - x - 4 + c_{1} x

y = x^{2} - 4

\frac{d}{d x} (4^{- 3} (\frac{2}{x ^{7}} + 3 x)^{4})

\frac{d}{d x} ((9 x + 4) x^{2})

(40 q^{2} + q + 500)^{^{'}}

x \to - 1 + lim (\frac{3}{x + 1})

What is the solution for (4x^2+3y^2)dx-(2xy)dy=0 ?
The solution for (4x^2+3y^2)dx-(2xy)dy=0 is y=xsqrt(-4+c_{1)x},y=-xsqrt(-4+c_{1)x}