What is the solution for (y^4-x^2)(dy)/(dx)=-x*y ?

The solution for (y^4-x^2)(dy)/(dx)=-x*y is

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

(y^{4} - x^{2}) \frac{d y}{d x} = - x \cdot y

y = c_{1} + - x^{2} + c_{1}^{2}, y = - c_{1} + - x^{2} + c_{1}^{2}, y = c_{1} - - x^{2} + c_{1}^{2}, y = - c_{1} - - x^{2} + c_{1}^{2}

Solution steps

(y^{4} - x^{2}) \frac{d y}{d x} = - x \cdot y

Solve exact ODE:

y = c_{1} + - x^{2} + c_{1}^{2}, y = - c_{1} + - x^{2} + c_{1}^{2}, y = c_{1} - - x^{2} + c_{1}^{2}, y = - c_{1} - - x^{2} + c_{1}^{2}

y = c_{1} + - x^{2} + c_{1}^{2}, y = - c_{1} + - x^{2} + c_{1}^{2}, y = c_{1} - - x^{2} + c_{1}^{2}, y = - c_{1} - - x^{2} + c_{1}^{2}

(x^{2} + 1) (\frac{d y}{d x}) + 3 x (y - 1) = 0

\frac{\partial}{\partial x} (7 x ln (x y))

\frac{d}{d x} (1 3^{x})

\int_{2}^{4} \frac{1}{∣ x - 3 ∣ ^{\frac{1}{2}}} d x

f (x) = x^{2} + x, at x = - 4, - 1

What is the solution for (y^4-x^2)(dy)/(dx)=-x*y ?
The solution for (y^4-x^2)(dy)/(dx)=-x*y is