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

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

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

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

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

\int \frac{9 + t + t ^{2}}{t} d t

\int sin (x) \cdot e^{c o s (x)} d x

\frac{d}{d x} (\frac{30}{x ^{4}})

y = x^{2} (4 x^{4} + 5 x^{3} - 16)

\frac{\partial}{\partial y} (1 + x^{3} + y^{3})

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