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

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

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

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

y = \frac{x c _{1} - x ^{2}}{2}, y = - \frac{x c _{1} - x ^{2}}{2}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x c _{1} - x ^{2}}{2}, y = - \frac{x c _{1} - x ^{2}}{2}

y = \frac{x c _{1} - x ^{2}}{2}, y = - \frac{x c _{1} - x ^{2}}{2}

tan (7 x)

\int 4 (x - 1) ln (x - 1) d x

y^{^{'}} - \frac{3}{x} y = \frac{y ^{4}}{x ^{7}}

\frac{d}{d x} (4 sin^{2} (3 x))

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{- 4 x} (x^{2} + 2 x - 4)

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