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

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

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(dy)/(dx)=((x^2+3y^2))/(2xy)

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

ln (x^{2} + 1) + e^{- x^{2}}

\frac{100}{x}

\int \frac{x ^{2}}{( x + 1 ) ^{3}} d x

5 y

\frac{d}{d x} (\frac{d y}{d x}) + y = 7 sin (x)

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