What is the solution for y^'=(4x^2+y^2)/(xy) ?

The solution for y^'=(4x^2+y^2)/(xy) is y=xsqrt(8ln(x)+c_{1)},y=-xsqrt(8ln(x)+c_{1)}

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y^'=(4x^2+y^2)/(xy)

y^{^{'}} = \frac{4 x ^{2} + y ^{2}}{x y}

y = x 8 ln (x) + c_{1}, y = - x 8 ln (x) + c_{1}

Solution steps

y^{'} (x) = \frac{4 x ^{2} + y ^{2}}{x y}

Solve Bernoulli ODE:

y = x 8 ln (x) + c_{1}, y = - x 8 ln (x) + c_{1}

y = x 8 ln (x) + c_{1}, y = - x 8 ln (x) + c_{1}

\frac{d}{d x} (5 x^{3})

y^{^{'}} = \frac{x ^{10}}{y}

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

\int (- 2 x^{- 1}) d x

cos (2 \cdot t)

What is the solution for y^'=(4x^2+y^2)/(xy) ?
The solution for y^'=(4x^2+y^2)/(xy) is y=xsqrt(8ln(x)+c_{1)},y=-xsqrt(8ln(x)+c_{1)}