What is the solution for xyy^'=4x^2+y^2 ?

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

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xyy^'=4x^2+y^2

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

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

Solution steps

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

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}

- 3 x^{\frac{2}{3}}

\frac{d}{d x} (\frac{sin ( x ) - 1}{cos ( x ) + 1})

\int \frac{x + 4}{x ^{2} - 2 x + 1} d x

\frac{d}{d x} (arctan (\frac{x ^{2}}{y}))

11 x^{\frac{1}{2}} + x^{\frac{3}{2}}

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