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

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

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

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

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

Solution steps

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

Solve Bernoulli ODE:

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

x \to x lim (tan (x))

\frac{4}{( s - 4 ) ^{3}}

\frac{\partial}{\partial x} (3 cos (2 x))

h (x) = \frac{1}{x ^{2}} + \frac{1}{x} + x

\int \frac{cos ( a x )}{b - a sin ( a x )} d x

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