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

The solution for x^2y^'=2x^2+xy+2y^2 is y=tan(2ln(x)+c_{1})x

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x^2y^'=2x^2+xy+2y^2

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

y = tan (2 ln (x) + c_{1}) x

Solution steps

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

Solve using substitution:

y = tan (2 ln (x) + c_{1}) x

y = tan (2 ln (x) + c_{1}) x

\int \frac{( 5 - 4 x ^{2} )}{( 3 + 2 x )} d x

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

(y^{- 1})^{^{'}}

x \to - 1 lim (\frac{x + 1}{x ^{3} + 1})

y^{^{'}} = - y + \frac{1}{y}

What is the solution for x^2y^'=2x^2+xy+2y^2 ?
The solution for x^2y^'=2x^2+xy+2y^2 is y=tan(2ln(x)+c_{1})x