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

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

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

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

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

Solution steps

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

Solve using substitution:

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

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

\int x sin (\frac{x}{2}) d x

f (t) = 4 t^{2} + 5

3 - \frac{x ^{2}}{2}, 1, 2

y^{^{'}} + 20 y = 40 sin (60 x)

\frac{\partial}{\partial x} (sin (8 x) cos (3 y))

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