What is the solution for 3*y^'=(3x*y+18x^2+2y^2)/(x^2),y(1)=3 ?

The solution for 3*y^'=(3x*y+18x^2+2y^2)/(x^2),y(1)=3 is y=3tan(2ln(x)+pi/4+pin)x

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3y^'=(3xy+18x^2+2y^2)/(x^2),y(1)=3

3 \cdot y^{^{'}} = \frac{3 x \cdot y + 18 x ^{2} + 2 y ^{2}}{x ^{2}}, y (1) = 3

y = 3 tan (2 ln (x) + \frac{π}{4} + πn) x

Solution steps

3 \cdot y^{'} (x) = \frac{3 x \cdot y + 18 x ^{2} + 2 y ^{2}}{x ^{2}}

Solve using substitution:

y = 3 tan (2 ln (x) + \frac{π}{4} + πn) x

y = 3 tan (2 ln (x) + \frac{π}{4} + πn) x

f (x) = \frac{- 8 x}{x ^{2} + 1}, at x = 0

\int \frac{1}{x} - \frac{3}{x ^{2} + 1} d x

\int (2 sin (x)) d x

\frac{\partial}{\partial x} (e^{x} (x^{2} - y^{2}))

What is the solution for 3*y^'=(3x*y+18x^2+2y^2)/(x^2),y(1)=3 ?
The solution for 3*y^'=(3x*y+18x^2+2y^2)/(x^2),y(1)=3 is y=3tan(2ln(x)+pi/4+pin)x