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

The solution for 3y^'=(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^'=(3x*y+18x^2+2y^2)/(x^2),y(1)=3

3 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 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) = x^{4} (3 - x)^{2}, at x = 1

\frac{d y}{d x} = 200 - \frac{3 y}{100 - 10 x}

f (x) = 4 x^{2} + 7 x

\frac{d}{d x} ((x - 2) (x + 3))

\frac{d}{d x} ((4 x)^{l n (4 x)})

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