What is the solution for (x^2-3)y^'=x(y^2+16),y(2)=4 ?

The solution for (x^2-3)y^'=x(y^2+16),y(2)=4 is y=4tan(2ln(x^2-3)+pi/4)

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(x^2-3)y^'=x(y^2+16),y(2)=4

(x^{2} - 3) y^{^{'}} = x (y^{2} + 16), y (2) = 4

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

Solution steps

(x^{2} - 3) y^{'} (x) = x (y^{2} + 16)

Solve separable ODE:

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

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

\frac{d y}{d x} = 3 y^{2} - y^{2} sin (x)

\frac{d y}{d x} = 3 x e^{3 x + y}

y e^{2 x} - (1 + e^{2 x}) \frac{d y}{d x} = 0

\frac{d y}{d x} + y cos (x) = 8 cos (x)

y^{^{'}} = y - (\frac{1}{1000}) y^{2}

What is the solution for (x^2-3)y^'=x(y^2+16),y(2)=4 ?
The solution for (x^2-3)y^'=x(y^2+16),y(2)=4 is y=4tan(2ln(x^2-3)+pi/4)