What is the solution for x(dy)/(dx)=y+(x^2-4)^2 ?

The solution for x(dy)/(dx)=y+(x^2-4)^2 is y=-x^4-8x^2+c_{1}x+(4x^4)/3-16

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x(dy)/(dx)=y+(x^2-4)^2

x \frac{d y}{d x} = y + (x^{2} - 4)^{2}

y = - x^{4} - 8 x^{2} + c_{1} x + \frac{4 x ^{4}}{3} - 16

Solution steps

x \frac{d y}{d x} = y + (x^{2} - 4)^{2}

Solve linear ODE:

y = - x^{4} - 8 x^{2} + c_{1} x + \frac{4 x ^{4}}{3} - 16

y = - x^{4} - 8 x^{2} + c_{1} x + \frac{4 x ^{4}}{3} - 16

\frac{d}{d x} (sinh (3 x))

y = x^{3}, y = 4 x

\int - 9 x^{2} + 9 x + 2 d x

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

\int 3 x ln (x) d x

What is the solution for x(dy)/(dx)=y+(x^2-4)^2 ?
The solution for x(dy)/(dx)=y+(x^2-4)^2 is y=-x^4-8x^2+c_{1}x+(4x^4)/3-16