What is the solution for (dy)/(dx)-(2x^3y)/(x^4+3)=x^3 ?

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

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(dy)/(dx)-(2x^3y)/(x^4+3)=x^3

\frac{d y}{d x} - \frac{2 x ^{3} y}{x ^{4} + 3} = x^{3}

y = \frac{1}{2} (x^{4} + 3) + c_{1} x^{4} + 3

Solution steps

\frac{d y}{d x} - \frac{2 x ^{3} y}{x ^{4} + 3} = x^{3}

Solve linear ODE:

y = \frac{1}{2} (x^{4} + 3) + c_{1} x^{4} + 3

y = \frac{1}{2} (x^{4} + 3) + c_{1} x^{4} + 3

\frac{d}{d x} (2 A x + B)

\frac{d}{d x} (x^{6} ln (5 x))

\frac{d}{d x} (\frac{1}{x ^{2} + 2 y ^{2} - 5})

d er i v a t i v e f (x) = 5 x + 6 x

d er i v a t i v e f (x) = 20 x

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