What is the solution for (1+x^2)y^'-2xy=x^2+x^4 ?

The solution for (1+x^2)y^'-2xy=x^2+x^4 is y=x(1+x^2)-arctan(x)(1+x^2)+c_{1}(1+x^2)

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(1+x^2)y^'-2xy=x^2+x^4

(1 + x^{2}) y^{^{'}} - 2 x y = x^{2} + x^{4}

y = x (1 + x^{2}) - arctan (x) (1 + x^{2}) + c_{1} (1 + x^{2})

Solution steps

(1 + x^{2}) y^{'} (x) - 2 x y = x^{2} + x^{4}

Solve linear ODE:

y = x (1 + x^{2}) - arctan (x) (1 + x^{2}) + c_{1} (1 + x^{2})

y = x (1 + x^{2}) - arctan (x) (1 + x^{2}) + c_{1} (1 + x^{2})

y = x^{4} + x^{3}, y = x^{4} + x^{2} + 6 x

\frac{d}{d x} (e^{x} sin (4 x))

\frac{d}{d x} (A e^{4 x})

\frac{\partial}{\partial v} (2 u + v)

ln (7 x + 9)

What is the solution for (1+x^2)y^'-2xy=x^2+x^4 ?
The solution for (1+x^2)y^'-2xy=x^2+x^4 is y=x(1+x^2)-arctan(x)(1+x^2)+c_{1}(1+x^2)