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

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

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

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

y = \frac{c _{1}}{4 ( x ^{4} + 2 x ^{2} + 1 )} + \frac{1}{4}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{4 ( x ^{4} + 2 x ^{2} + 1 )} + \frac{1}{4}

y = \frac{c _{1}}{4 ( x ^{4} + 2 x ^{2} + 1 )} + \frac{1}{4}

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

y^{^{'}} = \frac{1 + y}{tan ( x )}

(x^{2} + 1) y^{^{'}} + 4 x y = 0

\frac{d y}{d x} = \frac{4 y}{x}, y = 2, x = 1

y^{2} sec (3 x) \frac{d y}{d x} = 1, y (0) = 0

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