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

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

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

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

y = \frac{e ^{2 c_{1}} x ^{2}}{2} + \frac{e ^{2 c_{1}}}{2} - \frac{1}{2}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{2 c_{1}} x ^{2}}{2} + \frac{e ^{2 c_{1}}}{2} - \frac{1}{2}

y = \frac{e ^{2 c_{1}} x ^{2}}{2} + \frac{e ^{2 c_{1}}}{2} - \frac{1}{2}

f (x) = (x^{2} + 3 x) (x + 4)

f (x) = x^{2} - 10 x + 4, at x = 0

8 x^{3} cot (x)

f (x) = \frac{- 2 x}{( x ^{2} + 1 )}, at x = - 1

\int \frac{x ^{7}}{x ^{4} - 2} d x

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