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

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

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

y^{^{'}} + \frac{x}{x ^{2} + 1} y = - 4 x

y = - \frac{4 ( x ^{2} + 1 )}{3} + \frac{c _{1}}{x ^{2} + 1}

Solution steps

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

Solve linear ODE:

y = - \frac{4 ( x ^{2} + 1 )}{3} + \frac{c _{1}}{x ^{2} + 1}

y = - \frac{4 ( x ^{2} + 1 )}{3} + \frac{c _{1}}{x ^{2} + 1}

y = (\frac{( x + 2 )}{( x - 3 )})^{2}, at x = 4

\frac{d}{d x} (a (e^{x} + e^{x} x))

h \to 0 lim (h sin (\frac{1}{h}))

\int \frac{1}{6} x^{2} d x

\int_{0}^{4} 8 e^{- 3 x} d x

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