What is the solution for y^'+x/(x^2-9)y=2x ?

The solution for y^'+x/(x^2-9)y=2x is y=(2(x^2-9))/3+(c_{1})/(sqrt(x^2-9))

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y^'+x/(x^2-9)y=2x

y^{^{'}} + \frac{x}{x ^{2} - 9} y = 2 x

y = \frac{2 ( x ^{2} - 9 )}{3} + \frac{c _{1}}{x ^{2} - 9}

Solution steps

y^{'} (x) + \frac{x}{x ^{2} - 9} y = 2 x

Solve linear ODE:

y = \frac{2 ( x ^{2} - 9 )}{3} + \frac{c _{1}}{x ^{2} - 9}

y = \frac{2 ( x ^{2} - 9 )}{3} + \frac{c _{1}}{x ^{2} - 9}

x \to - 9 lim (\frac{x ^{2} - 4}{x - 9})

x \to \infty lim (x^{4} e^{- x^{3}})

n = 0 \sum \infty \frac{1}{n + 1}

\int \frac{x ^{2} - 16}{x ^{3}} d x

f (x) = \frac{2 ln ( x )}{x}, at x = 1

What is the solution for y^'+x/(x^2-9)y=2x ?
The solution for y^'+x/(x^2-9)y=2x is y=(2(x^2-9))/3+(c_{1})/(sqrt(x^2-9))