What is the solution for (x-9)y^'+y=x^2-1 ?

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

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(x-9)y^'+y=x^2-1

(x - 9) y^{^{'}} + y = x^{2} - 1

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

Solution steps

(x - 9) y^{'} (x) + y = x^{2} - 1

Solve linear ODE:

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

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

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