What is the solution for (x-9)y^'+y=x^2-10 ?

The solution for (x-9)y^'+y=x^2-10 is y=(x^3)/(3(x-9))-(10x)/(x-9)+(c_{1})/(x-9)

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(x-9)y^'+y=x^2-10

(x - 9) y^{^{'}} + y = x^{2} - 10

y = \frac{x ^{3}}{3 ( x - 9 )} - \frac{10 x}{x - 9} + \frac{c _{1}}{x - 9}

Solution steps

(x - 9) y^{'} (x) + y = x^{2} - 10

Solve linear ODE:

y = \frac{x ^{3}}{3 ( x - 9 )} - \frac{10 x}{x - 9} + \frac{c _{1}}{x - 9}

y = \frac{x ^{3}}{3 ( x - 9 )} - \frac{10 x}{x - 9} + \frac{c _{1}}{x - 9}

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What is the solution for (x-9)y^'+y=x^2-10 ?
The solution for (x-9)y^'+y=x^2-10 is y=(x^3)/(3(x-9))-(10x)/(x-9)+(c_{1})/(x-9)