What is the solution for (x-6)y^'+y=x^2-20 ?

The solution for (x-6)y^'+y=x^2-20 is y=(x^3)/(3(x-6))-(20x)/(x-6)+(c_{1})/(x-6)

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(x-6)y^'+y=x^2-20

(x - 6) y^{^{'}} + y = x^{2} - 20

y = \frac{x ^{3}}{3 ( x - 6 )} - \frac{20 x}{x - 6} + \frac{c _{1}}{x - 6}

Solution steps

(x - 6) y^{'} (x) + y = x^{2} - 20

Solve linear ODE:

y = \frac{x ^{3}}{3 ( x - 6 )} - \frac{20 x}{x - 6} + \frac{c _{1}}{x - 6}

y = \frac{x ^{3}}{3 ( x - 6 )} - \frac{20 x}{x - 6} + \frac{c _{1}}{x - 6}

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What is the solution for (x-6)y^'+y=x^2-20 ?
The solution for (x-6)y^'+y=x^2-20 is y=(x^3)/(3(x-6))-(20x)/(x-6)+(c_{1})/(x-6)