What is the solution for (x-8)y^'+y=x^2-6 ?

The solution for (x-8)y^'+y=x^2-6 is y=(x^3)/(3(x-8))-(6x)/(x-8)+(c_{1})/(x-8)

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(x-8)y^'+y=x^2-6

(x - 8) y^{^{'}} + y = x^{2} - 6

y = \frac{x ^{3}}{3 ( x - 8 )} - \frac{6 x}{x - 8} + \frac{c _{1}}{x - 8}

Solution steps

(x - 8) y^{'} (x) + y = x^{2} - 6

Solve linear ODE:

y = \frac{x ^{3}}{3 ( x - 8 )} - \frac{6 x}{x - 8} + \frac{c _{1}}{x - 8}

y = \frac{x ^{3}}{3 ( x - 8 )} - \frac{6 x}{x - 8} + \frac{c _{1}}{x - 8}

z \to 0 lim (\frac{z}{tan ( z )})

\frac{d}{d t} (2 t + 7)

\frac{d}{d x} (2 - \frac{x}{4})

\int_{1}^{2} \frac{4}{( x - 1 ) ^{\frac{1}{4}}} d x

\frac{d}{d x} (\frac{1}{( - x ^{2} + 3 x + 3 ) ^{4}})

What is the solution for (x-8)y^'+y=x^2-6 ?
The solution for (x-8)y^'+y=x^2-6 is y=(x^3)/(3(x-8))-(6x)/(x-8)+(c_{1})/(x-8)