What is the solution for (x-6)y^'+y=x^2+18 ?

The solution for (x-6)y^'+y=x^2+18 is y=(x^3)/(3(x-6))+(18x)/(x-6)+(c_{1})/(x-6)

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(x-6)y^'+y=x^2+18

(x - 6) y^{^{'}} + y = x^{2} + 18

y = \frac{x ^{3}}{3 ( x - 6 )} + \frac{18 x}{x - 6} + \frac{c _{1}}{x - 6}

Solution steps

(x - 6) y^{'} (x) + y = x^{2} + 18

Solve linear ODE:

y = \frac{x ^{3}}{3 ( x - 6 )} + \frac{18 x}{x - 6} + \frac{c _{1}}{x - 6}

y = \frac{x ^{3}}{3 ( x - 6 )} + \frac{18 x}{x - 6} + \frac{c _{1}}{x - 6}

\frac{\partial}{\partial y} (10 y^{2} e^{10 x y})

t \to 0 lim (\frac{tan ( 3 t )}{t})

\frac{\partial}{\partial x} ((3 - x y) e^{- x y})

n = 1 \sum \infty (100) (\frac{1}{2})^{n}

- 1, 8, 4 e^{- 2 x} - 5

What is the solution for (x-6)y^'+y=x^2+18 ?
The solution for (x-6)y^'+y=x^2+18 is y=(x^3)/(3(x-6))+(18x)/(x-6)+(c_{1})/(x-6)