What is the solution for (x-4)y^'+y=x^2+8 ?

The solution for (x-4)y^'+y=x^2+8 is y=(x^3)/(3(x-4))+(8x)/(x-4)+(c_{1})/(x-4)

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(x-4)y^'+y=x^2+8

(x - 4) y^{^{'}} + y = x^{2} + 8

y = \frac{x ^{3}}{3 ( x - 4 )} + \frac{8 x}{x - 4} + \frac{c _{1}}{x - 4}

Solution steps

(x - 4) y^{'} (x) + y = x^{2} + 8

Solve linear ODE:

y = \frac{x ^{3}}{3 ( x - 4 )} + \frac{8 x}{x - 4} + \frac{c _{1}}{x - 4}

y = \frac{x ^{3}}{3 ( x - 4 )} + \frac{8 x}{x - 4} + \frac{c _{1}}{x - 4}

3 x^{3}

\frac{d}{d y} (ln (y))

x = - 2 + 4 y, x = y^{2} + 1

\frac{\partial}{\partial x} (10 x cos (4 x y))

x \to 0 lim (\frac{( x + 1 ) ^{4} - 1}{x})

What is the solution for (x-4)y^'+y=x^2+8 ?
The solution for (x-4)y^'+y=x^2+8 is y=(x^3)/(3(x-4))+(8x)/(x-4)+(c_{1})/(x-4)