What is the solution for xy^'+2y=x^2-x+1 ?

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

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xy^'+2y=x^2-x+1

x y^{^{'}} + 2 y = x^{2} - x + 1

y = \frac{x ^{2}}{4} - \frac{x}{3} + \frac{1}{2} + \frac{c _{1}}{x ^{2}}

Solution steps

x y^{'} (x) + 2 y = x^{2} - x + 1

Solve linear ODE:

y = \frac{x ^{2}}{4} - \frac{x}{3} + \frac{1}{2} + \frac{c _{1}}{x ^{2}}

y = \frac{x ^{2}}{4} - \frac{x}{3} + \frac{1}{2} + \frac{c _{1}}{x ^{2}}

f (x) = ln (x^{2} - 9 x + 1), (9, 0)

(- 2, 2), (1, - 2)

6 t^{2}

\int \frac{1}{484 - x ^{2}} d x

\frac{d}{d x} (I n^{2} x)

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