## Calculus, 10th Edition (Anton)

$$y=x^2-2x$$
Given $$x \frac{dy}{dx} - y =x^2,\ \ \ \ \ y(1)=-1$$ Rewriting the equation $$\frac{dy}{dx} -\frac{1}{x} y=x$$ Since \begin{align*} \mu(x)&=e^{\int \frac{-1}{x} dx}\\ &=e^{-\ln x } \\ &=\frac{1}{x} \end{align*} Then \begin{align*} y\mu(x)&=\int \mu(x) q(x)dx\\ \frac{1}{x}y&=\int dx\\ &= x +c \end{align*} since $y(1)=-1,$ then $c=-2$ Hence $$y=x^2-2x$$