How should I approach this ivp problem with two differentiation about x?

Question

The question given is as followed:

$$x dy - y dx - (1-x²)dx = 0, \\ y(1)=1$$

How should I approach this question? I tried to start this problem by finding an integrating factor but what should I do when 2 dx appears on the equation?

Answer 1

$$xdy = (1+y-x^2)dx \implies \frac{dy}{dx} = -x+\frac1x + \frac yx \implies \frac{dy}{dx} \color{blue}{- \frac{1}{x}}y = \color{green}{\frac1x-x}$$

$$\color{blue}{IF} = e^{\int\color{blue}{\frac{-1}{x}}dx} = e^{-\ln x} = \color{blue}{\frac 1x}$$

Now,

$$\color{blue}{IF} \cdot y = \int \color{green}{\left(\frac1x-x\right)}\color{blue}{IF} dx+c \implies \frac yx = \int\left(\frac 1{x^2} -1\right)dx +c$$

Could you proceed further?

Answer 2

$$x dy - y dx - (1-x²)dx = 0, y(1)=1$$ $$x dy - y dx - (1-x²)dx = 0$$ Divide by $x^2$ $$ \left (\dfrac {x dy - y dx }{x^2} \right)- \dfrac {(1-x²)}{x^2}dx = 0$$ $$d \left (\dfrac yx \right)+dx - \dfrac {1}{x^2}dx = 0$$ Integrate.