Answer
See answers below
Work Step by Step
We are given:
$$y'=2x(x+y)^2-1$$
a) When we change variable $y$ into:
$y(x)=w(x)-x$
we will get
$$\frac{dy}{dx}=\frac{dw}{dx}-1$$
Since $\frac{dy}{dx}=2x(x+y)^2-1$
so $$\frac{dw}{dx}-1=2x(x+y)^2-1$$
$$\frac{dw}{dx}-1=2x(x+w-x)^2-1$$
$$\frac{dw}{dx}=2x w^2$$
b) We can obtain:
$$\frac{dw}{w^2}=2xdx$$
Integrating left side by $w$ and right side by $x$:
$$\frac{dw}{w^2}=2xdx$$
$$-\frac{1}{w}=x^2+C$$
where C is a constant of integration
$$w=-\frac{1}{x^2+C}$$
We have from exercise $a$:
$$y=-\frac{1}{x^2+C}-x$$
$$y=\frac{-1-x^3+Cx}{x^2+C}$$
Since $y(0)=1 \rightarrow 1=\frac{-1}{C} \rightarrow C=-1$
The solution is:
$$y=\frac{-1-x(x^2+1)}{x^2-1}$$
